A rocket is fired vertically upward with a speed of $v = \frac{v_e}{\sqrt{2}}$ from the Earth's surface, where

Question

A rocket is fired vertically upward with a speed of $v = \frac{v_e}{\sqrt{2}}$ from the Earth's surface, where $v_e$ is escape velocity on the surface of Earth. The distance from the surface of Earth upto which the rocket can go before returning to the Earth is: (given, the radius of Earth $R = 6400$ km)

Accepted Answer

1. **Principle of Conservation of Energy:** The total mechanical energy of the rocket is conserved. Let $m$ be the mass of the rocket and $M$ be the mass of the Earth. The initial point is the surface ($r=R$) and the final point is the maximum height ($r=R+h$) where velocity becomes zero.
   $$ K_i + U_i = K_f + U_f $$
   $$ \frac{1}{2}mv^2 - \frac{GMm}{R} = 0 - \frac{GMm}{R+h} $$
2. **Substitute Velocity:** Given $v = \frac{v_e}{\sqrt{2}}$. We know escape velocity $v_e = \sqrt{\frac{2GM}{R}}$.
   Therefore, $v^2 = \frac{v_e^2}{2} = \frac{1}{2} \left( \frac{2GM}{R} \right) = \frac{GM}{R}$.
3. **Solve for h:** Substitute $v^2$ into the energy equation:
   $$ \frac{1}{2}m \left( \frac{GM}{R} \right) - \frac{GMm}{R} = -\frac{GMm}{R+h} $$
   $$ \frac{GMm}{2R} - \frac{GMm}{R} = -\frac{GMm}{R+h} $$
   $$ \frac{GMm}{R} \left( \frac{1}{2} - 1 \right) = -\frac{GMm}{R+h} $$
   $$ -\frac{GMm}{2R} = -\frac{GMm}{R+h} $$
   $$ \frac{1}{2R} = \frac{1}{R+h} $$
   $$ 2R = R + h \implies h = R $$
4. **Calculation:** Given $R = 6400$ km, the height reached is $h = 6400$ km.

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