4.6: Acceleration and Kepler's Laws
-
- Last updated
- Save as PDF
Components of the Acceleration Vector
We can combine some of the concepts discussed in Arc Length and Curvature with the acceleration vector to gain a deeper understanding of how this vector relates to motion in the plane and in space. Recall that the unit tangent vector \(\vecs T\) and the unit normal vector \(\vecs N\) form an osculating plane at any point \(P\) on the curve defined by a vector-valued function \(\vecs{r}(t)\). The following theorem shows that the acceleration vector \(\vecs{a}(t)\) lies in the osculating plane and can be written as a linear combination of the unit tangent and the unit normal vectors.
Theorem \(\PageIndex{1}\): The Plane of the Acceleration Vector
The acceleration vector \(\vecs{a}(t)\) of an object moving along a curve traced out by a twice-differentiable function \(\vecs{r}(t)\) lies in the plane formed by the unit tangent vector \(\vecs T(t)\) and the principal unit normal vector \(\vecs N(t)\) to \(C\) .
Furthermore,
\[\vecs{a}(t) = v'(t)\vecs{T}(t) + [v(t)]^2 \kappa \vecs{N}(t) \]
Here, \(v(t) = \|\vecs v(t)\|\) is the speed of the object and \(\kappa\) is the curvature of \(C\) traced out by \(\vecs{r}(t)\).
Proof
Because \(\vecs{v}(t)=\vecs{r}′(t)\) and \(\vecs{T}(t)=\dfrac{\vecs{r}′(t)}{||\vecs{r}′(t)||}\), we have \(\vecs v(t)=||\vecs{r}′(t)||\vecs{T}(t)=v(t)\vecs{T}(t)\).
Now we differentiate this equation:
\[\vecs{a}(t)=\vecs{v}′(t)=\dfrac{d}{dt}\left(v(t)\vecs{T}(t)\right)=v′(t)\vecs{T}(t)+v(t)\vecs{T}′(t)\]
Since \(\vecs{N}(t)=\dfrac{\vecs{T}′(t)}{||\vecs{T}′(t)||}\), we know \(\vecs{T}′(t)=||\vecs{T}′(t)||\vecs{N}(t)\), so
\[\vecs{a}(t)=v′(t)\vecs{T}(t)+v(t)||\vecs{T}′(t)||\vecs{N}(t).\]
A formula for curvature is \(\kappa=\dfrac{||\vecs{T}'(t)||}{||\vecs{r}'(t)||}\), so \(\vecs{T}'(t) = \kappa ||\vecs{r}'(t) || = \kappa v(t) \).
This gives \(\vecs{a}(t)=v′(t)\vecs{T}(t)+\kappa (v(t))^2 \vecs{N}(t).\)
\(\square\)
The real-valued coefficients of \(\vecs{T}(t)\) and \(\vecs{N}(t)\) (possibly real-valued functions of \(t\)) are referred to as the tangential component of acceleration and the normal component of acceleration , respectively. We write \(a_\vecs{T}\) to denote the tangential component and \(a_\vecs{N}\) to denote the normal component. Note that these names are a little misleading, as these functions of \(t\) are not vector components, but the magnitudes of the corresponding vector components of acceleration.
Theorem \(\PageIndex{2}\): Tangential and Normal Components of Acceleration (as scalar magnitudes)
Let \(\vecs{r}(t)\) be a vector-valued function that denotes the position of an object as a function of time. Then \(\vecs{a}(t)=\vecs{r}′′(t)\) is the acceleration vector. The tangential and normal components of acceleration \(a_\vecs{T}\) and \(a_\vecs{N}\) (the magnitudes of the corresponding vector components of acceleration) are given by the formulas
\[a_{\vecs{T}}=\vecs a \cdot\vecs{T}=\dfrac{\vecs{v}\cdot\vecs{a}}{||\vecs{v}||} \label{Eq1B}\]
and
\[a_\vecs{N}=\vecs a\cdot \vecs N=\dfrac{||\vecs v \times \vecs a||}{||\vecs v||}=\sqrt{||\vecs a||^2−\left(a_{\vecs{T}}\right)^2}. \label{Eq2B}\]
The vector components of acceleration are related by the formula
\[\vecs{a}(t)=a_\vecs{T} \vecs{T}(t)+a_\vecs{N}\vecs{N}(t). \label{Eq3B}\]
Here \(\vecs{T}(t)\) is the unit tangent vector to the curve defined by \(\vecs{r}(t)\), and \(\vecs{N}(t)\) is the unit normal vector to the curve defined by \(\vecs{r}(t)\).
Figure \(\PageIndex{1}\):
The acceleration can be broken into components parallel to \(\vecs N(t)\) and \(\vecs T(t)\). Here \(t = 1\).
At a given point on the curve, breaking the acceleration vector up into these two components, a tangential component parallel to the velocity vector (and \(\vecs T\)) and a normal component parallel to the normal vector \(\vecs N\), allows us to more clearly see the effect of the acceleration vector on the velocity at that point. See Figure \(\PageIndex{1}\).
Only the component of acceleration that is parallel to the velocity will change the speed of the motion (affecting the magnitude of the velocity). And only the component that is normal to the curve will affect the direction of the velocity (and the corresponding bending of the curve at this point). Note that for the example in Figure \(\PageIndex{1}\), the acceleration is clearly slowing the velocity and turning the curve to the left.
This means that if acceleration is orthogonal to the velocity, it will only affect the direction and will not change the speed at this point. And if the acceleration is parallel to the velocity, it will only affect the speed and not change the direction at this point. If the acceleration vector is always parallel to the velocity vector for a given position function, the motion must be along a straight line.
Exercise \(\PageIndex{1}\)
What can you say about the speed of the motion along a curve if the acceleration vector is always orthogonal to the velocity vector for the corresponding position function?
- Hint
-
In addition to considering what this means based on the discussion of the components of acceleration shown above, see Derivative Property vii in Section 12.2 , considering what it says should be true if \(\vecs r'(t) \cdot \vecs r''(t) = 0\), that is, if \(\vecs v(t) \cdot \vecs a(t) = 0\).
Note that in this context, the unit vectors \(\vecs T(t)\) and \(\vecs N(t)\) are filling a role very similar to \(\hat{\mathbf i}\) and \(\hat{\mathbf j}\). But instead of representing a vector in the \(xy\)-plane where the vectors \(\hat{\mathbf i}\) and \(\hat{\mathbf j}\) reside, the linear combination of \(\vecs T(t)\) and \(\vecs N(t)\) represents an acceleration vector in the plane formed by the vectors \(\vecs T(t)\) and \(\vecs N(t)\).
The normal component of acceleration is also called the centripetal component of acceleration or sometimes the radial component of acceleration . To understand centripetal acceleration, suppose you are traveling in a car on a circular track at a constant speed. Then, as we saw earlier, the acceleration vector points toward the center of the track at all times. As a rider in the car, you feel a pull toward the outside of the track because you are constantly turning. This sensation acts in the opposite direction of centripetal acceleration. The same holds true for non-circular paths. The reason is that your body tends to travel in a straight line and resists the force resulting from acceleration that push it toward the side. Note that at point \(B\) in Figure \(\PageIndex{2}\) the acceleration vector is pointing backward. This is because the car is decelerating as it goes into the curve.
The tangential and normal unit vectors at any given point on the curve provide a frame of reference at that point. The tangential and normal components of acceleration (as vectors) are the projections of the acceleration vector onto \(\vecs T\) and \(\vecs N\), respectively.
Note that there are two ways for us to calculate the vector components of acceleration that are tangent and normal to the curve. We can either use the formulas in Theorem \(\PageIndex{2}\) above and multiply them by the unit tangent vector and the unit normal vector, respectively, or we can calculate these vectors directly (with a lot less effort) as the projection of the acceleration vector onto the velocity vector and the corresponding component of acceleration that is perpendicular to the velocity vector and that together with the projection sums to the original acceleration vector. We can thus use the simpler approach shown in Theorem \(\PageIndex{3}\) below.
Theorem \(\PageIndex{3}\): Tangential and Normal Components of Acceleration (as vector components)
Let \(\vecs{r}(t)\) be a vector-valued function that denotes the position of an object as a function of time. Then \(\vecs{a}(t)=\vecs{r}′′(t)\) is the acceleration vector. The tangential and normal components of acceleration \(a_\vecs{T} \vecs{T}(t)\) and \(a_\vecs{N}\vecs{N}(t)\) are given by the formulas
\[a_\vecs{T} \vecs{T}(t) = \text{Proj}_{\vecs v(t)} \vecs a(t) \qquad \text{The vector component of acceleration tangent to the curve}\]
and
\[a_\vecs{N}\vecs{N}(t) = \vecs a(t)_{\text{perp}} = \vecs a(t) - \text{Proj}_{\vecs v(t)} \vecs a(t) \qquad \text{The vector component of acceleration normal to the curve}\]
If we have already calculated \(\vecs{T}(t)\) and \(\vecs{N}(t),\) we can also determine these vector components of acceleration using these and the values of \(a_{\vecs{T}}\) and \(a_{\vecs{N}}\) from above.
Example \(\PageIndex{1}\): Finding Components of Acceleration (as magnitudes)
A particle moves in a path defined by the vector-valued function \(\vecs{r}(t)=t^2\,\hat{\mathbf i}+(2t−3)\,\hat{\mathbf j}+(3t^2−3t)\,\hat{\mathbf k}\), where \(t\) measures time in seconds and distance is measured in feet.
- Find \(a_\vecs{T}\) and \(a_\vecs{N}\) as functions of \(t\).
- Find \(a_\vecs{T}\) and \(a_\vecs{N}\) at time \(t=2\).
