Q.3 (i) 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 ?
Best answer
Answer: About 43,000 s, i.e. nearly 12 hours — half the original period.
Orbital velocity \(v = \sqrt{\dfrac{GM}{r}}\), so if the mass of the earth becomes 4M, the velocity becomes \(\sqrt{4} = 2\) times its original value.
The period \(T = \dfrac{2\pi r}{v}\), and r = 6400 + 35780 = 42180 km is unchanged. With v doubled, T becomes half of the original period.
\[T = \frac{2\pi r}{v} = \frac{2 \times 3.142 \times 42180}{2 \times 3.08} = \frac{265{,}000}{6.16} \approx 43{,}020\ \text{s} \approx 12\ \text{hours}\]
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