Question

A projectile of mass $m$ is fired from the surface of the earth at an angle $\alpha = 6 0^{^\circ }$ from the vertical. The initial speed $v_{0}$ is equal to $\sqrt{\frac{\textit{GM}_{\text{e}}}{\textit{R}_{\text{e}}}}$ . The maximum height that projectile can rise from surface of earth is $\frac{\textit{R}_{\text{e}}}{\textit{x}}$ . Find the value of $x$ ?
$G$ = Universal Gravitational constant
$M_{e}$ = Mass of earth
$R_{e}$ = Radius of earth

• A. 2
• B. 4
• C. 6
• D. 8

Answer 1

Answer: (A)

By conservation of angular momentum about C at position P and F
$\text{mv}_{0} \frac{\sqrt{3}}{2} \text{R}_{\text{e}} = \text{mv} r_{\text{max}}$
$⇒ \, \, \, \text{v} = \frac{\sqrt{3} \text{R}_{\text{e}} \text{v}_{0}}{2 r_{\text{max}}}$ ...(i)
By conservation of mechanical energy between P and F
$\frac{1}{2} \text{mv}_{0}^{2} - \frac{\text{GM}_{\text{e}} \text{m}}{\text{R}_{\text{e}}} = \frac{1}{2} \text{mv}^{2} - \frac{\text{GM}_{\text{e}} \text{m}}{\text{r}_{\text{max}}}$ ...(ii)
Put (i) in (ii)
$\frac{1}{2} \text{mv}_{0}^{2} - \frac{\text{GM}_{\text{e}} \text{m}}{\text{R}_{\text{e}}} = \frac{1}{2} \text{m} \frac{3 \text{R}^{2} \text{v}_{0}^{2}}{4 r_{\text{max}}^{2}} - \frac{\text{GM}_{\text{e}} \text{m}}{\text{r}_{\text{max}}}$
$\text{v}_{0} = \sqrt{\frac{\text{GM}_{\text{e}}}{\text{R}}}$
$⇒ \, \, \, \frac{1}{2} \text{m} \frac{\text{GM}_{\text{e}}}{\text{R}_{\text{e}}} - \frac{\text{GM}_{\text{e}} \text{m}}{\text{R}_{\text{e}}} = \frac{3}{8} \frac{\text{mR}_{\text{e}}^{2}}{\text{r}_{\text{max}}^{2}} \frac{\text{GM}_{\text{e}}}{\text{R}} - \frac{\text{GM}_{\text{e}} \text{m}}{\text{r}_{\text{max}}}$
$⇒ \, \, \, \frac{1}{2 \text{R}_{\text{e}}} - \frac{1}{\text{R}_{\text{e}}} = \frac{3 \text{R}}{8 r_{\text{max}}^{2}} - \frac{1}{\text{r}_{\text{max}}}$
$⇒ \, \, \, \frac{- 1}{2 \text{R}_{\text{e}}} = \frac{3 \text{R}_{\text{e}} - 8 r_{\text{max}}}{8 r_{\text{max}}^{2}}$
$⇒ \, \, \, 4 \text{r}_{\text{max}}^{2} - 8 \text{R}_{\text{e}} r_{\text{max}} + 3 \text{R}_{\text{e}}^{2} = 0$
Solving the quadratic in r_max
we get
$\text{r}_{\text{max}} = \frac{8 \text{R}_{\text{e}} \pm \sqrt{1 6} \text{R}_{\text{e}}}{8} = \frac{8 \text{R}_{\text{e}} \pm 4 \text{R}_{\text{e}}}{8}$
$⇒ \, \, \, \text{r}_{\text{max}} = \frac{8 \text{R}_{\text{e}} + 4 \text{R}_{\text{e}}}{8} = \frac{1 2 \text{R}_{\text{e}}}{8} = \frac{3 \text{R}_{\text{e}}}{2}$
$∴ \, \, \, \text{h}_{\text{max}} = \text{r}_{\text{max}} - \text{R}_{\text{e}} = \frac{3 \text{R}_{\text{e}}}{2} - \text{R}_{\text{e}} = \frac{\text{R}_{\text{e}}}{2}$
$∴ \, \, \, \text{x} = 2$