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# When calculating the gravitational force between two objects, this force changes as a function of the radius as the other variables remain constant. However, when attempting to calculate the time it takes for an object to reach another when the net force is gravity, a problem arises in trying to calculate the net force, as the radius is always changing; this in turn causes the acceleration to change. Is there any way to calculate the time an object takes to reach another with given masses and a given initial velocity and radius?

Question #60550. Asked by StinkyCat. (Nov 14 05 11:19 PM)

hohohaha

Do you mean "radius" or "separation distance"?

 Nov 15 05, 5:38 PM
StinkyCat

"Radius" as in the distance from the center of one object to the center of the other object. This would be the radius if one object was traveling in a circle around the other. However, if you call it separation distance, then yes.

 Nov 15 05, 6:48 PM
gmackematix

Hmmm... I warn you that unless I'm missing something here (quite possible) some heavy mathing might take this outside the realm of "trivia"...

Well, the distance between their centres of gravity r is some function of time f(t).
If the masses of the objects are m and M, then the force between them is F=GMm/r^2 where G is the universal gravitational constant.
By Newton's Second Law, acceleration of m is MG/r^2 = MG/(f(t)^2).
But the acceleration is also the double derivative of r as a function of t.
Therefore f"(t) = MG/(f(t)^2)
If you like the look of that differential equation feel free to solve it.
I'm ducking out and heading for bed...

 Nov 15 05, 9:58 PM
gmackematix

Before I go, f'(t) should read f'(t) there.

Is it me or does the Stinkycat seem a little more intellectual than the Smellykat?

 Nov 15 05, 10:01 PM
gmackematix

OK, it doesn't like double apostrophes so lets try a space between them:
f' '(t). Or does that look silly?

 Nov 15 05, 10:03 PM
McGruff

f"(t)

Looks like you could use the double quotation mark, so I changed it to that. If you'd rather have the space, I can change it again.

 Nov 16 05, 2:44 AM
gmackematix

No, McG, that's fien and why on earth didn't using " dawn on me?
And I'm still not solving that equation.

 Nov 16 05, 5:26 PM
peasypod

Well gmack, I think if we knew what that darn function of time (f) was it'd be a piece of cake. ;)

 Nov 16 05, 6:23 PM

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