If the radius of a planet is $R$ and its density is $\rho$,the escape velocity from its surface will be

  • A
    $v_e \propto \rho R$
  • B
    $v_e \propto \sqrt{\rho} R$
  • C
    $v_e \propto \frac{\sqrt{\rho}}{R}$
  • D
    $v_e \propto \frac{1}{\sqrt{\rho} R}$

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$A$ body is projected up with a velocity equal to $\frac{3}{4}$ of the escape velocity from the surface of the earth. The height it reaches is (Radius of the earth $= R$)

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 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.

An object is thrown directly away from the surface of the earth with an initial speed $v$. The object reaches up to a height of $\frac{4}{5} R_E$ from the earth's surface,where $R_E$ is the radius of the earth. If the escape velocity of the object is $v_E$,then the value of $\frac{v}{v_E}$ is:

The escape velocity of a body depends upon its mass as:

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