I was wondering, what is the minimum amount of energy required to put 1 kg of payload into LEO?

  • $\begingroup$ To clarify, you mean just the 1 KG, with something like a rail gun? $\endgroup$ Apr 22 '14 at 10:44
  • $\begingroup$ Well I dont have any specific approach in mind, But I think something what space rockets need would also make picture clearer. $\endgroup$ Apr 22 '14 at 10:49
  • $\begingroup$ @JamesJenkins 1 Kg with a rail gun is an unusual method to expect someone to mean. I can't think of any instance where anything has been launched to orbit from a rail gun. $\endgroup$
    – ThePlanMan
    Apr 22 '14 at 13:48
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    $\begingroup$ @FraserOfSmeg I meant without consideration for lifting the fuel, the engine, the fuel tank, the cargo container, and all the other requirements for a self contained lift system (i.e. rocket) $\endgroup$ Apr 22 '14 at 14:29
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    $\begingroup$ @JamesJenkins also giving a mass an initial speed to raise it to LEO altitude would require a huge velocity since you have to overcome the atmospheric drag (which increases with V^2). Basically I'd imagine most of you're kg mass being ablated before you get anywhere near LEO altitude! $\endgroup$
    – ThePlanMan
    Apr 22 '14 at 17:53

Here's a simplistic and optimistic approach. First find the mass ratio.

$$ \frac{ m_0 }{ m_p } = e^{ \frac{ 10,000 }{ 4462 } } = 9.4 $$

This varies a great deal in practice, in both of those two numbers above. The delta_v to orbit fluctuates widely between about 9 km/s to 11 km/s, and the propellant velocity by a much larger margin. I also haven't taken into account take masses and staging. Basically, this is optimistically low.

To convert this information into energy, there are two approaches I want to follow. For one, you could apply the reaction energy of $232 kJ/mol$ for the reaction of liquid Hydrogen and Oxygen. The other, you could just calculate the kinetic energy of the propellant. The latter first, since it's simple (and more wrong):

$$ E_k = \frac{ 1}{2} \frac{ m_0 }{ m_p } v^2 \\ = \frac{ 1}{2} \frac{ 9.4 kg }{ 1 kg} \left( 4,462 \frac{m}{s} \right)^2 = 94 \frac{ MJ }{ 1 kg } $$

I hope the units help describe what the context is. This is the energy required per mass of payload.

Now, let's use the more chemistry-based approach.

$$ E = 232 \frac{ kJ }{ mol } \frac{ mol }{ 20 g } 9.4 \frac{ kg }{ 1 kg} = 109 \frac{ MJ }{ 1 kg} $$

Well this shows surprising parity. I thought that rocket engines would be less efficient, but I guess not.

In addition to this, manufacture of the propellant isn't thermodynamically trivial. A lot of chemical production processes will require a significant multiple of the stored enthalpy. I can't easily find the number for Hydrogen production. So suffice it to say that the energy required will probably be a good deal higher than the above number.

For reference, the above number of 100 MJ is about 28 kW-hours, which is about $3 of electricity. But that depends on where you live.

  • $\begingroup$ To be honest that number is surprisingly low... Let me do calculation again and I will accept answer :) $\endgroup$ Apr 22 '14 at 12:00
  • $\begingroup$ The main cost of space operations is that it takes very carefully manufactured equipment that mostly will be used only once. IIRC I've seen costs of a few $/kg if you could simply plug your spacecraft in rather than use rockets. $\endgroup$ Apr 22 '14 at 17:33
  • $\begingroup$ @LorenPechtel It is said that most forms of transportation cost roughly 3x the price of the fuel. However, reusable rockets would increase the mass fraction, increasing the energy cost too. To me, it sounds about right that today's lowest cost rockets are about 1,000 times the fuel cost. If LEO trips were a fully matured industry, it sounds somewhat plausible that costs could be 3x the fuel. Say the fuel becomes \$30/kg, total costs \$100/kg, and you can make something that looks vaguely like Musk's \$500k/person to Mars colony. Academically, I can see that. $\endgroup$
    – AlanSE
    Apr 22 '14 at 17:44
  • $\begingroup$ How did you arrive at the mass ratio in your first equation? You're making an assumption there (about the amount of support mass you need for your payload), it'd be good to have that visible. $\endgroup$
    – Hobbes
    Nov 12 '15 at 19:22
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    $\begingroup$ @AlanSE this is now on wolfram alpha. I'm not sure if they were supposed to credit you. wolframalpha.com/input/…" $\endgroup$ Feb 11 '18 at 7:02

32 Megajoules is the energy contained by an object in LEO that has a mass of 1kg, and is traveling at 8000 m/s (with respect to the stationary point around which the earth rotates, and extending a static vector outwards from that point to the orbit)

105.8 Kilojoules is the kinetic energy possessed by an object sitting on the earth's surface, roughly at the equator, in the same situation.

Thus the theoretical minimum energy required to get an object to 8000 m/s from the earth's surface is 31.89 Megajoules.

Rounded up, that leaves us with the figure of about 32 Megajoules.

Everything else about propellants, mass ratios, specific impulse, and so on is salient, but has nothing to do with your original question! This is the lower bound for any 1kg mass. You cannot get it into orbit with less than this much energy.

  • $\begingroup$ +1 nice approach, but I think you miss the potential energy of the object in orbit. $\endgroup$ Mar 14 '19 at 15:17
  • $\begingroup$ Also, you should add that this answer assumes the object gets into orbit by some other means than by rockets. The question was about how to "put" it there, not how much energy you could "get" out of the object. That said, your result is a lower bound, but much lower than a more realistic bound as discussed by @AlanSE $\endgroup$ Mar 14 '19 at 15:23

This seems a complicated way of doing it.

The theoretical minimum is surely the kinetic energy of a kg travelling at about 7.8km/sec, the speed required for a minimal LEO. This is given by 1/2.m.v^2 and comes out at 30 MegaJoules. Rather less than the thermal energy in a kg of gasoline (about 45MJ) If you want to go a bit higher and travel at 10,000km/sec it comes out at 50MJ.

If you want to escape from the earth completely you'll have to move at 40.27 km/sec and the theoretical minimum energy to get to that speed is 811MJ.

  • $\begingroup$ Firstly, 30 MJ leading to 7.8 km/s cannot possibly extrapolate to 50 MJ leading to 10000 km/s. Second, escape velocity from LEO is 10-11 km/s, escape velocity from the mean radius of Earth is 11.18 km/s. 40.27 km/s would correspond to escaping from orbit at 492 km radius (inside the Earth's core!). $\endgroup$ Nov 12 '15 at 20:17
  • $\begingroup$ Im not clear as to why this comment got downvoted, since it is the one that actually answered the question, exactly as put by the OP. Mr Lynch, he was assuming a near point mass as the theoretical limit. Also he was talking about the TOTAL energy required to achieve escape from that point mass when he stated 40.27 km/s as escape. Its just a way to get a theoretical ballpark answer, to a theoretical ballpark question. No it's not as specific as above, but the upvoted answer assumes reaction mass propulsion, which was also not stated in the original question. $\endgroup$
    – T. B.
    Jun 7 '17 at 22:49
  • $\begingroup$ See my comment to @T.B.'s answer, you miss the potential energy of the orbiting object. Not that it matters much, but it's still there. $\endgroup$ Mar 14 '19 at 15:28

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