# Thread: Physics challenge - hammer/feather revisited

1. 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:

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 distance r, moving only under their mutual gravitational force. And I'm only asking that you get the answer to within a multiplicative factor. I'll give you the answer (since I already posted it on another thread and don't want to be unfair to anyone):

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 as simply as 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  2.

3. You say that t~" ", but that means that you used some type of simplification along the way. I cant solve the differential equation so far, so are you willing to give any hints? The only simplifacation i can think o is M>>>m, but the result of that is that the larger body wont move, and that would make both of the objects fall at the same rate.  4. it will really hit the time the same time? how would u know? if u can slow time down to the single nanosecond we will prob see which one will hit first  5. Vroom wrote:
You say that t~" ", but that means that you used some type of simplification along the way. I cant solve the differential equation so far, so are you willing to give any hints? The only simplifacation i can think o is M>>>m, but the result of that is that the larger body wont move, and that would make both of the objects fall at the same rate.
Firstly, thanks for the attempt. I was starting to wonder if this thread was going to be lost forever....

Okay, the "t~' ' " usually means a simplification was made - an approximation if you will. But here I only mean for you to get the answer to within a multiplicative constant.

For example, using a simpler method, I get I

t = sqrt(2) * r^(3/2)/sqrt(G(m1+m2)) .

The exact solution (unless I goofed) is

t = (sqrt(2)*pi/4) * r^(3/2)/sqrt(G(m1+m2)) .

I'm only asking for the dependence on the masses (i.e., don't worry about the multiplicative constant).
(Why is the constant not important?)
Because all we need in order to answer the hammer/feather question is the dependence of the masses. We don't need to care about the exact time, only which will hit first.

For some more hints:
"-one simply uses the 'brute-force' approach"
(This is not the easiest way....)
You first use a change of variables and use the technique called 'separation of variables' (remember, there are two distinct techniques in diff-eqs that go by the same name... so choose wisely). Then apply the boundary conditions. That should get you to how the velocity depends on the masses. From here you just do an integration (a difficult one I might add) with the appropriate limits. I am purposely vague about this approach because it is not the approach I am hoping you use....

"-another dates back roughly 150 years"
(Think Clausius.)

"-and yet another dates back roughly 400 years"
(Think of a really famous astronomer of the time.)

And I thought of another way since the original post...

I hope these hints help.

Cheers,
william  6. Well, im gonna guess that one of them is a conservation of energy method... So i will try that and see what i get According to Wikipedia, Clausius did a lot of CoE and thermo work.

EDIT: i get -2Gm<sub>1</sub>m<sub>2</sub>/r +.5m<sub>1</sub>(v<sub>1</sub>)<sup>2</sup>+.5m<sub>2</sub>(v<sub>2</sub>)<sup>2</sup>=C... for CoE... dang the non-linearity of that v just kills you   Bookmarks
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