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Question

A sky laboratory of mass 2 ×103 kg has to be lifted from one circular orbit of radius 2R to another circular orbit of radius 3R, where R is radius of the earth equal to 6.37×106m. Acceleration due to gravity =9.8m/s2 lf the minimum energy required is n×1010J, find n.

Solution
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Let velocity of laboratory in a circular orbit of radius r be v. Then,

mv2r=GMmr2

Thus, kinetic energy is

K=12mv2=GMm2r

Thus, total energy in orbit is

E=12mv2GMmr=GMm2r

Hence, energy required to change orbit is

ΔE=GMm3RGMm6R=GMm4R=gmR12

ΔE=9.8(2000)(6.37X106)121X1010J

Answer is 1.

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