In the thread in this section called "Which hit first, the hammer or the feather" the question was asked (paraphrasing) "Which would hit first on the moon, a hammer or a feather, if dropped from equal distances." (I recommend reading that thread if you haven't already.)

I proposed a way in which to solve that problem. It went as follows:

Quote:

You have a mass m1 at rest a distance r away from (the center of) a much larger mass M also at rest. You let them move towards each other under only their mutual gravitational force and calculate the time it takes for them to collide (or not collide, but rather move a specified distance... and actual values are arbitrary as long as you are consistent throughout the calculation)...

Then you replace the mass m1 with a smaller (or larger) mass m2 and repeat the calculation (or thought experiment, or what-have-you) with all else remaining the same.

Will the collision times be equal, or will one be greater than the other?

...this has the beauty of allowing arbitrary values (especially masses). Either the collision time will be mass-dependent or not. If it turns out to be mass-dependent (dependent on m and M that is), then one of them will hit first.

If I were going to attempt the calculation, since we are only concerned with the "collision" time, I would simply solve the two-body problem for the collision time of two point masses m and M separated by a distance r. If m is in the result, then one of them will hit first (which one (i.e., the hammer or feather) depends on how the result depends on m). And since we are interested in the dependence on the smaller mass, m, make sure not to use the approximation m << M.

Actually, the approximation m << M is valid for Galileo's experiment, which is why he got the result he did....

I encourage you all to do this calculation. It turns out that m is in the answer....

So...

the challenge is to solve the two-body problem for the collision time between two point masses, m1 and m2, initially at rest, separated by a distancer, moving only under their mutual gravitational force.And I'm only asking that you get the answer toI'll give you the answer (since I already posted it on another thread and don't want to be unfair to anyone):within a multiplicative factor.

t ~ r^(3/2)/sqrt(G(m1+m2))

I can think of a few ways to solve this;

-one simply uses the 'brute-force' approach

-another dates back roughly 150 years

-and yet another dates back roughly 400 years

-plus, I'm sure there are other ways I haven't thought of....

Specifically, the challenge is to solve this assimplyas possible. The simplest (correct) solution gets the prize (whatever that may be...). But... you must show us enough steps to follow your work, explain what you are doing if it is not obvious, and state any theorems/laws you use.

I hope some of you find this fun and educational.

Cheers,

william