Question

mearth = 5.9742 x 1024 kg

rearth = 6.3781 x 106 m

mmoon = 7.36 x 1022 kg

rmoon = 1.7374 x 106 m

dearth to moon = 3.844 x 108 m (center to center)

G = 6.67428 x 10-11 N-m2/kg2

103 kg satellite is orbitting the earth in a circular orbit with an altitude of 1500 km.

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

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

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

4) What would this ratio be if the final altitude of the satellite were 4400 km?

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

Fully worked please.

Answer 1

The amount of energy need to get to this altitude would the sum of the initial potential and kinetic energies, So -GMm/r + .5mv^2 = total energy where r= Radius of earth + altiude (m) By definition, total energy is zero if you are on an escape trajectory. That means that any bound orbit will have a negative energy. To find the velocity, use the Vis-Viva equation, which relates energy, velocity, semi-major axis and orbital distance (and the escape velocity is drawn from the equation as well) ϵ = V 2 r − μ r = − μ 2 a ϵ=V2r−μr=−μ2a Where ϵ ϵ is total energy μ μ is the gravitational parameter, G*M r is the distance from the earth and a is the semi-major axis of the orbit (r for circular, infinity for parabolic or escape velocity)just put the values & get the answer s.