(A) The escape velocity $v$ is given by the energy conservation principle: $\frac{-GMm}{R} + \frac{1}{2}mv^2 = 0$.Solving for $v$,we get $v = \sqrt{\f

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$v = R\sqrt{\frac{8\pi}{3}G\rho}$

(A) The escape velocity $v$ is given by the energy conservation principle: $\frac{-GMm}{R} + \frac{1}{2}mv^2 = 0$.Solving for $v$,we get $v = \sqrt{\frac{2GM}{R}}$.The mass of the Earth $M$ in terms of its mean density $\rho$ and radius $R$ is $M = \rho \times \frac{4}{3}\pi R^3$.Substituting this value of $M$ into the escape velocity formula:$v = \sqrt{\frac{2G}{R} \times \rho \times \frac{4}{3}\pi R^3}$.Simplifying the expression:$v = \sqrt{\frac{8}{3}G\pi R^2 \rho} = R\sqrt{\frac{8\pi}{3}G\rho}$.

Class 11 Physics Gravitation Escape Velocity and Escape Energy

Medium

The escape velocity of an object from the Earth depends upon the mass of the Earth $(M)$,its mean density $(\rho)$,its radius $(R)$,and the gravitational constant $(G)$. Thus,the formula for escape velocity is:

A
$v = R\sqrt{\frac{8\pi}{3}G\rho}$
B
$v = M\sqrt{\frac{8\pi}{3}GR}$
C
$v = \sqrt{2GMR}$
D
$v = \sqrt{\frac{2GM}{R^2}}$

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Gravitation

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Escape Velocity and Escape Energy

$A$ particle released at a large distance from a planet reaches the planet only under gravitational attraction and passes through a smooth tunnel through its centre. If $v_e$ is the escape velocity of a body at the planet,then the particle's speed at the centre of the planet is

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The escape velocity of a body from the surface of the earth is $11.2 \,km/s$. The escape velocity of a body from a planet having the same mean density as the earth but twice the radius of earth is: (in $\,km/s$)

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$A$ body is projected vertically upwards from the surface of a planet of radius $R$ with a velocity equal to half the escape velocity of that planet. Then,the maximum height attained by the body is

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Masses and radii of Earth and Moon are $M_1, M_2$ and $R_1, R_2$ respectively. The distance between their centers is $d$. The minimum velocity given to a mass $m$ from the midpoint of the line joining their centers so that it escapes the gravitational field of both is:

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The escape velocity of an object from the Earth depends upon the mass of the Earth $(M)$,its mean density $(\rho)$,its radius $(R)$,and the gravitational constant $(G)$. Thus,the formula for escape velocity is:

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