A body is projected in vertically upward direction from the surface of the earth of radius ' $R$ ' into…

A body is projected in vertically upward direction from the surface of the earth of radius ' $R$ ' into space with velocity ' $n V_{\mathrm{c}}$ ' $(n \lt 1)$. The maximum height from the surface of earth to which a body can reach is
  1. $\frac{n^2 R}{\left(1-n^2\right)}$
  2. $\frac{\mathrm{n}^2 \mathrm{R}^2}{(1-\mathrm{n})}$
  3. $\frac{\mathrm{nR}^2}{\left(1+\mathrm{n}^2\right)}$
  4. $\frac{n^2 R^2}{(1+n)}$

Solution

$\begin{aligned} & \begin{array}{l}\text { As per the law of conservation of energy, } \\ \\ (K . E+P . E)_{\text {surface }}=(K . E+P . E)_{\text {max height }}\end{array} \\ \therefore \quad & \frac{-G M m}{R}+\frac{1}{2} m\left(n V_e\right)^2=-\frac{G M m}{R+h}+\frac{1}{2} m(0)^2 \\ \therefore \quad & \frac{1}{2} m\left(n V_e\right)^2=G M m\left(\frac{1}{R}-\frac{1}{R+h}\right) \\ \therefore \quad & \left(n V_e\right)^2=2 G M\left(\frac{R+h-R}{R(R+h)}\right) \\ \therefore \quad & \left(n V_e\right)^2=V_e^2\left(\frac{h}{(R+h)}\right)\end{aligned}$ $\begin{array}{ll}\therefore & \frac{1}{n^2}=\frac{R}{h}+1 \\ \therefore & \frac{R}{h}=\frac{1-n^2}{n^2} \\ \therefore & h=\frac{n^2 R}{\left(1-n^2\right)}\end{array}$

Asked in: MHT CET 2024 (11 May Shift 2)

Practice more Gravitation questions on Aicharya