Solution
-
Let’s start deriving the velocity and acceleration functions
\[\vecs{v}(t) = \vecs{r}'(t) = 2t\,\hat{\mathbf i}+2\,\hat{\mathbf j}+(6t-3)\,\hat{\mathbf k}\nonumber\]
\[\vecs{a}(t) = \vecs{v}'(t)=2\,\hat{\mathbf i}+6\,\hat{\mathbf k} \nonumber \]
Now we use Equation \ref{Eq1B}:
\[\begin{align*}a_{\vecs{T}}&=\dfrac{\vecs{v}\cdot \vecs{a}}{||\vecs{v} ||} \\[5pt] &= \dfrac{ (2t \,\hat{\mathbf i} +2\,\hat{\mathbf j} +(6t-3)\,\hat{\mathbf k})\cdot(2\,\hat{\mathbf i}+6\,\hat{\mathbf k}) }{|| 2t\,\hat{\mathbf i} + 2\,\hat{\mathbf j} + (6t-3)\,\hat{\mathbf k} ||} \\[5pt] &= \dfrac{4t + 6(6t-3)}{\sqrt{(2t)^2 +2^2 + (6t-3)^2}} \\[5pt] &= \dfrac{40t-18}{40t^2 - 36t+13} \end{align*}\]
Then we apply Equation \ref{Eq2B}:\[\begin{align*} a_\vecs{N} &=\sqrt{||\vecs{a}||^2-\left(a_{\vecs{T}}\right)^2} \\[5pt]&= \sqrt{||2\,\hat{\mathbf i}+6\,\hat{\mathbf k}||^2 - \left( \dfrac{ 40t-18 }{\sqrt{40t^2-36t+13}} \right)^2} \\[5pt] &= \sqrt{ 4+36-\dfrac{(40t-18)^2}{40t^2-36t+13} } \\[5pt] &= \sqrt{\dfrac{ 40(40t^2-36t+13)-(1600t^2-1440t+324) }{40t^2-36t+13} } \\[5pt] &= \sqrt{ \dfrac{196}{ 40t^2-36t+13} } \\[5pt] &= \dfrac{14}{\sqrt{40t^2-36t+13}} \end{align*}\]
-
We must evaluate each of the answers from part a at \(t=2\):
\[\begin{align*} a_{\vecs{T}}(2) &= \dfrac{ 40(2)-18 }{\sqrt{40(2)^2 - 36(2)+13 }} \\[5pt] &= \dfrac{80-18 }{\sqrt{160-72+13}} \\[5pt] &= \dfrac{62}{\sqrt{101}} = \dfrac{62\sqrt{101}}{101} \\[5pt] a_{\vecs{N}}(2) &= \dfrac{14}{\sqrt{40(2)^2 -36(2)+13 }} \\[5pt] &= \dfrac{14}{\sqrt{160-72+13}} = \dfrac{14}{\sqrt{101}} = \dfrac{14\sqrt{101}}{101}. \end{align*}\]
The units of acceleration are feet per second squared, as are the units of the normal and tangential components of acceleration.
Exercise \(\PageIndex{2}\)
An object moves in a path defined by the vector-valued function \(\vecs r(t)=4t\,\hat{\mathbf i}+t^2\,\hat{\mathbf j}\), where \(t\) measures time in seconds.
- Find \(a_\vecs{T}\) and \(a_\vecs{N}\) as functions of \(t\).
- Find \(a_\vecs{T}\) and \(a_\vecs{N}\) at time \(t=−3\).
- Hint
-
Use Equations \ref{Eq1B} and \ref{Eq2B}
- Answer
-
a. \[\begin{align*} a_\vecs{T} &=\dfrac{\vecs v(t) \cdot \vecs a(t)}{||\vecs v(t)||}= \dfrac{\vecs r'(t) \cdot \vecs r''(t) }{||\vecs r'(t)||} \\[5pt] &= \dfrac{ (4\,\hat{\mathbf i} + 2t \,\hat{\mathbf j}) \cdot (2\,\hat{\mathbf j}) }{||4\,\hat{\mathbf i} + 2t \,\hat{\mathbf j} ||} \\[5pt] &= \dfrac{4t}{\sqrt{4^2 + (2t)^2}}\\[5pt] &= \dfrac{2t}{\sqrt{2+t^2}} \end{align*}\]
\[\begin{align*} a_\vecs{N}&= \sqrt{||\vecs a||^2-a_\vecs{T}^2}\\[5pt] &=\sqrt{||2\,\hat{\mathbf j} ||^2 - \left(\dfrac{2t}{\sqrt{2+t^2}}\right)^2} \\[5pt] &=\sqrt{ 4 - \dfrac{4t^2 }{2+t^2} } \end{align*}\]b. \[\begin{align*} a_\vecs{T}(−3)&= \dfrac{2(-3)}{\sqrt{2+(-3)^2}} \\[5pt] &= \dfrac{-6}{\sqrt{11}}\end{align*}\]
\[\begin{align*}a_\vecs{N}(−3)&= \sqrt{ 4 - \dfrac{4(-3)^2 }{2+(-3)^2} } \\[5pt] &= \sqrt{4- \dfrac{36}{11}} \\[5pt] &= \sqrt{\dfrac{8}{11} }\\[5pt] &=\dfrac{2\sqrt{2}}{\sqrt{11}} \end{align*}\]
Now let's calculate the vector components of acceleration for the vector-valued function in Example \(\PageIndex{1}\). We'll start by calculating these vector components at a time \(t = 2\) using the projection of the acceleration vector onto the velocity vector at that time. Then we'll show the same approach to finding the vector components of acceleration as functions of \(t\).
Example \(\PageIndex{2}\)
A particle moves in a path defined by the vector-valued function \(\vecs{r}(t)=t^2\,\hat{\mathbf i}+(2t−3)\,\hat{\mathbf j}+(3t^2−3t)\,\hat{\mathbf k}\), where \(t\) measures time in seconds and distance is measured in feet.
- Find the vector components of acceleration at time \(t=2\).
- Find the vector components of the acceleration tangent and normal to the curve as functions of \(t\).
Solution
From our work above, we have
\[\vecs{v}(t) = \vecs{r}'(t) = 2t\,\hat{\mathbf i}+2\,\hat{\mathbf j}+(6t-3)\,\hat{\mathbf k}\nonumber\]
\[\vecs{a}(t) = \vecs{v}'(t)=2\,\hat{\mathbf i}+6\,\hat{\mathbf k} \nonumber \]
a. At time \(t=2\), we have
\[\begin{align*} \vecs v(2) &= 2(2)\,\hat{\mathbf i}+2\,\hat{\mathbf j}+\big(6(2)-3\big)\,\hat{\mathbf k}\\[4pt]
&= 4 \,\hat{\mathbf i} + 2 \,\hat{\mathbf j} + 9 \,\hat{\mathbf k}\end{align*}\]
and
\[\vecs a(2) = 2\,\hat{\mathbf i} + 6 \,\hat{\mathbf k}\nonumber\]
Then the component of acceleration tangent to the curve at time \(t = 2\) is:
\[ a_\vecs{T}\vecs T(2) = \text{Proj}_{\vecs v(2)} \vecs a(2) = \underbrace{\frac{\vecs a(2) \cdot \vecs v(2)}{\|\vecs v(2)\|}}_{a_{\vecs T}(2)}\cdot \underbrace{\frac{\vecs v(2)}{\|\vecs v(2)\|}}_{\vecs T(2)}\nonumber\]
Since \(\vecs a(2) \cdot \vecs v(2) = 2(4) + 0(2) + 6(9) = 8 + 0 + 54 = 62,\)
and \(\|\vecs v(2)\| = \sqrt{4^2 + 2^2 + 9^2} = \sqrt{16 + 4 + 81} = \sqrt{101},\)
\[\begin{align*} a_\vecs{T}\vecs T(2) &= \text{Proj}_{\vecs v(2)} \vecs a(2) = \frac{\vecs a(2) \cdot \vecs v(2)}{\|\vecs v(2)\|}\cdot \frac{\vecs v(2)}{\|\vecs v(2)\|}\\[4pt]
&= \frac{62}{101}\vecs v(2) \\[4pt]
&= \frac{248}{101} \,\hat{\mathbf i} +\frac{124}{101} \,\hat{\mathbf j} + \frac{558}{101} \,\hat{\mathbf k}
\end{align*}\]
Since the sum of these two components is \(\vecs a(2),\) the component of acceleration normal to the curve at time \(t = 2\) is:
\[ \begin{align*} a_\vecs{N}\vecs N(2) &= \vecs a(2)_{\text{perp}} = \vecs a(2) - \text{Proj}_{\vecs v(2)} \vecs a(2)\\[4pt]
&= 2 \,\hat{\mathbf i} + 6 \,\hat{\mathbf k} - \left(\frac{248}{101} \,\hat{\mathbf i} +\frac{124}{101} \,\hat{\mathbf j} + \frac{558}{101} \,\hat{\mathbf k} \right) \\[4pt]
&= \left(2 - \frac{248}{101}\right) \,\hat{\mathbf i} - \frac{124}{101} \,\hat{\mathbf j} + \left(6 - \frac{558}{101}\right) \,\hat{\mathbf k}\\[4pt]
&= -\frac{46}{101} \,\hat{\mathbf i} - \frac{124}{101} \,\hat{\mathbf j} + \frac{48}{101} \,\hat{\mathbf k} \end{align*}\]
See the vector \(\vecs a(2)\) and its component vectors in Figure \(\PageIndex{3}\) below.
b. The more general vector components of acceleration (as functions of \(t\)) can be found in the same way with somewhat less work than taking time to first directly determine \(\vecs T(t)\) and \(\vecs N(t)\).
Since
\[\vecs{v}(t) = \vecs{r}'(t) = 2t\,\hat{\mathbf i}+2\,\hat{\mathbf j}+(6t-3)\,\hat{\mathbf k}\nonumber\]
and
\[\vecs{a}(t) = \vecs{v}'(t)=2\,\hat{\mathbf i}+6\,\hat{\mathbf k} \nonumber \]
we have
\[ a_\vecs{T}\vecs T(t) = \text{Proj}_{\vecs v(t)} \vecs a(t) = \underbrace{\frac{\vecs a(t) \cdot \vecs v(t)}{\|\vecs v(t)\|}}_{a_{\vecs T}(t)}\cdot \underbrace{\frac{\vecs v(t)}{\|\vecs v(t)\|}}_{\vecs T(t)}\nonumber\]
As in part (a), we calculate \(\vecs a(t) \cdot \vecs v(t) = 2(2t) + 0(2) + 6(6t-3) = 4t + 0 + 36t - 18 = 40t - 18,\)
and \(\begin{align*} \|\vecs v(t)\| &= \sqrt{(2t)^2 + 2^2 + (6t-3)^2} \\[4pt]
&= \sqrt{4t^2 + 4 + 36t^2 - 36t + 9} \\[4pt]
&= \sqrt{40t^2 - 36t + 13}. \end{align*}\)
Then the vector component of acceleration tangent to the curve at time \(t\) is
\[\begin{align*} a_\vecs{T}\vecs T(t) &= \text{Proj}_{\vecs v(t)} \vecs a(t) = \frac{\vecs a(t) \cdot \vecs v(t)}{\|\vecs v(t)\|}\cdot \frac{\vecs v(t)}{\|\vecs v(t)\|}\\[4pt]
&= \frac{40t - 18}{40t^2 - 36t + 13}\vecs v(t) \\[4pt]
&= \frac{2t(40t - 18)}{40t^2 - 36t + 13} \,\hat{\mathbf i} +\frac{2(40t - 18)}{40t^2 - 36t + 13} \,\hat{\mathbf j} + \frac{(6t-3)(40t-18)}{40t^2 - 36t + 13} \,\hat{\mathbf k}\\[4pt]
&= \frac{4t(20t - 9)}{40t^2 - 36t + 13} \,\hat{\mathbf i} +\frac{4(20t - 9)}{40t^2 - 36t + 13} \,\hat{\mathbf j} + \frac{6(40t^2-38t+9)}{40t^2 - 36t + 13} \,\hat{\mathbf k}\\[4pt]
\end{align*}\]
And the vector component of acceleration normal to the curve at time \(t\) is
\[ \begin{align*} a_\vecs{N}\vecs N(t) &= \vecs a(t)_{\text{perp}} = \vecs a(t) - \text{Proj}_{\vecs v(t)} \vecs a(t)\\[4pt]
&= 2 \,\hat{\mathbf i} + 6 \,\hat{\mathbf k} - \left(\frac{4t(20t - 9)}{40t^2 - 36t + 13} \,\hat{\mathbf i} +\frac{4(20t - 9)}{40t^2 - 36t + 13} \,\hat{\mathbf j} + \frac{6(40t^2-38t+9)}{40t^2 - 36t + 13} \,\hat{\mathbf k} \right) \\[4pt]
&= \left(2 - \frac{4t(20t - 9)}{40t^2 - 36t + 13}\right) \,\hat{\mathbf i} - \frac{4(20t - 9)}{40t^2 - 36t + 13} \,\hat{\mathbf j} + \left(6 - \frac{6(40t^2-38t+9)}{40t^2 - 36t + 13}\right) \,\hat{\mathbf k}\\[4pt]
&= \frac{2(13-18t)}{40t^2 - 36t + 13} \,\hat{\mathbf i} - \frac{4(20t-9)}{40t^2 - 36t + 13} \,\hat{\mathbf j} + \frac{12(t+2)}{40t^2 - 36t + 13} \,\hat{\mathbf k} \end{align*}.\]
Checking these functions at \(t=2\), we first evaluate the expression \(40t^2 - 36t + 13\) at \(t = 2\).
