Question

A spacecraft of 90 kg mass is in a circular orbit about the Earth at a height h-2RE (a) What is the period of the spacecraft's orbit about the Earth? (b) What is the spacecraft's kinetic energy? (c) Express the angular momentum L of the spacecraft about the center of the Earth in terms of its kinetic energy K. (Use the following as necessary: RE for the radius of the Earth, K for the kinetic energy of the satellite, and m for the mass of the satellite.) (d) Find the numerical value of the angular momentum eBook +13 points Tipler6 11.P069 My Notes Ask Your Teach A thin spherical shell has a radius of 2 m and a mass of 305 kg. What is the gravitational field at each of the following distances from the center of the shell? (a) 0.5 m N/kg (b) 1.9 m N/kg (c) 2.4 m N/kg eBook

Answer 1

The time period of a satellite is given by

$T =2\pi \sqrt{\frac{(R_{E}+h)^{3}}{gR_{E}^{2}}}$

$T =2\pi \sqrt{\frac{(R_{E}+2R_{E})^{3}}{gR_{E}^{2}}}$

$T =2\pi \sqrt{\frac{(3R_{E})^{3}}{gR_{E}^{2}}}$

$T =2\pi \sqrt{\frac{27R_{E}^{3}}{gR_{E}^{2}}}$

$T =2\pi \sqrt{\frac{27R_{E}}{g}}$

R_E = 6400 Km = 6400 X 10³ m

$T =2\pi \sqrt{\frac{27X6400 X 1000}{9.81}}$

T = 2.34 X 10⁴ s

T = 7.33 hours

b)

$K = \frac{1}{2}mv_{o}^{2}$

$K = \frac{1}{2}m\left ( \frac{GM}{R_{E} + h} \right )$

$K = \frac{1}{2}90\left ( \frac{gR_{E}^{2}}{R_{E} + h} \right )$

$K = 45\left ( \frac{gR_{E}^{2}}{R_{E} + 2R_{E}} \right )$

$K = 45\left ( \frac{gR_{E}^{2}}{3R_{E}} \right )$

$K = 45\left ( \frac{gR_{E}}{3} \right )$

K = 9.42 X 10⁸ J

c) Angular momentum is

$K = \frac{1}{2}mv_{o}^{2}$

$v_{o}^{2} = \frac{2K}{m}$

$v_{o} = \sqrt{\frac{2K}{m}}$

$L = m \sqrt{\frac{2K}{m}}r$

$L = m \sqrt{\frac{2K}{m}}(R_{E} + h)$

$L = \sqrt{2mK}(R_{E} + 2R_{E})$

$L = 3R_{E} \sqrt{2mK}$

$L = 3(6400 X 1000) \sqrt{2(90)(9.42 X 10^{8})}$

L = 7.91 X 10¹² J-s

2) As per Newton's shell theorem, the gravitational field inside a shell is zero

a) r = 0.5 m is inside the shell so gravitational field = 0

b) r =1.9 m is inside the shell so gravitational field = 0

c) r = 2.4 m , the field is given by

$g = \frac{GM}{r^{2}}$

$g = \frac{6.67 X 10^{-11}X 305}{2.4^{2}}$

g = 3.53 X 10^-9 N/kg