How hard is space travel, in principle?

Our Earth-based intuitions about the difficulty of travelling long distances break down badly in space. It’s tempting to think that just because Saturn, say, is nearly 4,000 times further away than the moon, it must be 4,000 times more difficult to reach. After all, on Earth it takes about 10 times as much work to go 10 kilometers as it does to go 1 kilometer. But in space, where there is no friction, this intuition is entirely wrong. In fact, in this essay I’ll show that with some important caveats the situation is far more favourable, and it doesn’t take all that much more energy to get to the outer planets than it does to get to low-Earth orbit.

The numbers are striking. We’ll see that it takes about 10 times as much energy to get to the moon as it does to get to low-Earth orbit. About 4 times more energy than that will get you to Mars, despite the fact that Mars is more than 200 times further away than the moon. Tripling that energy gets you all the way out to Saturn! So in some sense, the gap between getting to the moon and getting to Saturn isn’t really all that much different from the difference between getting into low-Earth orbit and getting to the moon.

The calculations required to obtain these results are all simple calculations in Newtonian gravitation, and they’re done in the next section. If you want to skip the details of the calculations, you should move to the final section, where I discuss some important caveats. In particular, we’ll see that just because getting to Saturn requires only 12 times as much energy as getting to the moon doesn’t mean that we only need to build rockets twelve times as big. The situation is more complex than that, and depends on the details of the propulsion technologies used.


To see why the numbers I quoted above are true, we need to figure out how much energy is required to get into low-Earth orbit. I will assume that the only obstacle to getting there is overcoming the Earth’s gravitation. In actual fact, of course, we need to overcome many other forces as well, most notably atmospheric friction, which adds considerably to the energy cost. I’ll ignore those extra forces – because the cost of getting to low-Earth orbit is the yardstick I’m using to compare with other possibilities this will mean that my analysis is actually quite pessimistic.

Suppose we have a mass [tex]m[/tex] that we want to send into space. Let’s use [tex]m_E[/tex] to denote the mass of the Earth. We suppose the radius of the Earth is [tex]r_E[/tex] and that we want to send our mass to a radius [tex]d r_E[/tex] for some [tex]d > 1[/tex]. We’ll call [tex]d[/tex] the distance parameter – it’s the distance from the center of the Earth that we want to send our mass to, measured in units of the Earth’s radius. For instance, if we want to send the mass to a height of [tex]600[/tex] kilometers above the Earth’s surface, then [tex]d = 1.1[/tex], since the Earth’s radius is about [tex]6,000[/tex] kilometers. The energy cost to do this is [tex]Gm_Em/r_E-Gm_Em/ d r_E[/tex], where [tex]G[/tex] is Newton’s gravitational constant. We can rewrite this energy cost as:

[tex]e_d = \frac{Gm_em}{r_E} (1-1/d)[/tex]

We can use this formula to analyse both the energy cost to send an object to low-Earth orbit and also to the moon. (We can do this because in going to the moon the main barrier to overcome is also the Earth’s gravitation). Suppose [tex]d_l[/tex] is the distance parameter for low-Earth orbit, let’s say [tex]d_l = 1.1[/tex], as above, and [tex]d_m[/tex] is the distance parameter for the moon, [tex]d_m = 60[/tex]. Then the ratio of the energy cost to go to the moon as opposed to the energy cost for low-Earth orbit is:

[tex]\frac{e_{d_m}}{e_{d_l}} = \frac{1-1/d_m}{1-1/d_l} = 10[/tex].

That is, it takes [tex]10[/tex] times as much energy to get to the moon as to get to low-Earth orbit.

What about if we want to go further away, outside the influence of the Earth’s gravitational field, but still within the sun’s gravitational field? For instance, what if we want to go to Mars or Saturn? For those cases we’ll do a similar calculation, but in terms of parameters relevant to the gravitational field of the sun, rather than the Earth – to avoid confusion, I’ll switch to using upper case letters to denote those parameters, where before I used lower-case letters. In particular, let’s use [tex]M_s[/tex] to denote the mass of the sun, and [tex]R_e[/tex] to denote the radius at which the Earth orbits the sun. Let’s suppose we want to send a mass [tex]m[/tex] to a distance [tex]D R_e[/tex] from the sun, i.e., now we’re measuring distance in units of the radius of the Earth’s orbit around the sun, not the radius of the Earth itself. The energy cost to do this is [1]:

[tex]E_D = \frac{GM_sm}{R_e}(1-1/D)[/tex]

The ratio of the energy required to go to a distance [tex]DR_E[/tex] from the sun versus the energy required to reach low-Earth orbit is thus:

[tex]\frac{E_{D}}{e_{d_l}} = \frac{M_sr_e}{m_e R_e}\frac{1-1/D}{1-1/d_l}[/tex]

Now the sun is about [tex]330,000[/tex] times as massive as the Earth. And the radius of the Earth’s orbit around the sun is about [tex]25,000[/tex] times the radius of the Earth. Substituting those values we see that:

[tex]\frac{E_{D}}{e_{d_l}} = 13 \frac{1-1/D}{1-1/d_l}[/tex]

For Mars, [tex]D[/tex] is roughly [tex]1.5[/tex], and this tells us that the energy cost to get to Mars is roughly [tex]40[/tex] times the energy cost to get to low-Earth orbit (but see footnote [1], below, if you haven’t already). For Saturn, [tex]D[/tex] is roughly [tex]10[/tex], and so the energy cost to get to Saturn is roughly [tex]120[/tex] times the energy cost to get to low-Earth orbit. For the stars, [tex]D[/tex] is infinity, or close enough to make no difference, and so the energy cost to get to the stars is only about 10 percent more than the cost to get to Saturn! Of course, with current propulsion technologies it might take rather a long time to get to the stars.


There are important caveats to the results above. Just because sending a payload to Mars only requires giving it about four times more energy than sending it to the moon, it doesn’t follow that if you want to send it on a rocket to Mars then you’ll only need about four times as much rocket fuel. In fact, nearly all of the energy expended by modern rockets goes into lifting the rocket fuel itself, and only a small amount into the payload. An unfortunate consequence of this is that you need a lot more than four times as much fuel – either that, or a much more efficient propulsion system than the rockets we have today. In a way, we’ve been both very lucky and very unlucky with our rockets. We’ve been lucky because our propulsion systems are just good enough to be able to carry tiny payloads into space, using enormous quantities of fuel. And we’ve been unlucky because to give those payloads even a tiny bit more of the energy they need to go further requires a lot more fuel. See, e.g., ref,ref and the pointers therein for more discussion of these points.

The extent to which any of this is a problem depends on your launch technology. Ideas like space guns and the space elevator don’t require any fuel to be carried along with the payload, and so escape the above problem. Of course, they’re also still pretty speculative ideas at this point! Still, I think it’s an interesting observation that the energy required to get a large mass to a distant part of the solar system is not, in principle, all that far beyond what we’ve already achieved in getting to the moon.


[1] Note that the object needs to first escape out of the Earth’s gravitational field, and this imposes an extra energy cost. This extra cost is roughly 10 times the cost of getting to low Earth orbit, by a calculation similar to that we did for getting to the moon. Strictly speaking, this energy cost should be added on to the numbers we’ll derive later for Mars and Saturn. But for the rough-and-ready calculation I’m doing, I won’t worry about it – we’re trying to get a sense for orders of magnitude here, not really detailed numbers!

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