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Question
Solve the problem.
How much time a satellite in an orbit at height 35780 km above earth's surface would take, if the mass of the earth would have been four times its original mass?
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Solution
Given:
Height of the satellite, h = 35780 km
Let the original mass of Earth be M. Then its new mass will be 4M.
The time taken by the satellite to revolve around the Earth's orbit is given as
`T = (2piR)/(v_c)`
Now, vc is given as
`v_c = sqrt((GM)/(R + h))`
Thus,
`T = (2pi(R + h))/(sqrt((GM)/(R + h))) = (2pi(R + h)sqrt(R + h))/(sqrt(GM))` ...(i)
= `T propto 1/(sqrt(M))` ...(ii)
Thus, from equation (ii), we see that when the mass of the Earth becomes 4 times, the time period of revolution of the satellite should be halved.
i.e., `T_(4M) = T/2` ...(iii)
Now, h = 35780 km
M = 6 × 1024 kg
R = 6.4 × 105 m
G = 6.67 × 10−11 N m2/kg2
Putting the values of h, M and R first, we get
T = 24 h
Using (iii), we get
`T_(4M) = 24/2 = 12`h
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