Question

A body is projected from point $\text{A}$, that is at a distance $4R$ from the centre of the earth with speed $v_1$, in a direction making angle ${30}^{\circ }$ with the extension of line joining centre of the earth and point $\text{A}$. Find the speed $v_1\left(\text{in m/s}\right)$ of the body if the body passes earth by grazing its surface.
$[$ Neglect air resistance and rotation of earth, and use $\frac{GM}{R}=6.4\times {10}^7{\text{m}}^2/{\text{s}}^2$ for calculation. $]$

• A. $800\sqrt{2}$
• B. $\frac{8000}{\sqrt{2}}$
• C. $800$
• D. None of these

Answer 1

The correct option is B $\frac{8000}{\sqrt{2}}$

The body is moving under influence of gravitational force, which is a central force.

Thus, ${\tau }_{ext}=0$ about centre of earth on system of earth and body.


$\therefore L_i=L_f$

$L_i$ and $L_f$ be the angular momentum of body at the two situations shown.

Thus, $m\left(v_1\sin {30}^{\circ }\right)\left(4R\right)=m{v}_{2}R$

$\Rightarrow m{v}_1\left(1/2\right)4R=m{v}_{2}R$

$\Rightarrow \frac{v_2}{v_1}=2............\left(1\right)$

As we know that gravitational field is conservative, so we can apply conservation of mechanical energy on this system.

So, applying mechanical energy conservation

$U_1+K_1=U_2+K_2$

$\Rightarrow -\frac{GMm}{4R}+\frac{1}{2}m{v}_1^2=-\frac{GMm}{R}+\frac{1}{2}m{v}_2^2$

$($ here, $4R\&R$ be the distance between centre of earth and body in two shown situations $)$

$\Rightarrow \frac{1}{2}(v_2^2-v_1^2)=\frac{3}{4}\frac{GM}{R}$

Using eq. $\left(1\right)$, we get

$\Rightarrow \frac{3}{2}{v}_1^2=\frac{3}{4}\frac{GM}{R}$

$\Rightarrow v_1^2=\frac{1}{2}\frac{GM}{R}=\frac{1}{2}\times (6.4\times {10}^7)$

$\Rightarrow v_1=\frac{8000}{\sqrt{2}}\text{m/s}$

Hence, option $\left(b\right)$ is the correct choice.

Why this question ? Tip: In a problem involving motion of a satellite or body around earth, always remember that "conservation of angular momentum and mechanical energy of system at the tools that must be sought after."