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Question

(A) The acceleration due to gravity at the equator $(g_e)$ is related to the acceleration due to gravity at the poles $(g_p)$ by the formula: $g_e = g

Vedclass · Accepted Answer

$V_e = 2V$

Vedclass · Answer

(A) The acceleration due to gravity at the equator $(g_e)$ is related to the acceleration due to gravity at the poles $(g_p)$ by the formula: $g_e = g_p - R\omega^2$.Given that $g_e = g_p / 2$,we have: $g_p / 2 = g_p - R\omega^2$.This simplifies to: $R\omega^2 = g_p / 2$.Since the velocity of a point on the equator is $V = R\omega$,we have $V^2 = R^2\omega^2 = R(R\omega^2) = R(g_p / 2)$.Thus,$V^2 = g_p R / 2$,which implies $g_p R = 2V^2$.The escape velocity $V_e$ is given by the formula: $V_e = \sqrt{2g_p R}$.Substituting $g_p R = 2V^2$ into the escape velocity formula:$V_e = \sqrt{2(2V^2)} = \sqrt{4V^2} = 2V$.

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	$(b)$ $7.92 \, km \, s^{-1}$
	$(c)$ $11.2 \, km \, s^{-1}$

Similar Questions

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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 $(b)$ $7.92 \, km \, s^{-1}$

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

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For a planet having mass equal to the mass of the Earth but a radius that is one-fourth of the radius of the Earth,the escape velocity for this planet will be (in $km/s$): (in $km/s$)

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