MHT CET 2026 19th April Evening Shift | Gravitation Question 3 | Physics

A satellite is revolving around a planet in a circular orbit close to its surface. Let $\rho$ be mean density and $R$ be the radius of the planet; then the period of the satellite is
($G$ = Universal constant of gravitation).

$\sqrt{\dfrac{4\pi}{\rho G}}$

$\sqrt{\dfrac{2\pi}{\rho G}}$

$\sqrt{\dfrac{\pi}{\rho G}}$

$\sqrt{\dfrac{3\pi}{\rho G}}$

MHT CET 2026 19th April Morning Shift

MCQ (Single Correct Answer)

-0

A geostationary satellite is orbiting the earth at a height of $4R$ above the surface of the earth, where $R$ is the radius of the earth. Another satellite is orbiting the earth at a height $1.5R$ from the surface of the earth with periodic time 'T' in hour. The value of 'T' in hour is

$\dfrac{6}{\sqrt{2}}$

$6\sqrt{2}$

$4$

$8$

MHT CET 2025 5th May Evening Shift

MCQ (Single Correct Answer)

-0

A uniform sphere has radius ' $R$ ' and mass ' $M$ '. The magnitude of gravitational field at distances ' $\mathrm{r}_1$ ' and ' $\mathrm{r}_2$ ' from the centre of the sphere are ' $E_1$ ' and ' $E_2$ ' respectively. The ratio $E_1: E_2$ is $\left(r_1>R\right.$ and $\left.r_2

$\frac{R^2}{r_1^2 r_2}$

$\frac{R^3}{r_1 r_2}$

$\frac{R^3}{r_1^2 r_2}$

$\frac{R^3}{r_1 r_2^2}$

MHT CET 2025 26th April Evening Shift

MCQ (Single Correct Answer)

-0

The depth at which acceleration due to gravity becomes $\frac{g^{\prime}}{n}$ is ( $R=$ radius of earth, $\mathrm{g}=$ acceleration due to gravity) ( $\mathrm{n}=$ integer)

$\frac{R(n-1)}{n}$

$\frac{R(n+1)}{n}$

$\frac{R(n-1)^2}{n}$

$\frac{R(n+1)^2}{n}$

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