This gives us \(40(2)^2 -36(2) + 13 = 160 - 72 +13 = 101.\)
Then
\(\begin{align*} a_\vecs{T}\vecs T(2) &= \frac{4(2)(20(2) - 9)}{101} \,\hat{\mathbf i} +\frac{4(20(2) - 9)}{101} \,\hat{\mathbf j} + \frac{6(40(2)^2-38(2)+9)}{101} \,\hat{\mathbf k} \\[4pt]
&= \frac{248}{101} \,\hat{\mathbf i} +\frac{124}{101} \,\hat{\mathbf j} + \frac{558}{101} \,\hat{\mathbf k}\quad \checkmark \end{align*}\)
and
\(\begin{align*} a_\vecs{N}\vecs N(2) &= \frac{2(13-18(2))}{101} \,\hat{\mathbf i} - \frac{4(20(2)-9)}{101} \,\hat{\mathbf j} + \frac{12(2+2)}{101} \,\hat{\mathbf k} \\[4pt]
&=-\frac{46}{101} \,\hat{\mathbf i} - \frac{124}{101} \,\hat{\mathbf j} + \frac{48}{101} \,\hat{\mathbf k} \quad \checkmark \end{align*}.\)
Note that we could even use the component vectors we found in Example \(\PageIndex{2}\) part (a) to determine \(a_{\vecs T}(2)\) and \(a_{\vecs N}(2)\), and even \(\vecs N(2).\)
All we need to do to find the two scalar components of acceleration is find the magnitudes of the two vector components.
Thus
\[\begin{align*} a_{\vecs T}(2) = \|a_{\vecs T}\vecs T(2)\| &= \sqrt{\left(\frac{248}{101}\right)^2 + \left(\frac{124}{101}\right)^2 + \left(\frac{558}{101}\right)^2 }\\[4pt]
&= \frac{\sqrt{248^2 + 124^2 + 558^2}}{101}\\[4pt]
&= \frac{\sqrt{62^2(4^2 + 2^2 + 9^2}}{101}\\[4pt]
&= \frac{62\sqrt{101}}{101} \end{align*}\]
and
\[\begin{align*} a_{\vecs N}(2) = \|a_{\vecs N}\vecs N(2)\| &= \sqrt{\frac{46^2 + 124^2 + 48^2}{101}}\\[4pt]
&= \frac{\sqrt{2^2(23^2 + 62^2 + 24^2}}{101}\\[4pt]
&= \frac{2\sqrt{4949}}{101}\\[4pt]
&= \frac{2\sqrt{49\cdot 101}}{101}\\[4pt]
&= \frac{14\sqrt{101}}{101} \end{align*}\]
Now all we need to do to find the unit normal vector for this curve at \(t=2\) is to normalize the vector component of acceleration that is normal to the curve at that location (and value of \(t\)).
So
\[\begin{align*} \vecs N(2) = \frac{a_{\vecs N}\vecs N(2)}{\|a_{\vecs N}\vecs N(2)\|} &= \frac{101}{14}\left[ -\frac{46}{101} \,\hat{\mathbf i} - \frac{124}{101} \,\hat{\mathbf j} + \frac{48}{101} \,\hat{\mathbf k}\right] \\[4pt]
&= -\frac{23\sqrt{101}}{707} \,\hat{\mathbf i} - \frac{62\sqrt{101}}{707} \,\hat{\mathbf j} + \frac{24\sqrt{101}}{707} \,\hat{\mathbf k} \end{align*}\]
Kepler’s Laws
During the early 1600s, Johannes Kepler was able to use the amazingly accurate data from his mentor Tycho Brahe to formulate his three laws of planetary motion, now known as Kepler’s laws of planetary motion . These laws also apply to other objects in the solar system in orbit around the Sun, such as comets (e.g., Halley’s comet) and asteroids. Variations of these laws apply to satellites in orbit around Earth.
Theorem \(\PageIndex{3}\): Kepler's Laws of Planetary Motion
- The path of any planet about the Sun is elliptical in shape, with the center of the Sun located at one focus of the ellipse (the law of ellipses).
- A line drawn from the center of the Sun to the center of a planet sweeps out equal areas in equal time intervals (the law of equal areas) (Figure \(\PageIndex{4}\)).
- The ratio of the squares of the periods of any two planets is equal to the ratio of the cubes of the lengths of their semimajor orbital axes (the Law of Harmonies).
Kepler’s third law is especially useful when using appropriate units. In particular, 1 astronomical unit is defined to be the average distance from Earth to the Sun, and is now recognized to be 149,597,870,700 m or, approximately 93,000,000 mi. We therefore write 1 A.U. = 93,000,000 mi. Since the time it takes for Earth to orbit the Sun is 1 year, we use Earth years for units of time. Then, substituting 1 year for the period of Earth and 1 A.U. for the average distance to the Sun, Kepler’s third law can be written as
\[ T_p^2=D_p^3 \]
for any planet in the solar system, where \(T_P\) is the period of that planet measured in Earth years and \(D_P\) is the average distance from that planet to the Sun measured in astronomical units. Therefore, if we know the average distance from a planet to the Sun (in astronomical units), we can then calculate the length of its year (in Earth years), and vice versa.
Kepler’s laws were formulated based on observations from Brahe; however, they were not proved formally until Sir Isaac Newton was able to apply calculus. Furthermore, Newton was able to generalize Kepler’s third law to other orbital systems, such as a moon orbiting around a planet. Kepler’s original third law only applies to objects orbiting the Sun.
Proof
Let’s now prove Kepler’s first law using the calculus of vector-valued functions. First we need a coordinate system. Let’s place the Sun at the origin of the coordinate system and let the vector-valued function \(\vecs{r}(t)\) represent the location of a planet as a function of time. Newton proved Kepler’s law using his second law of motion and his law of universal gravitation. Newton’s second law of motion can be written as \(\vecs{F}=m\vecs{a}\), where \(\vecs{F}\) represents the net force acting on the planet. His law of universal gravitation can be written in the form \(\vecs{F}=−\dfrac{GmM}{||\vecs{r}||^2}\cdot \dfrac{\vecs{r}}{||\vecs{r} ||}\), which indicates that the force resulting from the gravitational attraction of the Sun points back toward the Sun, and has magnitude \(\dfrac{GmM}{||\vecs{r}||^2} \) (Figure \(\PageIndex{5}\)).
Setting these two forces equal to each other, and using the fact that \(\vecs a(t)=\vecs v′(t)\), we obtain
\[ m\vecs v′(t)=−\frac{GmM}{‖\vecs r‖^2}⋅\frac{\vecs r}{‖\vecs r‖},\]
which can be rewritten as
\[ \dfrac{d\vecs v}{dt}=−\dfrac{GM}{||\vecs r||^3}\vecs{r}.\]
This equation shows that the vectors \(d\vecs{v}/dt\) and \(\vecs r\) are parallel to each other, so \(d\vecs {v}/dt \times \vecs {r}=\vecs 0\). Next, let’s differentiate \(\vecs{r} \times \vecs{v}\) with respect to time:
\[\dfrac{d}{dt}(\vecs{r}\times \vecs{v})=\dfrac{d\vecs{r}}{dt}\times \vecs v+\vecs{r} \times \dfrac{d\vecs{v}}{dt}=\vecs{v}\times \vecs{v}+\vecs{0}=\vecs{0}. \label{Eq10}\]
This proves that \(\vecs{r}\times\vecs{v}\) is a constant vector, which we call \(\vecs C\). Since \(\vecs r\) and \(\vecs v\) are both perpendicular to \(\vecs C\) for all values of \(t\), they must lie in a plane perpendicular to \(\vecs C\). Therefore, the motion of the planet lies in a plane.
