5)
given
m = 4500 kg
h1 = 170 km
h2 = 390 km
we know,
Re = 6.37*10^6 m
Me = 5.98*10^24 kg
so,
r1 = Re + h1 = 6.37*10^6 + 0.17*10^6 = 6.54*10^6 m
r2 = Re + h2 = 6.37*10^6 + 0.39*10^6 = 6.76*10^6 m
we know total energy of saceship, E1 = -G*Me*m/(2*r)
Workdone = change in mechanical energy
W = E2 - E1
= -G*Me*m/(2*r2) - ( -G*Me*m/(2*r1))
= G*Me*m*(1/r1 - 1/r2)
= 6.67*10^-11*5.98*10^24*4500*(1/(6.54*10^6) - 1/(6.76*10^6) )/2
= 4.47*10^9 J
(13% ) Problem 5: A 4500-kg spaceship is in a circular orbit 170 km above the...
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