hw8-10--solutions

# hw8-10--solutions - P2207 - Fall 2010 Solutions Assignment...

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P2207 - Fall 2010 Solutions Assignment 8 1. Chapter 13, Problem 39 (a) The spherical asteroid has a radius r a = 500 km and an acceleration due to gravity of g a = 3 . 0 m/s 2 . To ±nd the escape velocity, we use 1 2 mv 2 e Gm a r a = 0 , or v e = r 2 Gm a r a . We can obtain the quantity Gm a from the fact that g a = Gm a /r 2 a . Then Gm a = g a r 2 a = 7 . 5 × 10 11 m 3 /s 2 . So v e = r 2 Gm a r a = r 2 × 7 . 5 × 10 11 5 × 10 5 = 1 . 7 km/s . (b) An object leaves the surface of the asteroid with an initial speed of v 0 = 1 × 10 3 m/s . The distance it will travel before stopping can be found using conservation of energy: K i + U g ( r a ) = K f + U g ( r ) or 1 2 mv 2 0 Gm a m r a = 0 Gm a m r . Solving for r and using the value for Gm a found in part (a) gives r = 7 . 5 × 10 5 m . Then, the distance from the surface of the asteroid is h = 750 km 500 km = 250 km . (c) To ±nd the speed of impact of an object at the surface of the asteroid if it falls a distance h , we again use conservation of energy: K i + U g ( r a + h ) = K f + U g ( r a ) or 0 Gm a m r a + h = 1 2 mv 2 f Gm a m r a . 1

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Then v 2 f = 2 Gm a p 1 r a 1 r a + h P = 2 × 10 6 m 2 /s 2 . Then v f = 1 . 4 km/s . 2. (a) The force of gravity on the satellite is F g = GMm r 2 where r = r earth + h and r earth = 6371 km. Assuming that the orbit is circular, the acceleration of the satellite is a = v 2 /r . Then ma = F g = GMm r 2 v 2 r = GM r 2 v = p GM r P 1 2 (1) = p (6 . 67 × 10 11 Nm 2 / kg)(5 . 97 × 10 24 kg) 300 × 10 3 m + 6371 × 10 3 m P 1 2 = 7726 m/s (b) Again we write v = p GM r P 1 2 = p (6 . 67 × 10 11 Nm 2 / kg)(5 . 97 × 10 24 kg) 35800 × 10 3 m + 6371 × 10 3 m P 1 2 = 3073 m/s The revolution period is T = 2 πr/v = 2 π (35800 × 10 3 m + 6371 × 10 3 m) / (3073 m / s) 24hrs. (c) The total energy of the satellite is the sum of kinetic and gravitational potential. Note that the total energy is less than zero. Total energy is less than zero as long as the velocity is less than escape velocity. E = GMm r + 1 2 mv 2 2
Substitute for v using Equation 1 and we Fnd that E = GMm r 2 + 1 2 m p GM r P = 1 2 GMm r = 1 2 U ( r ) = K ( r ) The ratios of energies of equal mass satellites at radii r 1 and r 2 is E 1 E 2 = 1 2 GMm/r 1 1 2 GMm/r 2 = r 2 r 1 = (35800 + 6371) km (300 + 6371) km = 6.3 Note that the energies of both satellites are negative, indicating that that the satellites are bound by gravity to the earth. Satellite 1, at the lower altitude is more tightly bound than satellite 2 at the higher altitude. (d) At the surface of the earth at launch the energy of the satellite is

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## This note was uploaded on 10/27/2010 for the course PHYS 2207 at Cornell.

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hw8-10--solutions - P2207 - Fall 2010 Solutions Assignment...

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