Question

A rocket is launched with velocity of $10km/\sec$ . If radius of Earth is R then maximum height attained by it will be:
A.2R
B.3R
C.4R
D.5R

Answer 1

Use the Law of conservation of Energy. At the maximum height, the potential energy will be equal to the kinetic energy of the rocket.

Complete Step by Step Answer:
We are given that the velocity of the rocket, v= $10km/\sec$ = $10000m/\sec$, radius of Earth=R. According to the law of conservation of energy, we have, total energy of rocket at surface of Earth=total energy of rocket at maximum height. At surface of Earth we have $K.{E_{surface}} = \dfrac{{m{v^2}}}{2}$ and $P.E{._{surface}} = \dfrac{{ - GMm}}{R}$ . At maximum height h, the rocket has velocity=0. So we have $K.{E_{\max }} = \dfrac{{m{v^2}}}{2} = 0$ and $P.E{._{\max }} = \dfrac{{ - GMm}}{x}$ , where x is the distance of rocket from centre of the Earth, i.e. $x = R + h$ .
Applying the law of conservation of energy ,
$\Rightarrow K.{E_{surface}} + P.E{._{surface}} = K.E{._{\max }} + P.{E_{\max }} \\\ \Rightarrow \dfrac{{m{v^2}}}{2} + ( - \dfrac{{GMm}}{R}) = 0 + ( - \dfrac{{GMm}}{{R + h}}) \\\$
For simplification we multiply R in both numerator and denominator of the term $( - \dfrac{{GM}}{R})$ and similarly ${R^2}$in numerator and denominator of the term $( - \dfrac{{GM}}{x})$ . Substituting values we get
$\Rightarrow \dfrac{{{v^2}}}{2} + [( - \dfrac{{GM}}{{{R^2}}}) \times R] = 0 + [( - \dfrac{{GM}}{{{R^2}x}}) \times {R^2}]$
$\Rightarrow \dfrac{{{{(10)}^2}}}{2} + [( - g) \times R] = ( - \dfrac{g}{x}) \times {R^2}$ , as we know $g = \dfrac{{GM}}{{{R^2}}}$ = $9.8 \times {10^{ - 3}}km/{\sec ^2}$
$\Rightarrow \dfrac{{{{(10)}^2}}}{2} + [( - 9.8 \times {10^{ - 3}}) \times R] = ( - \dfrac{{9.8 \times {{10}^{ - 3}}}}{x}) \times {R^2}$
On solving the above equation we get $x \approx 4R$ , where $R \approx 6371km$ .

Note: Here approximations play an important role. Keep in mind to approximate values to simplify equations. Make sure that all values have similar types of units.