The masses and radii of the earth and moon ar | vedclass

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Question

(B) To escape the gravitational influence of both masses,the particle must reach a point where the total gravitational potential energy is zero,which

Vedclass · Accepted Answer

$V = \sqrt{\frac{4G(M_1 + M_2)}{r}}$

Vedclass · Answer

(B) To escape the gravitational influence of both masses,the particle must reach a point where the total gravitational potential energy is zero,which is at infinity. By the law of conservation of energy,the total energy at the midpoint must be equal to the total energy at infinity (which is $0$).At the midpoint,the distance from each mass is $r/2$.The total energy at the midpoint is the sum of kinetic energy and gravitational potential energy:$E_i = \frac{1}{2}mV^2 - \frac{GM_1m}{r/2} - \frac{GM_2m}{r/2}$Setting the total energy to zero:$\frac{1}{2}mV^2 - \frac{2GM_1m}{r} - \frac{2GM_2m}{r} = 0$$\frac{1}{2}mV^2 = \frac{2Gm}{r}(M_1 + M_2)$$V^2 = \frac{4G(M_1 + M_2)}{r}$$V = \sqrt{\frac{4G(M_1 + M_2)}{r}}$

Column-$I$	Column-$II$
$(1)$ Value of escape velocity on the surface of the Earth	$(a)$ $2.38 \, km \, s^{-1}$
$(2)$ Value of escape velocity on the surface of the Moon	$(b)$ $7.92 \, km \, s^{-1}$
	$(c)$ $11.2 \, km \, s^{-1}$

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.

Similar Questions

The escape velocity from the surface of Earth of mass $M$ and radius $R$ is $V_{e}$. The escape velocity from the surface of a planet whose mass and radius are $3$ times that of the Earth will be:

The ratio of the radius of a planet $A$ to that of planet $B$ is $r$. The ratio of acceleration due to gravity on the planets is $x$. The ratio of the escape velocities from the two planets is

Match Column-$I$ with Column-$II$.

Column-$I$ Column-$II$

$(1)$ Value of escape velocity on the surface of the Earth $(a)$ $2.38 \, km \, s^{-1}$

$(2)$ Value of escape velocity on the surface of the Moon $(b)$ $7.92 \, km \, s^{-1}$

$(c)$ $11.2 \, km \, s^{-1}$

$A$ planet in a distant solar system is $10$ times more massive than the earth and its radius is $10$ times smaller. Given that the escape velocity from the earth is $11 \ km/s$,the escape velocity from the surface of the planet would be ........ $km/s$.

$A$ particle of mass $m_0$ is projected from the midpoint of the line joining two fixed particles,each of mass $m$. If the separation between the fixed particles is $l$,the minimum velocity of projection of the particle so as to escape is equal to:

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