Why does the Parker Solar Probe slow down as the distance from the Sun increases?

Image of probe's trajectory
Image credit: Wikipedia user Phoenix777, CC BY-SA 4.0

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    $\begingroup$ Hey, folks. If you have an answer, please post it as an answer, rather than a comment. Thanks! $\endgroup$ – HDE 226868 May 25 '20 at 3:21

Why does the “parker solar probe” lose speed as the distance from the sun increases?

Because energy and angular momentum are individually conserved quantities in the two body problem. Except for where the Parker Space Probe has close fly-bys with Venus, the gravitationally interactions between the Parker Space Probe and the solar system are very closely modeled as a two body problem (the Sun and the probe), plus very small perturbations from the planets.

One way to express conservation of energy in the two body problem is the vis-viva equation, $$v^2 = \mu\left(\frac2r - \frac1a\right)$$


  • $\mu = G(M+m)$ is the sum of the central body's standard gravitational parameter and that of the orbiting body,
  • $r$ is the distance between the two bodies,
  • $a$ is the semi-major axis length (a constant), and
  • $v$ is the magnitude of the velocity vector.

Note that the mass of the Parker Space Probe is so much less than that of the Sun that one can drop the Parker Space Probe's mass from the expression $\mu = G(M+m)$, resulting in $\mu = GM_{\text{sun}}$.

Note that the only variable on the right hand side of the vis-viva equation is the radial distance. As radial distance increases the square magnitude of the velocity vector (and thus the magnitude of the velocity vector) decreases.

Without mathematics, conservation of energy dictates that the sum of an orbiting body's kinetic energy and gravitational potential energy must remain constant. As the orbiting body moves further from the central body, the orbiting body's potential energy increases, which means its kinetic energy must correspondingly decrease. This in turn means the orbiting body's velocity must decrease.

  • $\begingroup$ And how does this relate to the gravitational assistance used in Voyager missions? From what I understand, when moving away from the massive object, the probes should lose speed? $\endgroup$ – user1785960 May 22 '20 at 7:58
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    $\begingroup$ @user1785960 - They have lost speed, a whole lot of speed. But now that the Voyager probes are 123.9 astronomical units from the Sun (Voyager 2) and 149.3 astronomical units from the Sun (Voyager 1), the $\frac2r$ terms in the vis-viva equation are very small compared to the $\frac1a$ terms for both vehicles. While they do continue to lose speed, the loss is now very small for both of them. $\endgroup$ – David Hammen May 22 '20 at 8:15
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    $\begingroup$ @user1785960 In fact this is how gravity assists work. If you pass by a planet, you first gain speed as you come closer and then lose the same amount as you move away. But this speed is with respect to the planet, which is itself moving. If you play around a bit with directions and velocities, you'll notice that there are even situations where losing speed with respect to the planet actually means gaining speed with respect to the sun. Now a gravity assist is just picking the right directions when approaching and leaving a planet to benefit from these effects. $\endgroup$ – mlk May 22 '20 at 9:00
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    $\begingroup$ @AdamBarnes - Personally, I'm not a big fan of Kerbal Space Program. It makes too many simplifying assumptions for my taste. But nonetheless I do see the teaching value. $\endgroup$ – David Hammen May 23 '20 at 12:02
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    $\begingroup$ @user11153 there is a slight difference – gravity is attractive (and works at distance), while rubber ball collision is repulsive and works only at the (short) time they are actually touching. $\endgroup$ – Paŭlo Ebermann May 23 '20 at 23:57

Quite simply, the sun's gravity is pulling on the space probe at all times. As the probe moves away from the sun, the force of gravity pulls it back in, slowing it down. As the probe moves toward the sun, the force of gravity continues to pull, speeding it up. Any object orbiting the sun is always accelerating toward the sun - when that acceleration opposes motion, the object slows down, and when it's in the same direction as motion, the object speeds up.

It's no different from throwing a ball into the air and seeing it slow down as it rises, reverse direction, and then speed up again as it falls, except the primary gravitational body in that case is the earth and not the sun.

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    $\begingroup$ Or like a pendulum being pulled by the Earth. Closer to the Earth = fastest. Farthest from the earth = slowest. The difference between the ball and the pendulum is the pendulum "misses" the Earth when falling towards it like things in orbit. $\endgroup$ – DKNguyen May 22 '20 at 20:08

The simplest explanation is that the satellite generally obeys Kepler's second law of orbital mechanics:


When the satellite is further away from the sun, the area swept out remains constant only because the satellite is moving more slowly.

The physical answer, as others have pointed out, is that the satellite trades out kinetic energy for gravitational potential energy as it travels "up" the Sun's gravity well. This is no different from gradually slowing as you coast up a hill on your bike (although you are climbing up the Earth's gravity well not the Sun's).


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