From the pole of the earth,a body of mass m i | vedclass

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Question

(A) According to the law of conservation of mechanical energy,the total energy at the surface of the earth must equal the total energy at the maximum

Vedclass · Accepted Answer

$\frac{R v_0^2}{2 g R - v_0^2}$

Vedclass · Answer

(A) According to the law of conservation of mechanical energy,the total energy at the surface of the earth must equal the total energy at the maximum height $h$.At the surface: $E_i = K_i + U_i = \frac{1}{2} m v_0^2 - \frac{G M m}{R}$.At maximum height $h$: $E_f = K_f + U_f = 0 - \frac{G M m}{R+h}$.Equating $E_i = E_f$:$\frac{1}{2} m v_0^2 - \frac{G M m}{R} = - \frac{G M m}{R+h}$.Dividing by $m$ and rearranging:$\frac{v_0^2}{2} = G M \left( \frac{1}{R} - \frac{1}{R+h} \right) = G M \left( \frac{h}{R(R+h)} \right)$.Since $g = \frac{G M}{R^2}$,we have $G M = g R^2$.Substituting $G M$:$\frac{v_0^2}{2} = g R^2 \left( \frac{h}{R(R+h)} \right) = \frac{g R h}{R+h}$.$v_0^2 (R+h) = 2 g R h$.$v_0^2 R + v_0^2 h = 2 g R h$.$v_0^2 R = h (2 g R - v_0^2)$.$h = \frac{R v_0^2}{2 g R - v_0^2}$.

From the pole of the earth,a body of mass $m$ is imparted a velocity $v_0$ directed vertically up. If $M$ is the mass of the earth,$R$ its radius and $g$ is the free-fall acceleration on its surface,then the height $h$ to which the body will ascend is (neglect air resistance).

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