Solution:
Kinetic energy of a satellite is \(K = \frac{GMm}{2r}\). Here \(r_1 = R + R = 2R\) and \(r_2 = R + 7R = 8R\). The ratio \(frac{K_1}{K_2} = \frac{r_2}{r_1} = \frac{8R}{2R} = 4\).
Two identical satellites are at height \(R\) and \(7R\) from earth surface, the ratio of their kinetic energies will be :
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