Question

A small body of mass 'm' is projected up from the surface of given planet as shown in the figure. If small body has velocity which is just sufficient to get out from the gravity of planet then time taken by small body to reach at height of 3R from the surface of planet is $K\sqrt{\frac{2R^3}{GM}}$.

Find the value of $K$.

Answer 1

7/3

Answer 2

The initial velocity of the small body is the escape velocity from the surface of the planet, which is $v_0 = \sqrt{\frac{2GM}{R}}$.

Let $r$ be the distance of the body from the center of the planet at time $t$. The initial position is $r_0 = R$. The final position is $r_f = R + 3R = 4R$.

By conservation of energy, the total energy of the body is constant.

Initial total energy $E_i = \frac{1}{2} m v_0^2 - \frac{GMm}{R} = \frac{1}{2} m \left(\sqrt{\frac{2GM}{R}}\right)^2 - \frac{GMm}{R} = \frac{1}{2} m \frac{2GM}{R} - \frac{GMm}{R} = \frac{GMm}{R} - \frac{GMm}{R} = 0$.

At any distance $r$ from the center, the velocity is $v = \frac{dr}{dt}$. The total energy is $E(r) = \frac{1}{2} m v^2 - \frac{GMm}{r}$.

Since energy is conserved, $E(r) = E_i = 0$.

$\frac{1}{2} m v^2 - \frac{GMm}{r} = 0$

$\frac{1}{2} m v^2 = \frac{GMm}{r}$

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

Since the body is moving upwards, $v = \frac{dr}{dt} > 0$.

$\frac{dr}{dt} = \sqrt{\frac{2GM}{r}}$

To find the time taken to reach $r_f = 4R$ from $r_0 = R$, we integrate:

$dt = \frac{dr}{\sqrt{\frac{2GM}{r}}} = \sqrt{\frac{r}{2GM}} dr$

$t = \int_{R}^{4R} \sqrt{\frac{r}{2GM}} dr = \frac{1}{\sqrt{2GM}} \int_{R}^{4R} r^{1/2} dr$

Evaluating the integral:

$\int r^{1/2} dr = \frac{r^{1/2+1}}{1/2+1} = \frac{r^{3/2}}{3/2} = \frac{2}{3} r^{3/2}$

$t = \frac{1}{\sqrt{2GM}} \left[ \frac{2}{3} r^{3/2} \right]_{R}^{4R} = \frac{1}{\sqrt{2GM}} \left( \frac{2}{3} (4R)^{3/2} - \frac{2}{3} R^{3/2} \right)$

$t = \frac{2}{3\sqrt{2GM}} \left( (4^{3/2} R^{3/2}) - R^{3/2} \right) = \frac{2}{3\sqrt{2GM}} \left( (2^2)^{3/2} R^{3/2} - R^{3/2} \right)$

$t = \frac{2}{3\sqrt{2GM}} \left( 2^3 R^{3/2} - R^{3/2} \right) = \frac{2}{3\sqrt{2GM}} \left( 8 R^{3/2} - R^{3/2} \right)$

$t = \frac{2}{3\sqrt{2GM}} (7 R^{3/2}) = \frac{14}{3\sqrt{2GM}} R^{3/2}$

We are given that the time taken is $K\sqrt{\frac{2R^3}{GM}}$. We need to express our result in this form.

$t = \frac{14}{3} \frac{R^{3/2}}{\sqrt{2GM}} = \frac{14}{3} \frac{\sqrt{R^3}}{\sqrt{2GM}} = \frac{14}{3} \sqrt{\frac{R^3}{2GM}}$

We want the term $\frac{2R^3}{GM}$ inside the square root.

$\sqrt{\frac{R^3}{2GM}} = \sqrt{\frac{1}{4} \frac{4R^3}{2GM}} = \sqrt{\frac{1}{4} \frac{2 \cdot 2R^3}{2GM}} = \sqrt{\frac{1}{4}} \sqrt{\frac{2R^3}{GM}} = \frac{1}{2} \sqrt{\frac{2R^3}{GM}}$

So, $t = \frac{14}{3} \times \frac{1}{2} \sqrt{\frac{2R^3}{GM}} = \frac{7}{3} \sqrt{\frac{2R^3}{GM}}$.

Comparing this with the given form $K\sqrt{\frac{2R^3}{GM}}$, we find $K = \frac{7}{3}$.