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I must be missing something - escape velocity is about 25,000mph. The distance to the moon about 240,000 miles. That computes to about a 9 hour flight. Why did it take 3 days for Apollo missions to reach the moon?

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marked as duplicate by Russell Borogove, peterh, David Hammen, DrSheldon, Sean Feb 18 at 3:56

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Escape velocity is the velocity at a given altitude (usually the surface) that is enough to leave the body's sphere of influence with a positive net velocity. But if you leave a body at exactly escape velocity, your velocity bleeds off as you climb in exchange for gaining gravitational potential energy, and your velocity tends to the limit of zero at sufficiently large distances. (That is, your $V_{inf}$ is 0.)

Apollo did not leave at escape velocity. Rather, the trans-lunar injection gave considerable extra speed in order to make the trip much faster, so the lowest velocity was reached a few tens of thousands of miles short of the moon (where the gravity from the moon and the earth are equal), in the low thousands of miles per hour.

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Developing on Nathan's answer, let's do some math. For simplicity we suppose here that we are not really going to the Moon, that only Earth's gravity us relevant.

We leave at escape velocity, 25,000 miles per hour from an altitude of 4000 miles above the Earth's center, and climb straight up. As we do so our velocity decreases against Earth's gravity but remains matched to escape velocity at that altitude above Earth's center. Thus at an altitude of 6250 miles (2250 miles above Earth's surface) Earth's gravity has slowed the rocket down to 20,000 miles per hour which is the escape velocity at that altitude.

Escape velocity is proportional to the $-1/2$ power of altitude above the center, and for 25,000 mph at 4000 miles the proportionality constant is (to three significant figures) $1.58×10^6$ (miles)$^{3/2}$/hr. So to get from an altitude of 4000 miles to 250,000 miles, at escape velocity, we need this much time:

$\int_{4000}^{240000} \dfrac{dz}{1.58×10^6z^{-1/2}}=49.5\text{ hr}$

This is roughly correct but misses the fact that at the Moon's altitude we would still be going several thousand miles per hour upwards and the Moon's gravity would not have been strong enough to catch us from such speed. Nathan correctly points out that we went off slower than escape velocity so that the Moon, which is still bound to Earth, could reel us in. Hence the extra day. Similarly, when taking off from the Moon we had to go slowly enough for Earth to pull us in rather than sending us off like a slingshot; technically blasting from the Moon was not up to its escape velocity either.

So ultimately the reason Apollo missions had such long transit times between the Earth and Moon was not any limit on rocket power but by the limits of gravity within which a lunar mission has to work.

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    $\begingroup$ But the Apollo 13 astronauts were lucky that it was possible to shorten the time for return by using the the rocket power of the Lunar Module's descent stage. Using the remaining fuel even more time may have been saved, but the landing would have been at the wrong location. $\endgroup$ – Uwe Feb 17 at 19:36
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    $\begingroup$ The return time can be faster than the outbound time -- you just need to be on a trajectory that intersects the Earth's atmosphere, going slow enough for your heatshield to handle. Apollo used a minimum-energy trajectory to keep the launch mass down. $\endgroup$ – Mark Feb 17 at 20:03
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As already explained in other answers, essentially all of the distance is covered by unpowered flight - gaining altitude away from Earth while under the influence of Earth's gravity, the "upward" velocity will gradually diminish. At LEO distance this is about 10m/s-2, although this gravitational acceleration does fall off with square of distance.

This is a good thing because the craft wants to arrive at the Moon at a speed which requires the least adjustment (least propellant) to enter lunar orbit - the Moon orbits the Earth at about 1km/s (round numbers) and a low altitude orbit around the Moon requires a speed of about 1.6km/s. To arrive in a retrograde orbit basically means the craft must end up at the far side of the Moon with zero radial velocity and about 600m/s retrograde to Earth. To accomplish this, it's a good thing that radial velocity away from Earth is being lost due to gravity almost the entire way out, but this means the trip will take much longer than the 9 hours suggested by simple arithmetic.

Another factor is that the path from Earth to Moon is not traveled in a straight line. Departure from LEO is essentially tangential to Earth's surface; Earth's gravity pulls the trajectory into a curve. The lowest energy transfer occurs if the craft aims to enter lunar orbit retrograde (as I previously mentioned), so the overall path is actually somewhat S-shaped, making total path a bit longer.

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