Solution 4

Method

Let $r$ be the distance from the center of the earth where the rocket is switched off. At this instant ($t=0$), the velocity is $v_0$.

1. Conservation of Energy

After the rocket is switched off, the spaceship moves only under gravity. The problem states it eventually acquires a constant velocity $v_{\infty} = v_0/5$. This implies the spaceship escapes to infinity (where potential energy is zero) with residual kinetic energy.

Conservation of Mechanical Energy (Initial = Final):

$$ K_i + U_i = K_{\infty} + U_{\infty} $$ $$ \frac{1}{2}m v_0^2 – \frac{GMm}{r} = \frac{1}{2}m \left( \frac{v_0}{5} \right)^2 + 0 $$

Solving for $\frac{GM}{r}$:

$$ \frac{GM}{r} = \frac{1}{2}v_0^2 – \frac{1}{2}\frac{v_0^2}{25} = \frac{1}{2}v_0^2 \left( 1 – \frac{1}{25} \right) $$ $$ \frac{GM}{r} = \frac{1}{2}v_0^2 \left( \frac{24}{25} \right) = \frac{12 v_0^2}{25} \quad \text{…(i)} $$

2. Graphical Interpretation (Acceleration)

At $t=0$, the only force acting on the spaceship is gravity, providing a deceleration $a$. The magnitude of this acceleration is given by the slope of the $v-t$ graph at $t=0$.

$$ a = \frac{GM}{r^2} $$

From the graph’s geometry, the tangent at $t=0$ starts at $v_0$ and intercepts the time axis at $5t_0$. The magnitude of the slope is:

$$ |Slope| = \frac{\Delta v}{\Delta t} = \frac{v_0 – 0}{5t_0 – 0} = \frac{v_0}{5t_0} $$

Equating the physical acceleration to the graphical slope:

$$ \frac{GM}{r^2} = \frac{v_0}{5t_0} \quad \text{…(ii)} $$

3. Solving for Distance $r$

We divide equation (i) by equation (ii):

$$ \frac{GM/r}{GM/r^2} = \frac{12 v_0^2 / 25}{v_0 / 5t_0} $$ $$ r = \frac{12 v_0^2}{25} \times \frac{5t_0}{v_0} $$ $$ r = \frac{12 v_0 t_0}{5} $$