Escape velocity at surface of earth is 11.2,km/sec . If radius of planet is double that of earth but

English

Gujarati

Hindi

7.Gravitation

normal

Escape velocity at the surface of earth is $11.2\,km/sec$ . If radius of planet is double that of earth but mean density same as that of earth then the escape velocity will be ........ $km/sec$

A

$11.2$

B

$5.5$

C

$15.5$

D

$22.4$

Solution

$V_{e}=\sqrt{2 g e}=\sqrt{2 \times \frac{4}{3} \pi R p g \times R}$

$\mathrm{V}_{\mathrm{e}} \times \mathrm{R}$

$\frac{V_{e}}{V_{e}^{1}}=\frac{R}{R^{1}} \Rightarrow V_{e}^{1}=2 V_{e}=2 \times 11.2$

$\mathrm{V}_{\mathrm{e}}^{\mathrm{l}}=22.4 \mathrm{km} / \mathrm{s}$

Standard 11

Physics

Share

Similar Questions

A particle is kept at rest at a distance $'R'$ from the surface of earth (of radius $R$). The minimum speed with which it should be projected so that it does not return is

normal

The dependence of acceleration due to gravity $g$ on the distance $r$ from the centre of the earth assumed to be a sphere of radius $R$ of uniform density is as shown figure below

The correct figure is

normal

A projectile is projected with velocity $k{v_e}$ in vertically upward direction from the ground into the space. (${v_e}$ is escape velocity and $k < 1)$. If air resistance is considered to be negligible then the maximum height from the centre of earth to which it can go, will be : (R = radius of earth)

normal

If $M$ is mass of a planet and $R$ is its radius then in order to become black hole [ $c$ is speed of light]

normal

The distance of neptune and saturn from the sun is nearly $10^{13}$ and $10^{12}$ meter respectively. Assuming that they move in circular orbits, their periodic times will be in the ratio

normal