MHT CET 2024 15th May Morning Shift | Gravitation Question 8 | Physics

A satellite of mass ' $m$ ' is revolving around the earth of mass ' $M$ ' in an orbit of radius ' $r$ ' with constant angular velocity ' $\omega$ '. The angular momentum of satellite is ( $\mathrm{G}=$ Universal constant of gravitation)

$\mathrm{m}(\mathrm{GMr})^{3 / 2}$

$\mathrm{m}(\mathrm{GMr})$

$\mathrm{m}(\mathrm{GMr})^{1 / 2}$

$ \mathrm{m}(\mathrm{GMr})^{-1 / 2}$

MHT CET 2024 11th May Evening Shift

MCQ (Single Correct Answer)

-0

For a satellite moving in an orbit around the earth at height ' $h$ ' the ratio of kinetic energy to potential energy is

$2: 1$

$1: 2$

$1: \sqrt{2}$

$\sqrt{2}: 1$

MHT CET 2024 11th May Evening Shift

MCQ (Single Correct Answer)

-0

A body is projected in vertically upward direction from the surface of the earth of radius ' $R$ ' into space with velocity ' $n V_{\mathrm{e}}$ ' $(\mathrm{n}<1)$. The maximum height from the surface of earth to which a body can reach is

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

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

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

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

MHT CET 2024 11th May Morning Shift

MCQ (Single Correct Answer)

-0

The acceleration due to gravity at the surface of the planet is same as that at the surface of the earth, but the density of planet is thrice that of the earth. If 'R' is the radius of the earth, the radius of the planet will be

$\frac{\mathrm{R}}{9}$

$\frac{\mathrm{R}}{3}$

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