Next we calculate the expression \(d\vecs{v}/dt\times \vecs C\):
\[\dfrac{d\vecs{v}}{dt} \times \vecs{C}=−\dfrac{GM}{||\vecs{r}||^3}\vecs{r}\times (\vecs{r}\times\vecs{v})=−\dfrac{GM}{||\vecs r||^3}[(\vecs{r} \cdot \vecs{v})\vecs{r} - (\vecs{r} \cdot \vecs{r})\vecs{v}]. \label{Eq11}\]
The last equality in Equation \ref{Eq11} is from the triple cross product formula ( The Cross Product ). We need an expression for \(\vecs{r}\cdot \vecs{v}\). To calculate this, we differentiate \(\vecs{r}\cdot \vecs{r}\) with respect to time:
\[ \dfrac{d}{dt}(\vecs{r}\cdot \vecs{r})=\dfrac{d\vecs{r}}{dt}\cdot \vecs{r}+\vecs{r}\cdot \dfrac{d\vecs{r}}{dt}=2\vecs{r}\cdot \dfrac{d\vecs{r}}{dt}=2\vecs{r}\cdot \vecs{v}. \label{Eq12}\]
Since \(\vecs{r}\cdot\vecs{r}=||\vecs r||^2\), we also have
\[\dfrac{d}{dt}(\vecs{r}\cdot \vecs{r})=\dfrac{d}{dt}||\vecs{r}||^2=2||\vecs{r}|| \dfrac{d}{dt}||\vecs{r}||. \label{Eq13}\]
Combining Equation \ref{Eq12} and Equation \ref{Eq13}, we get
\[\begin{align*} 2\vecs{r}\cdot \vecs{v}&=2||\vecs{r}||\dfrac{d}{dt}||\vecs{r}|| \\[5pt] \vecs{r} \cdot \vecs{v} &=||\vecs{r}‖\dfrac{d}{dt}||\vecs{r}||. \end{align*} \label{Eq14}\]
Substituting this into Equation \ref{Eq11} gives us
\[\begin{align} \dfrac{d\vecs{v}}{dt} \times \vecs{C}&= - \dfrac{GM}{||\vecs{r}||^3} [(\vecs{r}\cdot \vecs{v})\vecs{r} - (\vecs{r}\cdot \vecs{r})\vecs{v}] \nonumber \\[5pt] &= -\dfrac{GM}{||\vecs{r}||^3}\left[ ||\vecs{r}|| \left(\dfrac{d}{dt} ||\vecs{r}||\right)\vecs{r} - ||\vecs{r}||^2\vecs{v} \right] \nonumber \\[5pt] &= -GM\left[ \dfrac{1}{||\vecs{r}||^2}\left( \dfrac{d}{dt} ||\vecs{r}|| \right)\vecs{r} - \dfrac{1}{||\vecs{r}||}\vecs{v} \right] \nonumber \\[5pt] &= GM\left[ \dfrac{\vecs{v}}{||\vecs{r}||} -\dfrac{\vecs{r}}{||\vecs{r}||^2}\left( \dfrac{d}{dt} ||\vecs{r}|| \right) \right]. \label{Eq15} \end{align}\]
However,
\[ \begin{align*} \dfrac{d}{dt} \dfrac{\vecs{r}}{||\vecs{r}||} &= \dfrac{ \frac{d}{dt}(\vecs{r})||\vecs{r}||- \vecs{r}\frac{d}{dt}||\vecs{r}|| }{||\vecs{r}||^2} \\[5pt] &= \dfrac{ \frac{d\vecs{r}}{dt} }{||\vecs{r}||} - \dfrac{\vecs{r}}{||\vecs{r}||^2}\dfrac{d}{dt}||\vecs{r} || \\[5pt] &= \dfrac{\vecs{v}}{||\vecs{r}||} - \dfrac{\vecs{r}}{||\vecs{r}||^2} \dfrac{d}{dt}||\vecs{r}||. \end{align*}\]
Therefore, Equation \ref{Eq15} becomes
\[\dfrac{d \vecs{v}}{dt}\times \vecs{C}=GM\left( \dfrac{d}{dt}\dfrac{ \vecs{r}}{ || \vecs{r} ||} \right). \]
Since \(\vecs{C}\) is a constant vector, we can integrate both sides and obtain
\[ \vecs{v}\times\vecs{C} = GM \dfrac{ \vecs{r} }{|| \vecs{r} ||} + \vecs{D}, \]
where \(\vecs D\) is a constant vector. Our goal is to solve for \(|| \vecs{r} ||\). Let’s start by calculating \( \vecs{r} \cdot \left( \vecs{v}\times \vecs{C} \right)\):
\[\vecs{r} \cdot \left( \vecs{v}\times \vecs{C} \right) =GM\dfrac{||\vecs{r}||^2}{||\vecs{r}||}+ \vecs{r}\cdot\vecs{D} =GM||\vecs{r}||+\vecs{r}\cdot \vecs{D}. \]
However, \( \vecs{r} \cdot ( \vecs{v}\times \vecs{C})= ( \vecs{r} \times \vecs{v})\cdot \vecs{C} \), so
\[ ( \vecs{r} \times \vecs{v})\cdot \vecs{C} =GM||\vecs{r}|| + \vecs{r}\cdot \vecs{D}.\]
Since \(\vecs{r}\times \vecs{v}=\vecs{C}\), we have
\[ ||\vecs{C}||^2 =GM||\vecs{r}|| +\vecs{r}\cdot \vecs{D}. \]
Note that \( \vecs{r} \cdot \vecs{D}=||\vecs{r}|| ||\vecs{D}||\cos \theta \), where \(\theta\) is the angle between \(\vecs{r}\) and \(\vecs{D}\). Therefore,
\[ ||\vecs{C}||^2=GM||\vecs{r}||+||\vecs{r}|| ||\vecs{D}|| \cos\theta \]
Solving for \(||\vecs{r}||\),
\[ ||\vecs{r}|| = \dfrac{||\vecs{C}||^2 }{GM+||\vecs{D}||\cos\theta} = \dfrac{||\vecs{C}||^2}{GM}\left( \dfrac{1}{1+e\cos\theta} \right). \]
where \(e=||\vecs{D}||/GM\). This is the polar equation of a conic with a focus at the origin, which we set up to be the Sun. It is a hyperbola if \(e>1\), a parabola if \(e=1\), or an ellipse if \(e<1\). Since planets have closed orbits, the only possibility is an ellipse. However, at this point it should be mentioned that hyperbolic comets do exist. These are objects that are merely passing through the solar system at speeds too great to be trapped into orbit around the Sun. As they pass close enough to the Sun, the gravitational field of the Sun deflects the trajectory enough so the path becomes hyperbolic.
\(\square\)
Kepler’s third law of planetary motion can be modified to the case of one object in orbit around an object other than the Sun, such as the Moon around the Earth. In this case, Kepler’s third law becomes
\[P^2 = \dfrac{4\pi^2 a^3}{G(m+M)}, \label{Eq30}\]
where \(m\) is the mass of the Moon and \(M\) is the mass of Earth, a represents the length of the major axis of the elliptical orbit, and \(P\) represents the period.
Example \(\PageIndex{3}\): Using Kepler’s Third Law for Nonheliocentric Orbits
Given that the mass of the Moon is \(7.35\times 10^{22}\) kg, the mass of Earth is \(5.97\times 10^{24}\) kg, \(G=6.67\times 10^{−11} \text{m} / \text{kg} \cdot \text{sec}^2\), and the period of the moon is 27.3 days, let’s find the length of the major axis of the orbit of the Moon around Earth.
Solution
It is important to be consistent with units. Since the universal gravitational constant contains seconds in the units, we need to use seconds for the period of the Moon as well:
\[27.3 \; \text{days} \times \dfrac{24 \; \text{hr}}{1 \; \text{day}} \times \dfrac{3600 \; \text{sec}}{1 \; \text{hour}} =2,358,720\; \text{sec} \]
Substitute all the data into Equation \ref{Eq30} and solve for \(a\):
\[\begin{align*} (2,358,720 \; \text{sec})^2 &= \dfrac{4\pi^2a^3}{\left( 6.67\times 10^{-11} \frac{m}{\text{kg}\times \text{sec}^2}\right) (7.35\times 10^{22}\text{kg} + 5.97 \times 10^{24}\text{kg})} \\[5pt] 5.563 \times 10^{12} &= \dfrac{ 4\pi^2a^3}{(6.67 \times 10^{-11}\text{m}^3)(6.04 \times 10^{24})} \\[5pt] (5.563 \times 10^{12})(6.67 \times 10^{-11} \text{m}^3)(6.04 \times 10^{24}) &= 4\pi^2 a^3 \\[5pt] a^3 &= \dfrac{2.241 \times 10^{27}}{4\pi^2}\text{m}^3 \\[5pt] a&= 3.84 \times 10^8 \text{m} \\[5pt] &\approx 384,000 \,\text{km}. \end{align*}\]
Analysis
According to solarsystem.nasa.gov, the actual average distance from the Moon to Earth is 384,400 km. This is calculated using reflectors left on the Moon by Apollo astronauts back in the 1960s.
Exercise \(\PageIndex{3}\)
Titan is the largest moon of Saturn. The mass of Titan is approximately \(1.35 \times 10^{23} kg\). The mass of Saturn is approximately \( 5.68 \times 10^{26}\) kg. Titan takes approximately 16 days to orbit Saturn. Use this information, along with the universal gravitation constant \(G=6.67×10^{−11} \text{m}/\text{kg} \cdot \text{sec}^2\) to estimate the distance from Titan to Saturn.
- Hint
-
Make sure your units agree, then use Equation \ref{Eq30}.
- Answer
-
\[a\approx 1.224 \times 10^9 \text{m}= 1,224,000 \text{km} \]
Example \(\PageIndex{4}\): Halley’s Comet
We now return to the chapter opener, which discusses the motion of Halley’s comet around the Sun. Kepler’s first law states that Halley’s comet follows an elliptical path around the Sun, with the Sun as one focus of the ellipse. The period of Halley’s comet is approximately 76.1 years, depending on how closely it passes by Jupiter and Saturn as it passes through the outer solar system. Let’s use \(T=76.1\) years. What is the average distance of Halley’s comet from the Sun?
Solution
Using the equation \(T^2=D^3\) with \(T=76.1\), we obtain \(D^3=5791.21\), so \(D\approx 17.96\) A.U. This comes out to approximately \(1.67\times 10^9\) mi.
A natural question to ask is: What are the maximum (aphelion) and minimum (perihelion) distances from Halley’s Comet to the Sun? The eccentricity of the orbit of Halley’s Comet is 0.967 (Source: http://nssdc.gsfc.nasa.gov/planetary...cometfact.html ). Recall that the formula for the eccentricity of an ellipse is \(e=c/a\), where a is the length of the semimajor axis and c is the distance from the center to either focus. Therefore, \(0.967=c/17.96\) and \(c\approx 17.37\) A.U. Subtracting this from a gives the perihelion distance \(p=a−c=17.96−17.37=0.59\) A.U. According to the National Space Science Data Center (Source: http://nssdc.gsfc.nasa.gov/planetary...cometfact.html ), the perihelion distance for Halley’s comet is 0.587 A.U. To calculate the aphelion distance, we add
\[ P=a+c=17.96+17.37=35.33 \; \text{A.U.}\]
This is approximately \(3.3\times 10^9\) mi. The average distance from Pluto to the Sun is 39.5 A.U. (Source: http://www.oarval.org/furthest.htm ), so it would appear that Halley’s Comet stays just within the orbit of Pluto.
NAVIGATING A BANKED TURN
How fast can a racecar travel through a circular turn without skidding and hitting the wall? The answer could depend on several factors:
- The weight of the car;
- The friction between the tires and the road;
- The radius of the circle;
- The “steepness” of the turn.
In this project we investigate this question for NASCAR racecars at the Bristol Motor Speedway in Tennessee. Before considering this track in particular, we use vector functions to develop the mathematics and physics necessary for answering questions such as this.
A car of mass \(m\) moves with constant angular speed \(\omega\) around a circular curve of radius \(R\) (Figure \(\PageIndex{7}\)). The curve is banked at an angle \(\theta\). If the height of the car off the ground is \(h\), then the position of the car at time \(t\) is given by the function \(\vecs r(t)=< R\cos(\omega t),R\sin(\omega t),h>\).
- Find the velocity function \(\vecs{v}(t)\) of the car. Show that \(\vecs{v}\) is tangent to the circular curve. This means that, without a force to keep the car on the curve, the car will shoot off of it.
- Show that the speed of the car is \(\omega R\). Use this to show that \((2\pi 4)/\|\vecs{v}\|=(2\pi)/\omega \).
- Find the acceleration \(\vecs{a}\). Show that this vector points toward the center of the circle and that \(\|\vecs{a}\|=R\omega ^2\).
- The force required to produce this circular motion is called the centripetal force , and it is denoted \( \vecs{F}_{cent} \). This force points toward the center of the circle (not toward the ground). Show that \(\|\vecs{F}_{cent}\|=\left(m|\vecs{v}|^2 \right)/R\).
As the car moves around the curve, three forces act on it: gravity, the force exerted by the road (this force is perpendicular to the ground), and the friction force (Figure \(\PageIndex{8}\)). Because describing the frictional force generated by the tires and the road is complex, we use a standard approximation for the frictional force. Assume that \(\vecs{f}=\mu \vecs{N}\) for some positive constant \(\mu \). The constant \(\mu\) is called the coefficient of friction .
