Why was the rocket to go to the moon so big? How do they get home?
May 18, 2004 11:06 AM   Subscribe

Something I've never understood. It takes a giant rocket to achieve escape velocity and leave the Earth. The Apollo mission made it all the way to the moon. But in all the pictures I've ever seen, the only thing they've got on the moon is a lander, a small capsule on legs. Where's the rocket they used to get off the moon? How did they get off the moon? I know the moon is much smaller and has almost no atmosphere, but is leaving it really as easy as firing a few booster rockets? Why is the rocket to leave the Earth so much bigger?
posted by scarabic to Travel & Transportation (26 answers total) 1 user marked this as a favorite
wasn't there something orbiting the moon that they went back to? it had the re-entry cone thingummy on the end and a rocket to get them back. so the lander only had to get back to that orbiting rocket, which could be in a very low orbit, presumably, because there's no atmosphere to slow it down.
(it's amazing it all worked!)
posted by andrew cooke at 11:11 AM on May 18, 2004

Response by poster: Yeah, the mission page lists a command module and a landing module. I'm still just shocked at how little energy, comparatively, it takes to achieve orbit on the moon. Is there some exponential relationship I'm missing here?
posted by scarabic at 11:14 AM on May 18, 2004

1) Look at this big diagram. They leave the bottom part (descent stage) on the moon when they leave. The top part (ascent stage) has a rocket on the bottom of it, that you can see when the two parts are connected..

2) You only need to get from the moon back into orbit. Then you meet up with the Service Module which was orbiting the whole time. That has its own rocket that sends you back to earth.
posted by smackfu at 11:15 AM on May 18, 2004 [1 favorite]

posted by rushmc at 11:30 AM on May 18, 2004

they're moving everything from earth. at take-off they have to shift all the fuel they'll use in both directions. by the timethey're coming back, they are pushing much less. that's either exponential or quadratic, and i'm too lazy to work out which.
posted by andrew cooke at 11:33 AM on May 18, 2004

The Moon's gravity is 1/6th that of Earth's. According to Wikipedia, the ascent module is only 4,547kg. On the moon, that's like lifting just 757kg (plus your fuel, people, and all those moon rocks, again 1/6th their Earth weight). Here's a nice diagram showing where the ascent engine and fuel tanks are.
posted by zsazsa at 11:45 AM on May 18, 2004

And here's the wikipedia entry for a theoretical space elevator, which would dramatically cut down the costs of escaping Earth's gravity well.
posted by mookieproof at 11:46 AM on May 18, 2004

Just to hammer the point being danced around: the exponational/quardratic factor is that higher the escape velocity, the more fuel you need, and the more fuel you need, the more weight you have, the more fuel you need, etc.
posted by Tlogmer at 11:51 AM on May 18, 2004

To summarize what people are linking to:

The escape velocity to get off the moon is about 20% of that needed to get off the Earth.

Thats not enough in itself though. The launch from Earth had to carry everything. The Saturn V Launch vehicle alone weighed 2.8 Million Kg! By the time you're on the moon all those ascent rockets and fuel have been discarded. By the time it needed to take off, the lunar module itself only weighed about 5,000 Kg.

The lander, by the way, weighed more on the way down (about 15,000 Kg) since it also had to do a controlled descent.
posted by vacapinta at 12:06 PM on May 18, 2004

Response by poster: Okay, so adding up all of the following it's starting to make sense:

1) To leave the moon they simply had to achieve orbit then dock with the command module
2) The moon is 1/4 the mass of the Earth so escape velocity is much less
3) The gravity on the moon is 1/6 the Earth
4) The rocket that took them away from Earth had to carry a lot more, including the command module

Okay. It's starting to seem reasonable that the lunar ascent module doesn't resemble the giant rocket. Thanks for all the responses!
posted by scarabic at 12:30 PM on May 18, 2004

2) and 3) are really the same thing.
posted by falconred at 1:01 PM on May 18, 2004

Oh, and the moon has no atmosphere, and thus no wind resistance.
posted by falconred at 1:09 PM on May 18, 2004

Response by poster: 2) and 3) are really the same thing.

Not to a bone-stupid physics idiot who wears his ass for a hat!

Just kidding, falconred. Right you are!
posted by scarabic at 1:10 PM on May 18, 2004

Another answer to this might be to suggest that no-one's ever landed on the moon.
posted by skylar at 1:39 PM on May 18, 2004

Astronaut Michael Collins also piloted a command module (the Columbia) in orbit around the moon while the Eagle landed with Armstrong and Aldrin. So the Eagle, in addition to taking off from a body with a lower escape velocity, also only had to get into orbit rather than all the way back to the earth.
posted by alphanerd at 1:59 PM on May 18, 2004

Details on the development of the LM ascent engine in case anyone is interested.
posted by vacapinta at 2:16 PM on May 18, 2004

Yikes, no and no. The Moon has about 1/8 Earth's mass, not 1/4. (It's not only smaller, but only about half as dense.) The gravity at the surface of the Moon is about 1/6 that at the surface of the Earth, more than 1/8, because you are closer to the center of Moon when you are on its surface.

