This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Astronomy 210 Fall 2010 Homework Set #3: Solutions 1. We are given the central force law F = − m 1 m 2 f ( r ) ˆ r . (1) with f ( r ) an unknown function of the distance r . We are also to assume that a particle of mass m 1 has a circular orbit centered on the origin of this force with radius r . The acceleration of this particle is the centripetal acceleration a = a c = − v 2 r ˆ r . (2) Since the orbit is circular, v = 2 πr/P , where P is the orbital period. Thus we have a = a c = − 4 π 2 r P 2 ˆ r . (3) Newton’s second law says F = m 1 a . Usign (1) and (3), we have m 1 m 2 f ( r ) = 4 π 2 m 1 r P 2 (4) which gives f ( r ) = 4 π 2 m 2 r P 2 (5) Note that this gives f ( r ) in terms of not only r but also P , so we still need to eliminate P to get f in terms of r only. But Alternative Kepler has found the relationship between circular radius r = a and period P : P ∝ a 2 (6) So, setting r = a in eq. (5) and using eq. (6) to eliminate P for a , we have f ( a ) = 4 π 2 m 2 a P 2 ∝ a a 4 = 1 a 3 (7) Thus we find that f ( r ) ∝ 1 /r 3 : the force law in this alternative universe is an inverse cube law, rather than the inverse square law as we find in our real universe. Thus we see that Kepler’s Third Law encodes the dependence of the gravitational force law on distance, i.e., from Newton’s laws of circular motion and Kepler’s Third Law, we can find f ( r ) from the exponents in the relationship between P and a . 2. Light and Strong Gravity (a) To escape from a mass M at a distance R , the needed escape speed is v esc = radicalBigg 2 GM R (8) Note that, depending on the values of M and R , v...
View
Full
Document
This note was uploaded on 10/06/2011 for the course ASTRO 210 taught by Professor Fields during the Spring '08 term at University of Illinois, Urbana Champaign.
 Spring '08
 Fields
 Astronomy

Click to edit the document details