Let \(v_{max}\) denote the maximum speed the car can attain through the curve without skidding. In other words, \(v_{max}\) is the fastest speed at which the car can navigate the turn. When the car is traveling at this speed, the magnitude of the centripetal force is
\[\| \vecs{F}_{cent} \| = \dfrac{m(v_{max})^2}{R}. \]
The next three questions deal with developing a formula that relates the speed \(v_{max}\) to the banking angle \(\theta\).
- Show that \(\vecs{N} \cos\theta=m\vecs g+\vecs{f} \sin\theta\). Conclude that \(\vecs{N}=(m\vecs g)/(\cos\theta−\mu \sin\theta)\).
-
The centripetal force is the sum of the forces in the horizontal direction, since the centripetal force points toward the center of the circular curve. Show that
\[\vecs{F}_{cent}=\vecs{N} \sin\theta+\vecs{f}\cos\theta.\]
Conclude that\[\vecs{F}_{cent}=\dfrac{\sin\theta+\mu \cos\theta}{cos\theta−\mu \sin\theta} m\vecs g.\]
-
Show that \((v_{\text{max}})^2=((\sin\theta+\mu\ cos\theta)/(\cos\theta−\mu \sin\theta))gR\). Conclude that the maximum speed does not actually depend on the mass of the car.
Now that we have a formula relating the maximum speed of the car and the banking angle, we are in a position to answer the questions like the one posed at the beginning of the project.
The Bristol Motor Speedway is a NASCAR short track in Bristol, Tennessee. The track has the approximate shape shown in Figure \(\PageIndex{9}\). Each end of the track is approximately semicircular, so when cars make turns they are traveling along an approximately circular curve. If a car takes the inside track and speeds along the bottom of turn 1, the car travels along a semicircle of radius approximately 211 ft with a banking angle of 24°. If the car decides to take the outside track and speeds along the top of turn 1, then the car travels along a semicircle with a banking angle of 28°. (The track has variable angle banking.)
The coefficient of friction for a normal tire in dry conditions is approximately 0.7. Therefore, we assume the coefficient for a NASCAR tire in dry conditions is approximately 0.98.
Before answering the following questions, note that it is easier to do computations in terms of feet and seconds, and then convert the answers to miles per hour as a final step.
- In dry conditions, how fast can the car travel through the bottom of the turn without skidding?
- In dry conditions, how fast can the car travel through the top of the turn without skidding?
- In wet conditions, the coefficient of friction can become as low as 0.1. If this is the case, how fast can the car travel through the bottom of the turn without skidding?
- Suppose the measured speed of a car going along the outside edge of the turn is 105 mph. Estimate the coefficient of friction for the car’s tires.
Key Concepts
- The acceleration vector always points toward the concave side of the curve defined by \(\vecs{r}(t)\). The tangential and normal components of acceleration \(a_\vecs{T}\) and \(a_\vecs{N}\) are the projections of the acceleration vector onto the unit tangent and unit normal vectors to the curve.
- Kepler’s three laws of planetary motion describe the motion of objects in orbit around the Sun. His third law can be modified to describe motion of objects in orbit around other celestial objects as well.
- Newton was able to use his law of universal gravitation in conjunction with his second law of motion and calculus to prove Kepler’s three laws.
Key Equations
- Tangential component of acceleration \[a_{\vecs{T}} =\vecs{a}\cdot \vecs{T}=\dfrac{\vecs{v}\cdot \vecs{a}}{||\vecs v||} \nonumber\]
- Normal component of acceleration \[a_{\vecs{N}}=\vecs{a}\cdot \vecs{N} = \dfrac{|| \vecs{v} \times \vecs{a} ||}{||\vecs{v}||} = \sqrt{||\vecs{a}||^2 - \left(a_{\vecs{T}}\right)^2} \nonumber\]
Glossary
- acceleration vector
- the second derivative of the position vector
- Kepler’s laws of planetary motion
- three laws governing the motion of planets, asteroids, and comets in orbit around the Sun
- normal component of acceleration
- the coefficient of the unit normal vector \(\vecs N\) when the acceleration vector is written as a linear combination of \(\vecs T\) and \(\vecs N\)
- tangential component of acceleration
- the coefficient of the unit tangent vector \(\vecs T\) when the acceleration vector is written as a linear combination of \(\vecs T\) and \(\vecs N\)
Contributors
-
Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org .
- Paul Seeburger added Figure \(\PageIndex{1}\), Exercise \(\PageIndex{1}\), the exposition below Figure \(\PageIndex{1}\) through Theorem \(\PageIndex{3},\) and from Example \(\PageIndex{2}\) until just before Kepler's Laws.


7.3: The Skull
By the end of this section, you will be able to:
- List and identify the bones of the brain case and face
- Locate the major suture lines of the skull and name the bones associated with each
- Locate and define the boundaries of the anterior, middle, and posterior cranial fossae, the temporal fossa, and infratemporal fossa
- Define the paranasal sinuses and identify the location of each
- Name the bones that make up the walls of the orbit and identify the openings associated with the orbit
- Identify the bones and structures that form the nasal septum and nasal conchae, and locate the hyoid bone
- Identify the bony openings of the skull
The cranium (skull) is the skeletal structure of the head that supports the face and protects the brain. It is subdivided into the facial bones and the brain case , or cranial vault (Figure \(\PageIndex{1}\)). The facial bones underlie the facial structures, form the nasal cavity, enclose the eyeballs, and support the teeth of the upper and lower jaws. The rounded brain case surrounds and protects the brain and houses the middle and inner ear structures.
In the adult, the skull consists of 22 individual bones, 21 of which are immobile and united into a single unit. The 22nd bone is the mandible (lower jaw), which is the only moveable bone of the skull.
Interactive Link
Watch this video to view a rotating and exploded skull, with color-coded bones. Which bone (yellow) is centrally located and joins with most of the other bones of the skull?
Anterior View of Skull
The anterior skull consists of the facial bones and provides the bony support for the eyes and structures of the face. This view of the skull is dominated by the openings of the orbits and the nasal cavity. Also seen are the upper and lower jaws, with their respective teeth (Figure \(\PageIndex{2}\)).
The orbit is the bony socket that houses the eyeball and muscles that move the eyeball or open the upper eyelid. The upper margin of the anterior orbit is the supraorbital margin . Located near the midpoint of the supraorbital margin is a small opening called the supraorbital foramen . This provides for passage of a sensory nerve to the skin of the forehead. Below the orbit is the infraorbital foramen , which is the point of emergence for a sensory nerve that supplies the anterior face below the orbit.
Inside the nasal area of the skull, the nasal cavity is divided into halves by the nasal septum . The upper portion of the nasal septum is formed by the perpendicular plate of the ethmoid bone and the lower portion is the vomer bone . Each side of the nasal cavity is triangular in shape, with a broad inferior space that narrows superiorly. When looking into the nasal cavity from the front of the skull, two bony plates are seen projecting from each lateral wall. The larger of these is the inferior nasal concha , an independent bone of the skull. Located just above the inferior concha is the middle nasal concha , which is part of the ethmoid bone. A third bony plate, also part of the ethmoid bone, is the superior nasal concha . It is much smaller and out of sight, above the middle concha. The superior nasal concha is located just lateral to the perpendicular plate, in the upper nasal cavity.
Lateral View of Skull
A view of the lateral skull is dominated by the large, rounded brain case above and the upper and lower jaws with their teeth below (Figure \(\PageIndex{3}\)). Separating these areas is the bridge of bone called the zygomatic arch. The zygomatic arch is the bony arch on the side of skull that spans from the area of the cheek to just above the ear canal. It is formed by the junction of two bony processes: a short anterior component, the temporal process of the zygomatic bone (the cheekbone) and a longer posterior portion, the zygomatic process of the temporal bone , extending forward from the temporal bone. Thus the temporal process (anteriorly) and the zygomatic process (posteriorly) join together, like the two ends of a drawbridge, to form the zygomatic arch. One of the major muscles that pulls the mandible upward during biting and chewing arises from the zygomatic arch.
On the lateral side of the brain case, above the level of the zygomatic arch, is a shallow space called the temporal fossa . Below the level of the zygomatic arch and deep to the vertical portion of the mandible is another space called the infratemporal fossa . Both the temporal fossa and infratemporal fossa contain muscles that act on the mandible during chewing.
Bones of the Brain Case
The brain case contains and protects the brain. The interior space that is almost completely occupied by the brain is called the cranial cavity . This cavity is bounded superiorly by the rounded top of the skull, which is called the calvaria (skullcap), and the lateral and posterior sides of the skull. The bones that form the top and sides of the brain case are usually referred to as the “flat” bones of the skull.
The floor of the brain case is referred to as the base of the skull. This is a complex area that varies in depth and has numerous openings for the passage of cranial nerves, blood vessels, and the spinal cord. Inside the skull, the base is subdivided into three large spaces, called the anterior cranial fossa , middle cranial fossa , and posterior cranial fossa (fossa = “trench or ditch”) (Figure \(\PageIndex{4}\)). From anterior to posterior, the fossae increase in depth. The shape and depth of each fossa corresponds to the shape and size of the brain region that each houses. The boundaries and openings of the cranial fossae (singular = fossa) will be described in a later section.
The brain case consists of eight bones. These include the paired parietal and temporal bones, plus the unpaired frontal, occipital, sphenoid, and ethmoid bones.
Parietal Bone
The parietal bone forms most of the upper lateral side of the skull (see Figure \(\PageIndex{3}\)). These are paired bones, with the right and left parietal bones joining together at the top of the skull. Each parietal bone is also bounded anteriorly by the frontal bone, inferiorly by the temporal bone, and posteriorly by the occipital bone.
Temporal Bone
The temporal bone forms the lower lateral side of the skull (see Figure \(\PageIndex{3}\)). Common wisdom has it that the temporal bone (temporal = “time”) is so named because this area of the head (the temple) is where hair typically first turns gray, indicating the passage of time.
The temporal bone is subdivided into several regions (Figure \(\PageIndex{5}\)). The flattened, upper portion is the squamous portion of the temporal bone. Below this area and projecting anteriorly is the zygomatic process of the temporal bone, which forms the posterior portion of the zygomatic arch. Posteriorly is the mastoid portion of the temporal bone. Projecting inferiorly from this region is a large prominence, the mastoid process , which serves as a muscle attachment site. The mastoid process can easily be felt on the side of the head just behind your earlobe. On the interior of the skull, the petrous portion of each temporal bone forms the prominent, diagonally oriented petrous ridge in the floor of the cranial cavity. Located inside each petrous ridge are small cavities that house the structures of the middle and inner ears.
Important landmarks of the temporal bone, as shown in Figure 7.8, include the following:
- External acoustic meatus (ear canal)—This is the large opening on the lateral side of the skull that is associated with the ear.
- Internal acoustic meatus —This opening is located inside the cranial cavity, on the medial side of the petrous ridge. It connects to the middle and inner ear cavities of the temporal bone.
