# From Earth to Moon

Ever wondered why it takes a tremendously huge rocket to launch people from the earth, but the Apollo astronauts managed to launch from the moon in a comparatively tiny lunar module? Easy, the whole thing was faked and NASA forgot to come up with a plausible explanation!

Wow, it was almost physically painful to type that even as a joke. The real reason is some very easy but still pretty cool physics. Shall we take a look?

For a given object, the gravitational potential energy per kilogram with respect to distant space is given by this easy little equation:

So to pick up one kilogram of mass from the earth and remove it from the earth’s gravitational influence requires energy given by that formula. Plugging in values for the mass of the earth and the radius of the earth gives U = 6.25 x 107 J/kg.

But the moon? Use the same equation and plug in the mass and radius of the moon. I get U = 2.8 x 106 J/kg.

You might remember that the surface gravity of the moon is about 1/6 that of the earth. But the potential energy of an object at the moon’s surface is (from the above equations) only about 1/23 that of the same object at the earth’s surface. It’s much, much easier to launch from the surface of the moon than it is from the surface of the earth. And since it’s so much easier, you don’t have as many kilograms of fuel and rocket to move to orbit in the first place. This is why it takes one of these to launch people from the earth into deep space:

And only one of these to launch from the moon:

1. #1 Tom
September 10, 2008

In addition, one must account for how much more mass one launched from the earth. It takes more energy to launch a kilogram, so you need more fuel, and so it all compounds. Plus, the launch from earth included the moon lander, and a lot of stuff was left on the moon (though they did bring some rocks back)

2. #2 Chris
September 10, 2008

Thanks, I’d been wondering about this lately but hadn’t gotten around to looking up any relevant equations. I’d assumed it was linked to escape velocities – am I wrong or is escape velocity somehow linked to your equation above?
Great blog by the way – I’m only a lowly biochemist, and it’s great to see some easy physics made relevant to the real world!

3. #3 Uncle Al
September 10, 2008

http://en.wikipedia.org/wiki/MythBusters_(season_7)#Episode_104_-_NASA_Moon_Landing

If only we could turn down inertial mass. Gravitation as spatial geomtry is a tougher assignment. Stochastic electrodynamics has (questionable) intervention with inertia,

http://www.calphysics.org/research.html

No other proposed large magnitude derailments are credible.

4. #4 nanoAl
September 12, 2008

Chris:
Its sort of related to the escape velocity, to find it , you just set U equal to the kinetic energy mv*v/2. Thats the amount of kinetic energy(and thus velocity) it needs to get from the surface to infinity and stop there(wherever that is). I think anyway.
I find it so amazing that some of the moon cranks think this is a convincing argument. “the moon is way smaller/lighter than the earth” would seem obvious but it isn’t apparently…