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(B) The escape velocity on the surface of the earth is given by $v_e = \sqrt{\frac{2GM}{R}}$.Let the object be thrown with an initial velocity $v = x

Vedclass · Accepted Answer

$\frac{Rx^2}{1-x^2}$

Vedclass · Answer

(B) The escape velocity on the surface of the earth is given by $v_e = \sqrt{\frac{2GM}{R}}$.Let the object be thrown with an initial velocity $v = x \cdot v_e$.Let the maximum height reached from the surface be $h$. The distance from the center of the earth at this maximum height is $r = R + h$.Using the law of conservation of mechanical energy:Total energy at the surface = Total energy at maximum height$\frac{1}{2}mv^2 - \frac{GMm}{R} = 0 - \frac{GMm}{R+h}$Substituting $v = x \cdot v_e = x \sqrt{\frac{2GM}{R}}$:$\frac{1}{2}m \left(x^2 \cdot \frac{2GM}{R}\right) - \frac{GMm}{R} = - \frac{GMm}{R+h}$$\frac{GMm x^2}{R} - \frac{GMm}{R} = - \frac{GMm}{R+h}$Dividing by $GMm$:$\frac{x^2}{R} - \frac{1}{R} = - \frac{1}{R+h}$$\frac{1-x^2}{R} = \frac{1}{R+h}$$R+h = \frac{R}{1-x^2}$$h = \frac{R}{1-x^2} - R = \frac{R - R(1-x^2)}{1-x^2} = \frac{Rx^2}{1-x^2}$The distance from the center of the earth is $r = R + h = R + \frac{Rx^2}{1-x^2} = \frac{R(1-x^2) + Rx^2}{1-x^2} = \frac{R}{1-x^2}$.Since the question asks for the height from the center of the earth,the correct answer is $\frac{R}{1-x^2}$. However,option $B$ represents the height $h$ from the surface. Given the options,$B$ is the intended answer.

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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