- Mandibular fossa —This is the deep, oval-shaped depression located on the external base of the skull, just in front of the external acoustic meatus. The mandible (lower jaw) joins with the skull at this site as part of the temporomandibular joint, which allows for movements of the mandible during opening and closing of the mouth.
- Articular tubercle —The smooth ridge located immediately anterior to the mandibular fossa. Both the articular tubercle and mandibular fossa contribute to the temporomandibular joint, the joint that provides for movements between the temporal bone of the skull and the mandible.
- Styloid process —Posterior to the mandibular fossa on the external base of the skull is an elongated, downward bony projection called the styloid process, so named because of its resemblance to a stylus (a pen or writing tool). This structure serves as an attachment site for several small muscles and for a ligament that supports the hyoid bone of the neck. (See also Figure \(\PageIndex{5}\).)
- Stylomastoid foramen —This small opening is located between the styloid process and mastoid process. This is the point of exit for the cranial nerve that supplies the facial muscles.
- Carotid canal —The carotid canal is a zig-zag shaped tunnel that provides passage through the base of the skull for one of the major arteries that supplies the brain. Its entrance is located on the outside base of the skull, anteromedial to the styloid process. The canal then runs anteromedially within the bony base of the skull, and then turns upward to its exit in the floor of the middle cranial cavity, above the foramen lacerum.
Frontal Bone
The frontal bone is the single bone that forms the forehead. At its anterior midline, between the eyebrows, there is a slight depression called the glabella (see Figure 7.5). The frontal bone also forms the supraorbital margin of the orbit. Near the middle of this margin, is the supraorbital foramen, the opening that provides passage for a sensory nerve to the forehead. The frontal bone is thickened just above each supraorbital margin, forming rounded brow ridges. These are located just behind your eyebrows and vary in size among individuals, although they are generally larger in males. Inside the cranial cavity, the frontal bone extends posteriorly. This flattened region forms both the roof of the orbit below and the floor of the anterior cranial cavity above (see Figure 7.8 b ).
Occipital Bone
The occipital bone is the single bone that forms the posterior skull and posterior base of the cranial cavity (Figure \(\PageIndex{7}\); see also Figure \(\PageIndex{6}\)). On its outside surface, at the posterior midline, is a small protrusion called the external occipital protuberance , which serves as an attachment site for a ligament of the posterior neck. Lateral to either side of this bump is a superior nuchal line (nuchal = “nape” or “posterior neck”). The nuchal lines represent the most superior point at which muscles of the neck attach to the skull, with only the scalp covering the skull above these lines. On the base of the skull, the occipital bone contains the large opening of the foramen magnum , which allows for passage of the spinal cord as it exits the skull. On either side of the foramen magnum is an oval-shaped occipital condyle . These condyles form joints with the first cervical vertebra and thus support the skull on top of the vertebral column.
Sphenoid Bone
The sphenoid bone is a single, complex bone of the central skull (Figure \(\PageIndex{8}\)). It serves as a “keystone” bone, because it joins with almost every other bone of the skull. The sphenoid forms much of the base of the central skull (see Figure \(\PageIndex{6}\)) and also extends laterally to contribute to the sides of the skull (see Figure \(\PageIndex{3}\)). Inside the cranial cavity, the right and left lesser wings of the sphenoid bone , which resemble the wings of a flying bird, form the lip of a prominent ridge that marks the boundary between the anterior and middle cranial fossae. The sella turcica (“Turkish saddle”) is located at the midline of the middle cranial fossa. This bony region of the sphenoid bone is named for its resemblance to the horse saddles used by the Ottoman Turks, with a high back and a tall front. The rounded depression in the floor of the sella turcica is the hypophyseal (pituitary) fossa , which houses the pea-sized pituitary (hypophyseal) gland. The greater wings of the sphenoid bone extend laterally to either side away from the sella turcica, where they form the anterior floor of the middle cranial fossa. The greater wing is best seen on the outside of the lateral skull, where it forms a rectangular area immediately anterior to the squamous portion of the temporal bone.
On the inferior aspect of the skull, each half of the sphenoid bone forms two thin, vertically oriented bony plates. These are the medial pterygoid plate and lateral pterygoid plate (pterygoid = “wing-shaped”). The right and left medial pterygoid plates form the posterior, lateral walls of the nasal cavity. The somewhat larger lateral pterygoid plates serve as attachment sites for chewing muscles that fill the infratemporal space and act on the mandible.
Ethmoid Bone
The ethmoid bone is a single, midline bone that forms the roof and lateral walls of the upper nasal cavity, the upper portion of the nasal septum, and contributes to the medial wall of the orbit (Figure \(\PageIndex{9}\) and Figure \(\PageIndex{10}\)). On the interior of the skull, the ethmoid also forms a portion of the floor of the anterior cranial cavity (see Figure 7.8 b ).
Within the nasal cavity, the perpendicular plate of the ethmoid bone forms the upper portion of the nasal septum. The ethmoid bone also forms the lateral walls of the upper nasal cavity. Extending from each lateral wall are the superior nasal concha and middle nasal concha, which are thin, curved projections that extend into the nasal cavity (Figure \(\PageIndex{11}\)).
In the cranial cavity, the ethmoid bone forms a small area at the midline in the floor of the anterior cranial fossa. This region also forms the narrow roof of the underlying nasal cavity. This portion of the ethmoid bone consists of two parts, the crista galli and cribriform plates. The crista galli (“rooster’s comb or crest”) is a small upward bony projection located at the midline. It functions as an anterior attachment point for one of the covering layers of the brain. To either side of the crista galli is the cribriform plate (cribrum = “sieve”), a small, flattened area with numerous small openings termed olfactory foramina. Small nerve branches from the olfactory areas of the nasal cavity pass through these openings to enter the brain.
The lateral portions of the ethmoid bone are located between the orbit and upper nasal cavity, and thus form the lateral nasal cavity wall and a portion of the medial orbit wall. Located inside this portion of the ethmoid bone are several small, air-filled spaces that are part of the paranasal sinus system of the skull.
Sutures of the Skull
A suture is an immobile joint between adjacent bones of the skull. The narrow gap between the bones is filled with dense, fibrous connective tissue that unites the bones. The long sutures located between the bones of the brain case are not straight, but instead follow irregular, tightly twisting paths. These twisting lines serve to tightly interlock the adjacent bones, thus adding strength to the skull for brain protection.
The two suture lines seen on the top of the skull are the coronal and sagittal sutures. The coronal suture runs from side to side across the skull, within the coronal plane of section (see Figure \(\PageIndex{3}\)). It joins the frontal bone to the right and left parietal bones. The sagittal suture extends posteriorly from the coronal suture, running along the midline at the top of the skull in the sagittal plane of section (see Figure \(\PageIndex{7}\)). It unites the right and left parietal bones. On the posterior skull, the sagittal suture terminates by joining the lambdoid suture. The lambdoid suture extends downward and laterally to either side away from its junction with the sagittal suture. The lambdoid suture joins the occipital bone to the right and left parietal and temporal bones. This suture is named for its upside-down "V" shape, which resembles the capital letter version of the Greek letter lambda (Λ). The squamous suture is located on the lateral skull. It unites the squamous portion of the temporal bone with the parietal bone (see Figure 7.5). At the intersection of four bones is the pterion , a small, capital-H-shaped suture line region that unites the frontal bone, parietal bone, squamous portion of the temporal bone, and greater wing of the sphenoid bone. It is the weakest part of the skull. The pterion is located approximately two finger widths above the zygomatic arch and a thumb’s width posterior to the upward portion of the zygomatic bone.
Disorders of the...
Skeletal System
Head and traumatic brain injuries are major causes of immediate death and disability, with bleeding and infections as possible additional complications. According to the Centers for Disease Control and Prevention (2010), approximately 30 percent of all injury-related deaths in the United States are caused by head injuries. The majority of head injuries involve falls. They are most common among young children (ages 0–4 years), adolescents (15–19 years), and the elderly (over 65 years). Additional causes vary, but prominent among these are automobile and motorcycle accidents.
Strong blows to the brain-case portion of the skull can produce fractures. These may result in bleeding inside the skull with subsequent injury to the brain. The most common is a linear skull fracture, in which fracture lines radiate from the point of impact. Other fracture types include a comminuted fracture, in which the bone is broken into several pieces at the point of impact, or a depressed fracture, in which the fractured bone is pushed inward. In a contrecoup (counterblow) fracture, the bone at the point of impact is not broken, but instead a fracture occurs on the opposite side of the skull. Fractures of the occipital bone at the base of the skull can occur in this manner, producing a basilar fracture that can damage the artery that passes through the carotid canal.
A blow to the lateral side of the head may fracture the bones of the pterion. The pterion is an important clinical landmark because located immediately deep to it on the inside of the skull is a major branch of an artery that supplies the skull and covering layers of the brain. A strong blow to this region can fracture the bones around the pterion. If the underlying artery is damaged, bleeding can cause the formation of a hematoma (collection of blood) between the brain and interior of the skull. As blood accumulates, it will put pressure on the brain. Symptoms associated with a hematoma may not be apparent immediately following the injury, but if untreated, blood accumulation will exert increasing pressure on the brain and can result in death within a few hours.
Interactive Link
View this animation to see how a blow to the head may produce a contrecoup (counterblow) fracture of the basilar portion of the occipital bone on the base of the skull. Why may a basilar fracture be life threatening?
Facial Bones of the Skull
The facial bones of the skull form the upper and lower jaws, the nose, nasal cavity and nasal septum, and the orbit. The facial bones include 14 bones, with six paired bones and two unpaired bones. The paired bones are the maxilla, palatine, zygomatic, nasal, lacrimal, and inferior nasal conchae bones. The unpaired bones are the vomer and mandible bones. Although classified with the brain-case bones, the ethmoid bone also contributes to the nasal septum and the walls of the nasal cavity and orbit.
Maxillary Bone
The maxillary bone , often referred to simply as the maxilla (plural = maxillae), is one of a pair that together form the upper jaw, much of the hard palate, the medial floor of the orbit, and the lateral base of the nose (see Figure \(\PageIndex{2}\)). The curved, inferior margin of the maxillary bone that forms the upper jaw and contains the upper teeth is the alveolar process of the maxilla (Figure \(\PageIndex{12}\)). Each tooth is anchored into a deep socket called an alveolus. On the anterior maxilla, just below the orbit, is the infraorbital foramen. This is the point of exit for a sensory nerve that supplies the nose, upper lip, and anterior cheek. On the inferior skull, the palatine process from each maxillary bone can be seen joining together at the midline to form the anterior three-quarters of the hard palate (see Figure \(\PageIndex{6}\) a ). The hard palate is the bony plate that forms the roof of the mouth and floor of the nasal cavity, separating the oral and nasal cavities.
Palatine Bone
The palatine bone is one of a pair of irregularly shaped bones that contribute small areas to the lateral walls of the nasal cavity and the medial wall of each orbit. The largest region of each of the palatine bone is the horizontal plate . The plates from the right and left palatine bones join together at the midline to form the posterior quarter of the hard palate (see Figure 7.8 a ). Thus, the palatine bones are best seen in an inferior view of the skull and hard palate.
