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
ID: 1396009 • Letter: M
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 1200 kg satellite is orbitting the earth in a circular orbit with an altitude of 1700 km.
How much energy does it take just to get it to this altitude?
____J
How much kinetic energy does it have once it has reached this altitude?
______J
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 4800 km?
_______m
What would this ratio be if the final altitude of the satellite were 3185 km?
Answer = 1
Please help! thank you
Explanation / Answer
A) Energy given to the satellite, E = U1 - U2
= -G*Me*m/(Re+h) -(-G*Me*m/Re)
= -6.67*10^-11*5.9742*10^24*1200/(6.3781*10^6 + 1700*10^3) + 6.67*10^-11*5.9742*10^24*1200/(6.3781*10^6)
= 1.578*10^10 J
B) orbital speed, vo = sqrt(G*Me/(Re+h))
= sqrt(6.67*10^-11*5.97*10^24/(6.378*10^6 + 1700*10^3))
= 7021 m/s
so, Kinetic enrgy of satellite, KE = 0.5*m*vo^2
= 0.5*1200*7021^2
= 2.957*10^10 J
C) require ratio = 1.578*10^10/(2.957*10^10)
= 0.534
d)
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