# And the Cow Jumped Over the Moon

Poking around on the internet looking for interesting problems to assign the students I’m about to have this semester, I came across this one:

How much energy must a 450 kg cow expend in jumping over the moon?

Fig. 1: Buzz Cowdrin?

Now hints and solutions are available on the linked site, so what I’d like to do is just talk conceptually for a moment. Leaving aside for the moment the derivation of the equation, it will turn out that the energy required for a cow (or anything else) to reach a particular elevation is given by a particular potential energy equation:

G is the universal gravitational constant, M the mass of the earth, m the mass of the cow (or whatever) and the two r’s are the initial and final heights of the jumping critter.

Compared to the initial height, the final height of the moon is vastly, vastly bigger. But notice that the equation scales as 1 divided by that final height. As that final height gets higher and higher, that 1/r factor will dwindle down to zero, eventually subtracting nothing off of the final number at all. The energy required to jump to an arbitrarily high height is not infinite, and that’s where the concept of escape velocity comes from. Leave the earth with higher kinetic energy than GMm divided by the radius of the earth and you’re not coming back. So the cow jumping over the moon can probably jump over Saturn pretty easily – the extra energy is going to be very small since r for the moon is already so big in the first place.

In the non-nursery-rhyme world things will be more difficult because after leaving the neighborhood of earth there’s still the Sun’s gravity to deal with, and so the Saturn-jumping cow is going to have to work harder for this reason.

Easy Bonus Question: If it takes “1.0 cowpower” units of energy to jump over the moon, how many cowpower units does it take to jump over Saturn, ignoring (for the purposes of this conceptual problem) the gravity of the Sun, Saturn, and the other heavenly bodies?

1. #1 Braxton Thomason
January 16, 2009

Hooray, I’ve always wanted to work a problem where I get to assume a spherical cow.

number of cowpowers = (1/r0 – 1/r2) / (1/r0 – 1/r1)

where r2 = distance of saturn from Earth. Have to pick a good time of the year to go though, otherwise he’d have to jump through the sun.

2. #2 andyb
January 17, 2009

Or Mooi Gagarin maybe…

If the cow has just enough energy to reach the moon, they might get trapped in the moon’s field, and be too puffed out to jump back. (Given the sparsity of lunar grazing)

They would have to carefully jump just over the moon, without overdoing it and ending up on Saturn. A tricky feat without some small rockets to tweak position. Perhaps the cow could belch methane in a controlled manner.

3. #3 paul
February 26, 2009

i just wanted to find out what is the total height of the moon? Does anyone know this? Thank-you…

Paul