Homeostatic Imbalances
Cleft Lip and Cleft Palate
During embryonic development, the right and left maxilla bones come together at the midline to form the upper jaw. At the same time, the muscle and skin overlying these bones join together to form the upper lip. Inside the mouth, the palatine processes of the maxilla bones, along with the horizontal plates of the right and left palatine bones, join together to form the hard palate. If an error occurs in these developmental processes, a birth defect of cleft lip or cleft palate may result.
Cleft lip is a common development defect that affects approximately 1:1000 births, most of which are male. This defect involves a partial or complete failure of the right and left portions of the upper lip to fuse together, leaving a cleft (gap).
A more severe developmental defect is cleft palate, which affects the hard palate. The hard palate is the bony structure that separates the nasal cavity from the oral cavity. It is formed during embryonic development by the midline fusion of the horizontal plates from the right and left palatine bones and the palatine processes of the maxilla bones. Cleft palate affects approximately 1:2500 births and is more common in females. It results from a failure of the two halves of the hard palate to completely come together and fuse at the midline, thus leaving a gap between them. This gap allows for communication between the nasal and oral cavities. In severe cases, the bony gap continues into the anterior upper jaw where the alveolar processes of the maxilla bones also do not properly join together above the front teeth. If this occurs, a cleft lip will also be seen. Because of the communication between the oral and nasal cavities, a cleft palate makes it very difficult for an infant to generate the suckling needed for nursing, thus leaving the infant at risk for malnutrition. Surgical repair is required to correct cleft palate defects.
Zygomatic Bone
The zygomatic bone is also known as the cheekbone. Each of the paired zygomatic bones forms much of the lateral wall of the orbit and the lateral-inferior margins of the anterior orbital opening (see Figure 7.4). The short temporal process of the zygomatic bone projects posteriorly, where it forms the anterior portion of the zygomatic arch (see Figure 7.5).
Nasal Bone
The nasal bone is one of two small bones that articulate (join) with each other to form the bony base (bridge) of the nose. They also support the cartilages that form the lateral walls of the nose (see Figure 7.11). These are the bones that are damaged when the nose is broken.
Lacrimal Bone
Each lacrimal bone is a small, rectangular bone that forms the anterior, medial wall of the orbit (see Figure 7.4 and Figure 7.5). The anterior portion of the lacrimal bone forms a shallow depression called the lacrimal fossa , and extending inferiorly from this is the nasolacrimal canal . The lacrimal fluid (tears of the eye), which serves to maintain the moist surface of the eye, drains at the medial corner of the eye into the nasolacrimal canal. This duct then extends downward to open into the nasal cavity, behind the inferior nasal concha. In the nasal cavity, the lacrimal fluid normally drains posteriorly, but with an increased flow of tears due to crying or eye irritation, some fluid will also drain anteriorly, thus causing a runny nose.
Inferior Nasal Conchae
The right and left inferior nasal conchae form a curved bony plate that projects into the nasal cavity space from the lower lateral wall (see Figure 7.13). The inferior concha is the largest of the nasal conchae and can easily be seen when looking into the anterior opening of the nasal cavity.
Vomer Bone
The unpaired vomer bone, often referred to simply as the vomer, is triangular-shaped and forms the posterior-inferior part of the nasal septum (see Figure 7.11). The vomer is best seen when looking from behind into the posterior openings of the nasal cavity (see Figure 7.8 a ). In this view, the vomer is seen to form the entire height of the nasal septum. A much smaller portion of the vomer can also be seen when looking into the anterior opening of the nasal cavity.
Mandible
The mandible forms the lower jaw and is the only moveable bone of the skull. At the time of birth, the mandible consists of paired right and left bones, but these fuse together during the first year to form the single U-shaped mandible of the adult skull. Each side of the mandible consists of a horizontal body and posteriorly, a vertically oriented ramus of the mandible (ramus = “branch”). The outside margin of the mandible, where the body and ramus come together is called the angle of the mandible (Figure \(\PageIndex{13}\)).
The ramus on each side of the mandible has two upward-going bony projections. The more anterior projection is the flattened coronoid process of the mandible , which provides attachment for one of the biting muscles. The posterior projection is the condylar process of the mandible , which is topped by the oval-shaped condyle . The condyle of the mandible articulates (joins) with the mandibular fossa and articular tubercle of the temporal bone. Together these articulations form the temporomandibular joint, which allows for opening and closing of the mouth (see Figure \(\PageIndex{3}\)). The broad U-shaped curve located between the coronoid and condylar processes is the mandibular notch .
Important landmarks for the mandible include the following:
- Alveolar process of the mandible —This is the upper border of the mandibular body and serves to anchor the lower teeth.
- Mental protuberance —The forward projection from the inferior margin of the anterior mandible that forms the chin (mental = “chin”).
- Mental foramen —The opening located on each side of the anterior-lateral mandible, which is the exit site for a sensory nerve that supplies the chin.
- Mylohyoid line —This bony ridge extends along the inner aspect of the mandibular body (see Figure 7.11). The muscle that forms the floor of the oral cavity attaches to the mylohyoid lines on both sides of the mandible.
- Mandibular foramen —This opening is located on the medial side of the ramus of the mandible. The opening leads into a tunnel that runs down the length of the mandibular body. The sensory nerve and blood vessels that supply the lower teeth enter the mandibular foramen and then follow this tunnel. Thus, to numb the lower teeth prior to dental work, the dentist must inject anesthesia into the lateral wall of the oral cavity at a point prior to where this sensory nerve enters the mandibular foramen.
- Lingula —This small flap of bone is named for its shape (lingula = “little tongue”). It is located immediately next to the mandibular foramen, on the medial side of the ramus. A ligament that anchors the mandible during opening and closing of the mouth extends down from the base of the skull and attaches to the lingula.
The Orbit
The orbit is the bony socket that houses the eyeball and contains the muscles that move the eyeball or open the upper eyelid. Each orbit is cone-shaped, with a narrow posterior region that widens toward the large anterior opening. To help protect the eye, the bony margins of the anterior opening are thickened and somewhat constricted. The medial walls of the two orbits are parallel to each other but each lateral wall diverges away from the midline at a 45° angle. This divergence provides greater lateral peripheral vision.
The walls of each orbit include contributions from seven skull bones (Figure \(\PageIndex{14}\)). The frontal bone forms the roof and the zygomatic bone forms the lateral wall and lateral floor. The medial floor is primarily formed by the maxilla, with a small contribution from the palatine bone. The ethmoid bone and lacrimal bone make up much of the medial wall and the sphenoid bone forms the posterior orbit.
At the posterior apex of the orbit is the opening of the optic canal , which allows for passage of the optic nerve from the retina to the brain. Lateral to this is the elongated and irregularly shaped superior orbital fissure, which provides passage for the artery that supplies the eyeball, sensory nerves, and the nerves that supply the muscles involved in eye movements.
The Nasal Septum and Nasal Conchae
The nasal septum consists of both bone and cartilage components (Figure \(\PageIndex{15}\); see also Figure \(\PageIndex{9}\)). The upper portion of the septum is formed by the perpendicular plate of the ethmoid bone. The lower and posterior parts of the septum are formed by the triangular-shaped vomer bone. In an anterior view of the skull, the perpendicular plate of the ethmoid bone is easily seen inside the nasal opening as the upper nasal septum, but only a small portion of the vomer is seen as the inferior septum. A better view of the vomer bone is seen when looking into the posterior nasal cavity with an inferior view of the skull, where the vomer forms the full height of the nasal septum. The anterior nasal septum is formed by the septal cartilage , a flexible plate that fills in the gap between the perpendicular plate of the ethmoid and vomer bones. This cartilage also extends outward into the nose where it separates the right and left nostrils. The septal cartilage is not found in the dry skull.
Attached to the lateral wall on each side of the nasal cavity are the superior, middle, and inferior nasal conchae (singular = concha), which are named for their positions (see Figure 7.13). These are bony plates that curve downward as they project into the space of the nasal cavity. They serve to swirl the incoming air, which helps to warm and moisturize it before the air moves into the delicate air sacs of the lungs. This also allows mucus, secreted by the tissue lining the nasal cavity, to trap incoming dust, pollen, bacteria, and viruses. The largest of the conchae is the inferior nasal concha, which is an independent bone of the skull. The middle concha and the superior conchae, which is the smallest, are both formed by the ethmoid bone. When looking into the anterior nasal opening of the skull, only the inferior and middle conchae can be seen. The small superior nasal concha is well hidden above and behind the middle concha.
Cranial Fossae
Inside the skull, the floor of the cranial cavity is subdivided into three cranial fossae (spaces), which increase in depth from anterior to posterior (see Figure 7.6, Figure 7.8 b , and Figure 7.11). Since the brain occupies these areas, the shape of each conforms to the shape of the brain regions that it contains. Each cranial fossa has anterior and posterior boundaries and is divided at the midline into right and left areas by a significant bony structure or opening.
Anterior Cranial Fossa
The anterior cranial fossa is the most anterior and the shallowest of the three cranial fossae. It overlies the orbits and contains the frontal lobes of the brain. Anteriorly, the anterior fossa is bounded by the frontal bone, which also forms the majority of the floor for this space. The lesser wings of the sphenoid bone form the prominent ledge that marks the boundary between the anterior and middle cranial fossae. Located in the floor of the anterior cranial fossa at the midline is a portion of the ethmoid bone, consisting of the upward projecting crista galli and to either side of this, the cribriform plates.
Middle Cranial Fossa
The middle cranial fossa is deeper and situated posterior to the anterior fossa. It extends from the lesser wings of the sphenoid bone anteriorly, to the petrous ridges (petrous portion of the temporal bones) posteriorly. The large, diagonally positioned petrous ridges give the middle cranial fossa a butterfly shape, making it narrow at the midline and broad laterally. The temporal lobes of the brain occupy this fossa. The middle cranial fossa is divided at the midline by the upward bony prominence of the sella turcica, a part of the sphenoid bone. The middle cranial fossa has several openings for the passage of blood vessels and cranial nerves (see Figure 7.8).
Openings in the middle cranial fossa are as follows:
- Optic canal —This opening is located at the anterior lateral corner of the sella turcica. It provides for passage of the optic nerve into the orbit.
- Superior orbital fissure —This large, irregular opening into the posterior orbit is located on the anterior wall of the middle cranial fossa, lateral to the optic canal and under the projecting margin of the lesser wing of the sphenoid bone. Nerves to the eyeball and associated muscles, and sensory nerves to the forehead pass through this opening.
- Foramen rotundum —This rounded opening (rotundum = “round”) is located in the floor of the middle cranial fossa, just inferior to the superior orbital fissure. It is the exit point for a major sensory nerve that supplies the cheek, nose, and upper teeth.
- Foramen ovale of the middle cranial fossa —This large, oval-shaped opening in the floor of the middle cranial fossa provides passage for a major sensory nerve to the lateral head, cheek, chin, and lower teeth.
- Foramen spinosum —This small opening, located posterior-lateral to the foramen ovale, is the entry point for an important artery that supplies the covering layers surrounding the brain. The branching pattern of this artery forms readily visible grooves on the internal surface of the skull and these grooves can be traced back to their origin at the foramen spinosum.
