JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Gravitation

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Given below are two statements:
Statement I: A satellite is moving around earth in the orbit very close to the earth surface. The time period of revolution of satellite depends upon the density of earth.
Statement II: The time period of revolution of the satellite is

T = 2 π \frac{R _{e}}{g}

(for satellite very close to the earth surface), where

R_{e}

radius of earth and

g

acceleration due to gravity.
In the light of the above statements, choose the correct answer from the options given below:

(A)Both Statement I and Statement II are false
(B)Both Statement I and Statement II are true✓ Correct
(C)Statement I is true but Statement II is false
(D)Statement I is false but Statement II is true

Answer: (B)

Solution

For a satellite skimming the surface, gravity supplies the centripetal force:

\frac{GM}{R _{e}^{2}} = \frac{4 π ^{2} R _{e}}{T ^{2}}

, which gives

T = 2 π \frac{R _{e}^{3}}{GM} = 2 π \frac{R _{e}}{g}

, so Statement II is correct. Writing

M = ρ \cdot \frac{4}{3} π R_{e}^{3}

turns this into

T = 2 π \frac{3}{4 π Gρ}

, which depends only on the mean density

ρ

, so Statement I is correct too. Both statements are true.

JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Single correct option

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JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Gravitation

Solution

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