ps8 - P207 - Fall 2007 Solutions Assignment 8 1. Chapter...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
P207 - Fall 2007 Solutions Assignment 8 1. Chapter 13, Problem 37 (a) The escape velocity v = p 2 GM/R . The gravitational acceleration at the surface is g = GM/R 2 . Therefore, v = 2 Rg = p 2(500 × 10 3 m)(3 m / s 2 ) = 1732 m / s. (b) The energy of a particle at rest, infinitely far from the asteroid, (far enough away that the force of gravity from the asteroid is essentially zero), is zero. The work done on the particle to move it from R = to R = R i is W = ~ F · ~ d = Z R i - GMm r 2 ˆ r · d~ r = GMm r | R i = GMm R i Since W = - Δ U , the potential energy at the surface radius R i is U = - GMm R i Now at the surface we give the particle a velocity v i , and the total energy (U+K) of the particle at the surface radius R i is E i = - GMm R i + 1 2 mv 2 i The energy when the particle stops at radial distance R f is E f = - GMm R f Energy is conserved so - GMm R f = - GMm R i + 1 2 mv 2 i (1) 1 R f = 1 R i - 1 2 v 2 i GM = 1 R i - 1 2 v 2 i R 2 i g = 1 (500 × 10 3 m) - 1 2 (1000 m / s) 2 (3 m / s 2 )(500 × 10 3 m) 2 = 1 . 33 × 10 - 6 m - 1 R f = 750 km 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
The height of the particle relative to the surface will be R f - R i = 250 km. (c) We use Equation 1 again. We solve for velocity in terms of R f the radial distance from the center of the asteroid at which the particle is dropped, and R i , the radius of the asteroid. We get that v 2 i = 2 GM ± 1 R i - 1 R f ² v i = ± 2 GM ± 1 R i - 1 R f ²² 1 2 = ± 2 R 2 i g ± 1 R i - 1 R f ²² 1 2 = ± 2(500 × 10 3 m) 2 (3 m / s 2 ) ± 1 500 × 10 3 m - 1 1500 × 10 3 m ²² 1 2 = 1414 m / s Note that R f = R i + h = 500 km+1000 km, where h is the height above the surface. 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 = ± GM r ² 1 2 (2) = ± (6 . 67 × 10 - 11 Nm 2 / kg)(5 . 97 × 10 24 kg) 300 × 10 3 m + 6371 × 10 3 m ² 1 2 = 7726 m / s 2
Background image of page 2
(b) Again we write v = ± GM r ² 1 2 = ± (6 . 67 × 10 - 11 Nm 2 / kg)(5 . 97 × 10 24 kg) 35800 × 10 3 m + 6371 × 10 3 m ² 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 2 + 1 2 mv 2 Substitute for v using Equation 2 and we find that E = - GMm r 2 + 1 2 m ± GM r ² = - 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
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/22/2009 for the course PHYS 207 taught by Professor Liepe during the Fall '08 term at Cornell University (Engineering School).

Page1 / 9

ps8 - P207 - Fall 2007 Solutions Assignment 8 1. Chapter...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online