A particle of mass $m$ is projected with a velocity, $v = kv_e$ ($k

Question

A particle of mass $m$ is projected with a velocity, $v = kv_e$ ($k < 1$) from the surface of the earth. The maximum height, above the surface, reached by the particle is: (Where $v_e = $ escape velocity, $R = $ the radius of the earth)

Accepted Answer

1. **Conservation of Energy:** The total mechanical energy of the particle is conserved. Let the maximum height reached be $h$. At this point, the velocity of the particle is zero.
   $$E_i = E_f$$
   $$K_i + U_i = K_f + U_f$$
2. **Initial Energy:** At the surface ($r=R$), velocity $v = kv_e$.
   Since escape velocity $v_e = \sqrt{\frac{2GM}{R}}$, then $v^2 = k^2 \frac{2GM}{R}$.
   $$K_i = \frac{1}{2}mv^2 = \frac{1}{2}m \left( k^2 \frac{2GM}{R} \right) = \frac{k^2 GMm}{R}$$
   $$U_i = -\frac{GMm}{R}$$
3. **Final Energy:** At maximum height ($r = R+h$), velocity is zero.
   $$K_f = 0$$
   $$U_f = -\frac{GMm}{R+h}$$
4. **Solve for h:**
   $$ \frac{k^2 GMm}{R} - \frac{GMm}{R} = -\frac{GMm}{R+h} $$
   $$ \frac{GMm}{R}(k^2 - 1) = -\frac{GMm}{R+h} $$
   $$ \frac{1-k^2}{R} = \frac{1}{R+h} $$
   $$ R+h = \frac{R}{1-k^2} $$
   $$ h = \frac{R}{1-k^2} - R = R \left( \frac{1 - (1-k^2)}{1-k^2} \right) = \frac{Rk^2}{1-k^2} $$

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