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JEE Main 2023 (10 Apr, Shift 2) · Satellites - Orbital Velocity & Period

Satellites - Orbital Velocity & Period question from JEE Main 2023 (10 Apr, Shift 2)

Q35JEE Main 2023 · 10 APR · SHIFT 2Medium

Satellites - Orbital Velocity & Period

The time period of a satellite, revolving above earth's surface at a height equal to $R$ will be (Given $g = \pi^2\text{ m/s}^2,\ R =$ radius of earth)

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Answer: (a)

At height $R$ , $r = 2R$ : $T = 2\pi\sqrt{\frac{(2R)^3}{gR^2}} = 2\pi\sqrt{\frac{8R}{g}}$
With $g = \pi^2$ : $T = 2\pi\sqrt{\frac{8R}{\pi^2}} = 2\sqrt{8R}$
$T = \sqrt{32R}$
So the answer is (a).

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