Question

A projectile of mass $m$ is fired from the surface of the earth at an angle $\theta ={60}^{\circ }$ with the vertical. The initial speed $v_o$ is equal to $\sqrt{\frac{G{M}_e}{R_e}}$. How high does the projectile rise$?$ Neglect air resistance and the earth's rotation.

• A. $\frac{Re}{2}$
• B. $\frac{Re}{5}$
• C. $\frac{Re}{4}$
• D. $\frac{Re}{8}$

Answer 1

The correct option is A $\frac{Re}{2}$


Let $v$ be the speed of the projectile at the highest point and $r_{max}$ its distance from the center of the Earth.

Applying conservation of angular momentum about the center of the Earth.

$m{v}_o{R}_e\sin {60}^{\circ }=mv{r}_{max}\sin {90}^{\circ }$

Since, at the farthest point velocity of the projectile will be perpendicular to the line joining it to the center of the earth.

$\Rightarrow v_o{R}_e\frac{\sqrt{3}}{2}=v{r}_{max}$

$\Rightarrow v=\frac{\sqrt{3}{v}_o{R}_e}{2r_{max}}...\left(1\right)$

Conserving mechanical energy between the initial point of projectile and the highest point.

$-\frac{G{M}_{e}m}{R_e}+\frac{1}{2}m{v}_0^2=-\frac{G{M}_{e}m}{r_{max}}+\frac{1}{2}m{v}^2$

As, $v_0=\sqrt{\frac{G{M}_e}{R_e}}$

$\Rightarrow -\frac{G{M}_{e}m}{R_e}+\frac{1}{2}m\frac{GM}{R_e}=-\frac{G{M}_{e}m}{r_{max}}+\frac{1}{2}m{v}^2$

From equation $\left(1\right)$,

$\Rightarrow -\frac{G{M}_e}{2R_e}=-\frac{G{M}_e}{r_{max}}+\frac{1}{2}\times \frac{3}{4}\frac{R_e^2{v}_0^2}{r_{max}^2}$

Substituting the value of $v_0$,

$\Rightarrow -\frac{G{M}_e}{2R_e}=-\frac{G{M}_e}{r_{max}}+\frac{1}{2}\times \frac{3}{4}\frac{R_e^2}{r_{max}^2}\times \frac{G{M}_e}{R_e}$

$\Rightarrow -\frac{1}{2R_e}=-\frac{1}{r_{max}}+\frac{3R_e}{8r_{max}^2}$

$\Rightarrow 4r_{max}^2-8R_e{r}_{max}+3R_e^2=0$

$\Rightarrow 4r_{max}^2-6R_e{r}_{max}-2R_e{r}_{max}+3R_e^2=0$

$\Rightarrow (2r_{max}-3R_e)(2r_{max}-R_e)=0$

Solving the quadratic equation $r_{max},$ we get

$r_{max}=\frac{R_e}{2},\frac{3R_e}{2}$

Neglecting $\frac{R_e}{2}$ as $r_{max}>R_e$

$\therefore$ Maximum height =$\frac{3R_e}{2}-R_e=\frac{R_e}{2}$

Hence, option (a) is the corrected answer.

Why this question? Note - The velocity of the particle at the closest or the farthest point is perpendicular to the line joining the particle and the other mass. This is because at these points the velocity of approach or separation will become zero.