m earth = 5.9742 x 10 24 kg r earth = 6.3781 x 10 6 m m moon = 7.36 x 10 22 kg r

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

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 4600 km?

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

Explanation / Answer

A)Energy =work done=G*Me*Ms/(Re+h)=7.985*10^10 J...........

B) V0=sqrt[G*Me/(R+h)]=7.06*10^3m/s....

K.E =0.5*Ms*Vo^2=3.99*10^10 J.........

C) A/B=2....

D) 2

E) 2...the ratio does not changes

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.