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This link describes the way back from the Moon surface to the orbiting spacecraft. But how could that spacecraft get back to the Earth?

It is a long journey. How much rocket fuel would be needed, in terms of tons?

Where is a technical description of this?

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    $\begingroup$ Welcome to space! Unfortunately Stack Exchange search is down for maintenance right now (I'd never seen that particular message before). A search of this site for "Apollo return trajectory" might return answers already on this site with helpful information on the "how did it get back" part. For the "How much fuel" part, that might be a new, unique question. Let's wait and see... $\endgroup$ – uhoh Aug 8 at 13:48
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    $\begingroup$ Note that energy used for orbital mechanism is usually accounted in delta V. For a given delta V, the quantity of propellant used depend on the mass of the ship (before and after the maneuver, the ship mass changing because of consumed propellant). $\endgroup$ – Manu H Aug 8 at 14:07
  • $\begingroup$ search is back! See 1, 2, 3, 4 5 $\endgroup$ – uhoh Aug 9 at 12:57
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    $\begingroup$ See the map in the delta-v budget wikipedia article for approximate delta-v costs of major destinations around you: en.wikipedia.org/wiki/… $\endgroup$ – njzk2 Aug 9 at 15:16
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It’s a long journey, but it’s all “downhill” — once the spacecraft leaves the moon’s gravitational sphere of influence, Earth’s gravity brings it home.

The process of leaving the moon is called “trans-Earth injection” or TEI; the rocket engine on the CSM fires for about two and a half minutes, adding about 1000 m/s to the spacecraft's speed in lunar orbit, burning about 10,000 lbs (4.5 metric tons) of propellant in the process.

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  • $\begingroup$ @hyde The Moon doesn't give a gravity assist to the spaceraft. The spacecraft just decelerates in the earths frame of reference, making the orbit elliptical and low enough that the perigee is inside the atmosphere. This deceleration wrt. earth is also an acceleration wrt. the moon, allowing the craft to escape the moons gravity. $\endgroup$ – Polygnome Aug 9 at 11:59
  • $\begingroup$ @Polygnome maybe, need to refresh my knowledge a bit. Earlier comment deleted. $\endgroup$ – hyde Aug 9 at 13:04
  • $\begingroup$ it's mostly downhill because of aero-braking. powered-landing on Earth would be quite expensive $\endgroup$ – njzk2 Aug 9 at 15:16
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This is an extended comment on jumpjack's answer, because it raises an interesting question and it's too long for a comment. The interesting question is:

how much energy does a spacecraft returning from the Moon need to lose on atmospheric entry, and how does this compare with the energy required to launch the spacecraft?

Well, we can answer this, and as is traditional I will take Apollo 11. Based on the Apollo 11 flight journal, the speed of the CM at the entry interface was $11045\,\mathrm{m/s}$. From NASA the Apollo 11 CM mass was $5557\,\mathrm{kg}$.

If we assume the CM was stationary after splashdown then the amount of energy lost is then $3.39\times 10^{11}\,\mathrm{J}$.

Well, the energy density of kerosene (from Wikipedia) is $43\times 10^6\,\mathrm{J/kg}$, so the energy lost by the CM corresponds to $7880\,\mathrm{kg}$ of RP-1.

So, then, the S1-C carried about $770\,\mathrm{m^3}$ of RP-1, and the density of RP-1 is about $850\,\mathrm{kg/m^3}$: in other words the S1-C carried about $654\times 10^3\,\mathrm{kg}$ of RP-1.

So the energy lost on the way back through the atmosphere is about $1.2\%$ of the energy available in the S1-C.

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  • $\begingroup$ I don't think that calculating energy of kerosene is the right method, not to talk about just taking into account first stage. We need to know the final kinetic energy impressed to the capsule at the moment of TLI, because this is the whole energy that it will carry when coming back. And probably also potential energy due to altitude must be taken into account (not sure). $\endgroup$ – jumpjack Aug 17 at 16:45
  • $\begingroup$ But being it a multi-stage configuration, and as not all the stages fly to the moon, this is a very complex problem. $\endgroup$ – jumpjack Aug 17 at 16:46
  • $\begingroup$ @jumpjack: kerosene is the right thing because that's what RP-1 is & I wanted to know how it compared with the S1-C. All I was trying to answer was how much energy the command module needed to dump & comparing it with the energy in the launch vehicle. In particular I wasn't trying to answer how large a vehicle you'd need to do a powered descent which is much more complicated (and I'm not disagreeing that you'd need almost the same fuel!) $\endgroup$ – tfb Aug 17 at 17:23
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The spacecraft must accelerate enough to have its centrifugal force overriding the Moon's gravity, thus escaping lunar orbit. With proper calculations, the orbit is left when the spacecraft is closer to the Earth, so it eventually get caught by Earth gravity and pulled back.

But this is just the "easy" part.

Then the spacecraft must lose all the energy that the rocket had put in it when launching from Earth! If an engine was used, it would require almost the same amount of fuel used to launch the spacecraft. So, rather than slowing down by engine, the spacecraft gives its energy back to Earth system, by using air friction: the "potential energy" (mass * 9.81 m/s2 * altitude) and the kinetic energy (0.5 * mass * velocity2) are converted into heat, which is dispersed in the atmosphere.

Some spacecrafts use only air friction to slow down, other ones (modern ones) also use retrorockets in the final part of the landing procedure.

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    $\begingroup$ "almost the same amount of fuel used to launch", no, not at all. The spacecraft is now much much lighter, because of all the spent fuel. The fuel required to land on rocket power is a tiny fraction of the fuel used to go up. $\endgroup$ – hyde Aug 9 at 10:24
  • $\begingroup$ This makes me now curious about actual numbers, but I have no idea where to find them. Full rocket mass with fuel? Spacecraft mass? There are so many different rockets and so many different spacecrafts... $\endgroup$ – jumpjack Aug 9 at 10:37
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    $\begingroup$ @jumpjack: see my answer, which is really an extended comment on yours. For Apollo 11, about 1.2% of the energy of the first stage was dumped into the atmosphere during reentry. $\endgroup$ – tfb Aug 9 at 13:56
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    $\begingroup$ @jumpjack you can get detailed mass numbers for different mission stages for Apollo at Apollo by the Numbers history.nasa.gov/SP-4029.pdf $\endgroup$ – Organic Marble Aug 9 at 13:57
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    $\begingroup$ The Trans-Earth Injection (TEI) burn happened when the vehicle was opposite the Earth rather than closest to the Earth. $\endgroup$ – David Hammen Aug 10 at 9:36

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