Given below are two statements : A satellite is moving around earth in the orbit very close to the earth surface
Q. 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 $$ T = 2\pi\sqrt{\frac{R_e}{g}} $$ (for satellite very close to the earth surface), where $R_e$ is radius of earth and $g$ is acceleration due to gravity.

In the light of the above statements, choose the correct answer from the options given below :

(A) Statement I is true but Statement II is false

(B) Statement I is false but Statement II is true

(C) Both Statement I and Statement II are true

(D) Both Statement I and Statement II are false

Correct Answer: Both Statement I and Statement II are true

Explanation

For a satellite moving very close to the surface of the earth, the time period of revolution is given by

$$ T = 2\pi\sqrt{\frac{R_e}{g}} $$

Thus, Statement II is correct.

Now express acceleration due to gravity in terms of earth’s density. We know that

$$ g = \frac{GM}{R_e^2} $$

Mass of earth,

$$ M = \frac{4}{3}\pi R_e^3 \rho $$

Substituting in expression of $g$,

$$ g = \frac{G \left(\frac{4}{3}\pi R_e^3 \rho\right)}{R_e^2} $$

$$ g = \frac{4}{3}\pi G R_e \rho $$

Substitute this value of $g$ in time period formula,

$$ T = 2\pi\sqrt{\frac{R_e}{\frac{4}{3}\pi G R_e \rho}} $$

$$ T = 2\pi\sqrt{\frac{3}{4\pi G \rho}} $$

Thus, the time period depends only on the density of earth.

Hence, Statement I is also true.

Therefore, the correct option is

Both Statement I and Statement II are true

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