"Escape velocity" is the speed to which you need to accelerate a projectile at some distance from a gravitational mass if you want to stop accelerating it at that point and have the projectile coast out into a natural ballistic orbit. If escape velocity at the surface of the Earth is 7 miles/sec (actually it's about 6.95 miles/sec) then that's the speed you'd need to fire a spacecraft with no engines into orbit from a buried cannon Jules-Verne-style.

Of course our space rockets accelerate up from the Earth's surface, burn their fuel as fast as possible, and cut off their engines after they reach a slightly lower escape velocity at some altitude above the surface and further from the planet's center of mass. From this cool NASA page of Apollo 11 crew narratives:

We started the [third-stage] burn at 100 miles altitude, and had reached only 180 at cutoff [...] At the instant of shutdown, Buzz recorded our velocity as 35,579 feet per second, more than enough to escape from the Earth's gravitational field.

That's 6.75 miles/sec and it was "more than enough". But the point is that you can get up out of a planet's gravity and into space at any speed you like, however slow, if you have unlimited fuel on board - you don't actually have to hit any particular escape velocity to get out.
posted by nicwolff at 2:37 PM on May 18, 2004

Which is not to say that all the points other people made here aren't correct! I'm just clearing up the numbers a little.
posted by nicwolff at 2:44 PM on May 18, 2004

I was under the impression that the escape velocity is irrelevant for powered objects. The only thing you need to worry is that you continue to move forward, rather than being dragged back by gravity. Escape velocity is only relevant for objects which are leaving the surface of the Earth (or wherever) and will not be powered after that point. Eg, a ball would need to be thrown at escape velocity to leave a planet, but a rocket would only need to accelerate faster than gravity pulls it the other way, so with regard to Earth a rocket accelerating at anything beyond 10m/s^2 would eventually get away from Earth.

On preview: curse you nicwolff
posted by biffa at 3:49 PM on May 18, 2004

<nelson>HA ha!</nelson>

The real-world consideration that brings escape velocity back into the calculation is that the longer you're carrying all that fuel with you up against gravity, the more work you have to do, so the more fuel you need, &c. So you really want to burn your fuel as quickly as you efficiently can just to get it off the ship, and you end up basically ballistic and do need enough speed to throw you into orbit.

Note from above that the Apollo launch cutoff the engines at 180 miles up which is just 5% of Earth's radius - really just above the surface, but still not far from low-Earth orbit. (The International Space Station is around 230 miles up, just outside the atmosphere; geosynchronous orbit, where the TV satellites are, is a bit over 22,000 miles up.) At each cutoff altitude you might select, there's a different escape velocity and a different natural orbital speed and altitude, so it's a complicated compromise to calculate.
posted by nicwolff at 4:23 PM on May 18, 2004

Another data point to add to the discussion: the Earth's atmosphere is thick enough and the main booster goes fast enough, that most of the atmosperic drag on the main booster is drag from "shocks", i.e. sonic booms. That's why rockets have the shape they do (blunt noses usually, although some older ones had spiky noses): to mitigate the shock wave (by spreading it around; the math gets complex fast), and that's why the Apollo crafts had to fit under a canopy/nose, that was ejected after passing through the atmosphere.

OTOH, the Moon's atmosphere is nearly non-existent, and there is no drag (regular by friction or from sonic booms) to overcome and thus no need for any sort of aerodynamical shape.

As an added bonus, the energy wasted in creating the sonic booms is roughly proportional to the mass of the launcher, so there is way more waste to overcome leaving the Earth than leaving the Moon.

That, plus the difference in gravity (and thus escape velocity) are the two main factors in the difference. I wouldn't be suprised if aero drag is more than half of the energy loss.

(yes, I was a rocket scientist :-)
posted by costas at 8:05 PM on May 18, 2004

And don't forget that to get back to the earth, you pretty much just have to get close enough to start falling towards the ocean, which requires less power than getting into orbit to begin with.
posted by bingo at 9:28 PM on May 18, 2004

Dammit, everyone beat me to the "no atmosphere" comment. Grr.
posted by Civil_Disobedient at 10:28 PM on May 18, 2004

and they left that little car down there, didn't they?
posted by headless at 10:50 PM on May 18, 2004

As an added bonus, the energy wasted in creating the sonic booms is roughly proportional to the mass of the launcher, so there is way more waste to overcome leaving the Earth than leaving the Moon.

why? i can see how cross-section area would be important. are you saying it scales as length too, or does mass come in in some other way?
posted by andrew cooke at 8:00 AM on May 19, 2004

The energy in a sonic boom (or, really, the total energy in the multiple booms a supersonic projectile creates) is determined by the volume of air being pushed out of the projectile's way, and therefore by the volume (not the mass!) of the projectile.
posted by nicwolff at 6:17 PM on May 19, 2004

« Older Help in New York, Part 2   |   What is first motion picture to climax with an... Newer »
This thread is closed to new comments.