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A sky laboratory of mass 3 x 103 kg has to be lifted from one circular orbit of radius 2R into another circular orbit of radius 3R. Calculate the minimum energy (in x 1010J ) required if the radius of earth is R=6.4×106m and g=10ms−2

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answer is 1.60.

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Detailed Solution

Total energy of sky laboratory, E=−GMm2rE1=−GMm2(2R) and E2=−GMm2(3R)Therefore, the energy required,ΔE=E2−E1=GMm14R−16R=GMm12R=R2gm12R=mgR12=1.60×1010J

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