Question

A satellite is orbiting around earth in a circular orbit of radius $r$. A particle of mass $m$ is projected from satellite in forward direction with velocity $v=\sqrt{\frac{2}{3}}$ times orbital velocity (this velocity is given with respect to earth). During subsequent motion of the particle, its minimum distance from the centre of earth is

• A. $\frac{r}{2}$
  
• B. $\frac{r}{3}$
• C. $\frac{2r}{3}$
• D. $\frac{4r}{5}$

Answer 1

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


As the velocity of particle is less than orbital velocity of satellite, the particle goes in elliptical orbit of semi-major axis less than $r$.

Let $r_1$ be the minimum distance and $v_1$ be velocity of particle at this position.
Then from the conservation of angular momentum,

$m_0\times \sqrt{\frac{2}{3}}{v}_0\times r=m_0{v}_1{r}_1$

where $m_0$ is the mass of particle and $v_0$ is orbital speed equal to $\sqrt{\frac{GM}{r}}$

$\Rightarrow v_1{r}_1=\sqrt{\frac{2}{3}}{v}_{0}r...\left(1\right)$

From energy conservation, we have

$\frac{1}{2}{m}_0{\left(\sqrt{\frac{2}{3}}{v}_0\right)}^2-\frac{GM{m}_0}{r}=\frac{m_0{v}_1^2}{2}-\frac{GM{m}_0}{r_1}$

Using equation $\left(1\right)$ and $v_0=\sqrt{\frac{GM}{r}}$,

$\Rightarrow \frac{1}{3}{m}_0{v}_0^2-\frac{v_0^{2}r{m}_0}{r}=\frac{m_0{\left(\sqrt{\frac{2}{3}}\frac{v_{0}r}{r_1}\right)}^2}{2}-\frac{r{v}_0^2{m}_0}{r_1}$

$\Rightarrow \frac{1}{3}-1=\frac{r^2}{3r_1^2}-\frac{r}{r_1}$

$\Rightarrow \frac{r^2}{3r_1^2}-\frac{r}{r_1}+\frac{2}{3}=0$

$\Rightarrow r^2-3r_{1}r+2r_1^2=0$

$\Rightarrow r(r-2r_1)-r_1(r-2r_1)=0$

Solving above equations, we get

$r_1=\frac{r}{2}$ and $r_1=0$

So, the acceptable value is, $r_1=\frac{r}{2}$

Hence, option (a) is the correct answer.