Question

m_earth = 5.9742 x 10²⁴ kg
r_earth = 6.3781 x 10⁶ m
m_moon = 7.36 x 10²² kg
r_moon = 1.7374 x 10⁶ m
d_{earth to moon} = 3.844 x 10⁸ m (center to center)
G = 6.67428 x 10^-11 N-m²/kg²

A 1600 kg satellite is orbitting the earth in a circular orbit with an altitude of 1600 km.

How much energy does it take just to get it to this altitude?

How much kinetic energy does it have once it has reached this altitude?

What is the ratio of the this change in potential energy to the change in kinetic energy? (i.e. what is (a)/(b)?)

What would this ratio be if the final altitude of the satellite were 4500 km?

What would this ratio be if the final altitude of the satellite were 3185 km?

Answer 1

a) Potential energy at surface of earth =

KE at surface = 0

Potential energy at height h is

potential energy at height 1.6*10⁶ m =

So energy required to take it to 1600 km = 10.002599 * 10¹⁰ - 7.99658*10¹⁰ = 2.006019*10¹⁰ J

b) KE at height h will be negative of potential energy/2 = 3.99829*10¹⁰ J

c) Change in potential energy/change in kinetic energy = (2.006019*10¹⁰)/(3.99829*10¹⁰) = 0.5017

d) potential energy at height 4.5*10⁶ m =

So energy required to take it to 4500 km = 10.002599 * 10¹⁰ - 5.864771*10¹⁰ = 4.137828*10¹⁰ J

KE at height h will be negative of potential energy/2 = 2.9323859*10¹⁰ J

So the ratio of change in potential energy to the change in kinetic energy = 4.137828*10¹⁰/ 2.9323859*10¹⁰ = 1.411

e) at height 3185 km

potential energy at height 3.185*10⁶ m =

So energy required to take it to 3185 km = 10.002599 * 10¹⁰ - 6.6712231*10¹⁰ = 3.331376*10¹⁰ J

KE at height h will be negative of potential energy/2 = 3.33561*10¹⁰ J

So ratio of the this change in potential energy to the change in kinetic energy = 3.331376*10¹⁰/3.33561*10¹⁰ = 0.99873