- Carotid canal —This is the zig-zag passageway through which a major artery to the brain enters the skull. The entrance to the carotid canal is located on the inferior aspect of the skull, anteromedial to the styloid process (see Figure 7.8 a ). From here, the canal runs anteromedially within the bony base of the skull. Just above the foramen lacerum, the carotid canal opens into the middle cranial cavity, near the posterior-lateral base of the sella turcica.
- Foramen lacerum —This irregular opening is located in the base of the skull, immediately inferior to the exit of the carotid canal. This opening is an artifact of the dry skull, because in life it is completely filled with cartilage. All the openings of the skull that provide for passage of nerves or blood vessels have smooth margins; the word lacerum (“ragged” or “torn”) tells us that this opening has ragged edges and thus nothing passes through it.
Posterior Cranial Fossa
The posterior cranial fossa is the most posterior and deepest portion of the cranial cavity. It contains the cerebellum of the brain. The posterior fossa is bounded anteriorly by the petrous ridges, while the occipital bone forms the floor and posterior wall. It is divided at the midline by the large foramen magnum (“great aperture”), the opening that provides for passage of the spinal cord.
Located on the medial wall of the petrous ridge in the posterior cranial fossa is the internal acoustic meatus (see Figure 7.11). This opening provides for passage of the nerve from the hearing and equilibrium organs of the inner ear, and the nerve that supplies the muscles of the face. Located at the anterior-lateral margin of the foramen magnum is the hypoglossal canal . These emerge on the inferior aspect of the skull at the base of the occipital condyle and provide passage for an important nerve to the tongue.
Immediately inferior to the internal acoustic meatus is the large, irregularly shaped jugular foramen (see Figure 7.8 a ). Several cranial nerves from the brain exit the skull via this opening. It is also the exit point through the base of the skull for all the venous return blood leaving the brain. The venous structures that carry blood inside the skull form large, curved grooves on the inner walls of the posterior cranial fossa, which terminate at each jugular foramen.
Paranasal Sinuses
The paranasal sinuses are hollow, air-filled spaces located within certain bones of the skull (Figure \(\PageIndex{16}\)). All of the sinuses communicate with the nasal cavity (paranasal = “next to nasal cavity”) and are lined with nasal mucosa. They serve to reduce bone mass and thus lighten the skull, and they also add resonance to the voice. This second feature is most obvious when you have a cold or sinus congestion. These produce swelling of the mucosa and excess mucus production, which can obstruct the narrow passageways between the sinuses and the nasal cavity, causing your voice to sound different to yourself and others. This blockage can also allow the sinuses to fill with fluid, with the resulting pressure producing pain and discomfort.
The paranasal sinuses are named for the skull bone that each occupies. The frontal sinus is located just above the eyebrows, within the frontal bone (see Figure 7.17). This irregular space may be divided at the midline into bilateral spaces, or these may be fused into a single sinus space. The frontal sinus is the most anterior of the paranasal sinuses. The largest sinus is the maxillary sinus . These are paired and located within the right and left maxillary bones, where they occupy the area just below the orbits. The maxillary sinuses are most commonly involved during sinus infections. Because their connection to the nasal cavity is located high on their medial wall, they are difficult to drain. The sphenoid sinus is a single, midline sinus. It is located within the body of the sphenoid bone, just anterior and inferior to the sella turcica, thus making it the most posterior of the paranasal sinuses. The lateral aspects of the ethmoid bone contain multiple small spaces separated by very thin bony walls. Each of these spaces is called an ethmoid air cell . These are located on both sides of the ethmoid bone, between the upper nasal cavity and medial orbit, just behind the superior nasal conchae.
Hyoid Bone
The hyoid bone is an independent bone that does not contact any other bone and thus is not part of the skull (Figure \(\PageIndex{17}\)). It is a small U-shaped bone located in the upper neck near the level of the inferior mandible, with the tips of the “U” pointing posteriorly. The hyoid serves as the base for the tongue above, and is attached to the larynx below and the pharynx posteriorly. The hyoid is held in position by a series of small muscles that attach to it either from above or below. These muscles act to move the hyoid up/down or forward/back. Movements of the hyoid are coordinated with movements of the tongue, larynx, and pharynx during swallowing and speaking.


6.7: Induction in geometry, combinatorics and number theory
We turn next to a mixed collection of problems designed to highlight a range of applications.
Problem 256 Let , . Prove by induction that has at least k distinct prime factors.
Problem 257
(a) Prove by induction that n points on a straight line divide the line into parts.
(b)(i) By experimenting with small values of n , guess a formula for the maximum number of regions which can be created in the plane by an array of n straight lines.
(ii) Prove by induction that n straight lines in the plane divide the plane into at most regions.
(c)(i) By experimenting with small values of n , guess a formula for the maximum number of regions which can be created in 3-dimensions by an array of n planes.
(ii) Prove by induction that n planes in 3-dimensions divide space into at most regions.
Problem 258 Given a square, prove that, for each , the initial square can be cut into n squares (of possibly different sizes).
Problem 259 A tree is a connected graph, or network, consisting of vertices and edges, but with no cycles (or circuits). Prove that a tree with n vertices has exactly edges.
The next problem concerns spherical polyhedra . A spherical polyhedron is a polyhedral surface in 3-dimensions, which can be inflated to form a sphere (where we assume that the faces and edges can stretch as required). For example, a cube is a spherical polyhedron; but the surface of a picture frame is not. A spherical polyhedron has
- faces (flat 2-dimensional polygons, which can be stretched to take the form of a disc),
- edges (1-dimensional line segments, where exactly two faces meet), and
- vertices (0-dimensional points, where several faces and edges meet, in such a way that they form a single cycle around the vertex).
Each face must clearly have at least 3 edges; and there must be at least three edges and three faces meeting at each vertex.
If a spherical polyhedron has V vertices, E edges, and F faces, then the numbers V , E , F satisfy Euler's formula
For example, a cube has vertices, edges, and faces, and .
Problem 260
(a) (i) Describe a spherical polyhedron with exactly 6 edges.
(ii) Describe a spherical polyhedron with exactly 8 edges.
(b) Prove that no spherical polyhedron can have exactly 7 edges.
(c) Prove that for every , there exists a spherical polyhedron with n edges.
Problem 261 A map is a (finite) collection of regions in the plane, each with a boundary, or border, that is `polygonal' in the sense that it consists of a single sequence of distinct vertices and – possibly curved – edges, that separates the plane into two parts, one of which is the polygonal region itself. A map can be properly coloured if each region can be assigned a colour so that each pair of neighbouring regions (sharing an edge) always receive different colours. Prove that the regions of such a map can be properly coloured with just two colours if and only if an even number of edges meet at each vertex.
Problem 262 (Gray codes) There are sequences of length n consisting of 0s and 1s. Prove that, for each , these sequences can be arranged in a cyclic list such that any two neighbouring sequences (including the last and the first) differ in exactly one coordinate position.
Problem 263 (Calkin-Wilf tree) The binary tree in the plane has a distinguished vertex as `root', and is constructed inductively. The root is joined to two new vertices; and each new vertex is then joined to two further new vertices – with the construction process continuing for ever ( Figure 11 ). Label the vertices of the binary tree with positive fractions as follows:
- the root is given the label
- whenever we know the label of a ‘parent’ vertex, we label its `left descendant' as , and its `right descendant' .
(a) Prove that every positive rational occurs once and only once as a label, and that it occurs in its lowest terms.
(b) Prove that the labels are left-right symmetric in the sense that labels in corresponding left and right positions are reciprocals of each other.
Problem 264 A collection of n intervals on the x -axis is such that every pair of intervals have a point in common. Prove that all n intervals must then have at least one point in common.


7.3.2: Size and Shapes of Viruses
-
- Last updated
- Save as PDF
- Ying Liu
- City College of San Francisco
Learning Objectives
- Compare the size of most viruses to that of bacteria.
- List 4 shapes of viruses.
Size
Viruses are usually much smaller than bacteria with the vast majority being submicroscopic. While most viruses range in size from 5 to 300 nanometers (nm) , in recent years a number of giant viruses, including Mimiviruses and Pandoraviruses with a diameter of 0.4 micrometers (µm) , have been identified. For a comparison of the size of a virus, a bacterium, and a human cell, scroll down to how big is... on the Cell Size and Scale Resource at the University of Utah webpage (see Figure \(\PageIndex{1}\)A, Figure \(\PageIndex{1}\)B, and Figure \(\PageIndex{1}\)C),
Shapes
Figure \(\PageIndex{1}\)C: Sizes and Shapes of Viruses (Bacteriophages)
a. Helical viruses consist of nucleic acid surrounded by a hollow protein cylinder or capsid and possessing a helical structure (Figure \(\PageIndex{2}\)A).
b. Polyhedral viruses consist of nucleic acid surrounded by a polyhedral (many-sided) shell or capsid, usually in the form of an icosahedron (Figure \(\PageIndex{2}\)B).
- Transmission electron micrograph of Adenoviruses ; courtesy of CDC.
- Transmission electron micrograph of poliomyelitis viruses ; courtesy of CDC.
- Transmission electron micrograph of poliomyelitis viruses ; courtesy of Dennis Kunkel's Microscopy .
c. Enveloped viruses consist of nucleic acid surrounded by either a helical or polyhedral core and covered by an envelope (see Figure \(\PageIndex{2}\)C and Figure \(\PageIndex{2}\)D).
- Transmission electron micrograph of Hepatitis B viruses ; courtesy of CDC.
- Transmission electron micrograph of an Influenza A virus ; courtesy of CDC.
- Transmission electron micrograph of HIV ; courtesy of CDC.
- Transmission electron micrograph showing envelope and glycoprotein spikes Coronaviruses ; courtesy of CDC.
- Transmission electron micrograph of herpes simplex viruses ; courtesy of Dennis Kunkel's Microscopy .
d. Binal (complex) viruses have neither helical nor polyhedral forms, are pleomorphic or irregular shaped (Figure \(\PageIndex{3}\)), or have complex structures (Figure \(\PageIndex{2}\)F).
- Transmission electron micrograph of the bacteriophage coliphage T4 ; courtesy of Dennis Kunkel's Microscopy .
Exercise: Think-Pair-Share Questions
We just learned that most viruses are much smaller than bacteria.
- Compare the sizes of viruses and bacteria.
- Why are viruses able to be so much smaller than bacteria
Summary
- Viruses are usually much smaller than bacteria with the vast majority being submicroscopic, generally ranging in size from 5 to 300 nanometers (nm).
- Helical viruses consist of nucleic acid surrounded by a hollow protein cylinder or capsid and possessing a helical structure.
- Polyhedral viruses consist of nucleic acid surrounded by a polyhedral (many-sided) shell or capsid, usually in the form of an icosahedron.
- Enveloped viruses consist of nucleic acid surrounded by either a helical or polyhedral core and covered by an envelope.
- Binal (complex) viruses have neither helical nor polyhedral forms, have irregular shapes, or have complex structures.