# 2-body central force ( gravity ) problem

by jobendorfer
Tags: 2body, central, force, gravity
 P: 3 1. The problem statement, all variables and given/known data A comet is released from rest at a distance r_max from the sun. Therefore it has no angular momentum. We can assume the sun is a point mass ( i.e. sun's radius is zero ). How long does it take for the comet to hit the sun? Let m = comet mass let M_s = sun's mass let G = gravitational constant r = distance of comet from the sun, with origin at the sun. 2. Relevant equations t = sqrt( m/2 ) * ∫ dr / sqrt( E - U(r) ) 3. The attempt at a solution Since the comet is released from rest, it initially has kinetic energy T = 0 and potential U = G*M_s*m/r_max. E = T + U = G*M_s*m/r_max. U(r) = G*M_s*m/r E-U(r) = (G*M_s*m)( 1/r_max - 1/r) After grinding a bit I got: t = 1/sqrt( 2*G*M_s) ∫ dr/sqrt( 1/r_max - 1/r ) with the limits of integration being r_max to 0. This is where my eyes glazed over, since it's been 25 years since I've been near a calculus course. Any hints about how to proceed? This is Taylor, _Classical Mechanics_, problem 8.21, part (c). I am NOT in a class. I'm a software grunt trying to educate myself, any help offered would be greatly appreciated. This problem has really had me tearing my hair out for about 3 days. Thanks, John (jobendorfer@cyberoptics.com)
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P: 10,763
Welcome to PF!

 Quote by jobendorfer Since the comet is released from rest, it initially has kinetic energy T = 0 and potential U = G*M_s*m/r_max.
The potential energy of the comet is negative. U=-G*M_s*m/r_max.

 Quote by jobendorfer E = T + U = -G*M_s*m/r_max. U(r) =-G*M_s*m/r E-U(r) = -(G*M_s*m)( 1/r_max - 1/r)
You see, the kinetic energy is all right now, greater than zero.

 Quote by jobendorfer After grinding a bit I got: t = 1/sqrt( 2*G*M_s) ∫ dr/sqrt(-1/r_max + 1/r ) with the limits of integration being r_max to 0.
Actually, v=-dr/dt as r decreases with t as the comet approaches the sun.

There is an easy way if you have problems with an integral. Try to put it into wolframalpha.com. But before that, remove irrelevant constants and denote the necessary ones (r_max) by a simple letter. It is also advisable to change the variable so you get a simpler function. This case you can use u=r/r_max.

When you want to do the integration yourself, it is a long way... start with substituting sqrt(r_max/r-1)=u.

ehild
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P: 10,763