(C) The escape velocity $v_e$ of an object from the surface of a planet is given by the formula $v_e = \sqrt{\frac{2 G M}{R}}$.$A$ black hole is defin

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$\sqrt{\frac{2 G M}{R}} \geq c$

(C) The escape velocity $v_e$ of an object from the surface of a planet is given by the formula $v_e = \sqrt{\frac{2 G M}{R}}$.$A$ black hole is defined as a celestial object from which even light cannot escape due to its intense gravitational field.For light to be unable to escape,the escape velocity of the planet must be greater than or equal to the speed of light $c$.Therefore,the condition is $v_e \geq c$.Substituting the expression for $v_e$,we get $\sqrt{\frac{2 G M}{R}} \geq c$.

Class 11 Physics Gravitation Escape Velocity and Escape Energy

Medium

If $M$ is the mass of a planet and $R$ is its radius,then in order to become a black hole (where $c$ is the speed of light),which condition must be satisfied?

A
$\sqrt{\frac{G M}{R}} \leq c$
B
$\sqrt{\frac{G M}{2 R}} \geq c$
C
$\sqrt{\frac{2 G M}{R}} \geq c$
D
$\sqrt{\frac{2 G M}{R}} \leq c$

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Gravitation

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Escape Velocity and Escape Energy

$A$ body is projected vertically upwards from the Earth's surface. If the velocity of projection is $\left(\frac{1}{3}\right)$ of the escape velocity,then the height up to which the body rises is $(R = \text{radius of Earth})$

MediumMHT CET 2020

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The masses and radii of the earth and moon are $(M_1, R_1)$ and $(M_2, R_2)$ respectively. Their centers are at a distance $r$ apart. Find the minimum escape velocity for a particle of mass $m$ to be projected from the midpoint between these two masses.

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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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An object is thrown vertically upwards from the surface of the earth with a velocity $x$ times the escape velocity on the earth $(x < 1)$. The maximum height to which it rises from the center of the earth is (radius of earth is $R$):

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Does the escape speed of a body from the Earth depend on:
$(a)$ the mass of the body,
$(b)$ the location from where it is projected,
$(c)$ the direction of projection,
$(d)$ the height of the location from where the body is launched?

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If $M$ is the mass of a planet and $R$ is its radius,then in order to become a black hole (where $c$ is the speed of light),which condition must be satisfied?

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$A$ body is projected vertically upwards from the Earth's surface. If the velocity of projection is $\left(\frac{1}{3}\right)$ of the escape velocity,then the height up to which the body rises is $(R = \text{radius of Earth})$

The masses and radii of the earth and moon are $(M_1, R_1)$ and $(M_2, R_2)$ respectively. Their centers are at a distance $r$ apart. Find the minimum escape velocity for a particle of mass $m$ to be projected from the midpoint between these two masses.

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:

An object is thrown vertically upwards from the surface of the earth with a velocity $x$ times the escape velocity on the earth $(x < 1)$. The maximum height to which it rises from the center of the earth is (radius of earth is $R$):

Does the escape speed of a body from the Earth depend on:$(a)$ the mass of the body,$(b)$ the location from where it is projected,$(c)$ the direction of projection,$(d)$ the height of the location from where the body is launched?

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Does the escape speed of a body from the Earth depend on:
$(a)$ the mass of the body,
$(b)$ the location from where it is projected,
$(c)$ the direction of projection,
$(d)$ the height of the location from where the body is launched?