MHT CET 2024 2nd May Evening Shift | Gravitation Question 44 | Physics

The gravitational potential energy required to raise a satellite of mass ' $m$ ' to height ' $h$ ' above the earth's surface is ' $\mathrm{E}_1$ '. Let the energy required to put this satellite into the orbit at the same height be ' $E_2$ '. If $M$ and $R$ are the mass and radius of the earth respectively then $E_1: E_2$ is

$\mathrm{h}: \mathrm{R}$

$\mathrm{h}: 2 \mathrm{R}$

$\mathrm{R}: \mathrm{h}$

$2 \mathrm{~h}: \mathrm{R}$

MHT CET 2024 2nd May Morning Shift

MCQ (Single Correct Answer)

-0

The height above the earth's surface at which the acceleration due to gravity becomes $\left(\frac{1}{n}\right)$ times the value at the surface is ( $R=$ radius of earth)

$\frac{\mathrm{R}}{\sqrt{\mathrm{n}}}$

$\mathrm{R} \cdot \sqrt{\mathrm{n}}$

$\quad(\sqrt{n}+1) R$

$(\sqrt{\mathrm{n}}-1) \mathrm{R}$

MHT CET 2024 2nd May Morning Shift

MCQ (Single Correct Answer)

-0

The magnitude of gravitational field at distance ' $r_1$ ' and ' $r_2$ ' from the centre of a uniform sphere of radius ' $R$ ' and mass ' $M$ ' are ' $F_1$ ' and ' $F_2$ ' respectively. The ratio ' $\left(F_1 / F_2\right)$ ' will be (if $r_1>R$ and $r_2

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

$\frac{\mathrm{R}^3}{\mathrm{r}_1 \mathrm{r}_2^2}$

$\frac{\mathrm{R}^3}{\mathrm{r}_1^2 \mathrm{r}_2}$

$\frac{\mathrm{R}^4}{\mathrm{r}_1^2 \mathrm{r}_2^2}$

MHT CET 2023 14th May Evening Shift

MCQ (Single Correct Answer)

-0

Earth is assumed to be a sphere of radius R. If '$$\mathrm{g}_\phi$$' is value of effective acceleration due to gravity at latitude $$30^{\circ}$$ and '$$g$$' is the value at equator, then the value of $$\left|g-g_\phi\right|$$ is ($$\omega$$ is angular velocity of rotation of earth, $$\cos 30^{\circ}=\frac{\sqrt{3}}{2}$$ )

$$\frac{1}{4} \omega^2 \mathrm{R}$$

$$\frac{3}{4} \omega^2 \mathrm{R}$$

$$\omega^2 \mathrm{R}$$

$$\frac{1}{2} \omega^2 \mathrm{R}$$

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