A spaceship orbits around a planet at a height of 20 km from its… | JEE Main 2019 (10 Apr, Shift 2)

Q59JEE Main 2019 · 10 APR · SHIFT 2Medium

Satellites - Orbital Velocity & Period

A spaceship orbits around a planet at a height of 20 km from its surface. Assuming that only gravitational field of the planet acts on the spaceship, what will be the number of complete revolutions made by the spaceship in 24 hours around the planet?

[Given; Mass of planet = 8 × 1022 kg, Radius of planet = 2 × 106 m, Gravitational constant G = 6.67 × 10–11 Nm2 /kg2]

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Answer: (d)

$r = R + h = 2\times10^6 + 2\times10^4 = 2.02\times10^6$ m
$T = 2\pi\sqrt{\frac{r^3}{GM}} = 2\pi\sqrt{\frac{8.24\times10^{18}}{6.67\times10^{-11}\times8\times10^{22}}} = 2\pi\sqrt{1.545\times10^6} \approx 7.81\times10^3$ s
Revolutions in 24 h $= \frac{86400}{7810} \approx 11.06$
Complete revolutions = 11, answer (d).

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