Question

A satellite is put in an orbit just above the earth's surface with a velocity $\sqrt{1.5}$ time the velocity for a circular orbit. The maximum distance of the satellite from the surface of the earth is $nR$ where $R$ is radius of earth. Find $n$

Answer 1

Let nearest point be $P$ and farthest point be $Q$.


Velocities at point $P$ and $Q$ are perpendicular to their distance $r_1\text{and}{r}_2$

Given, $v_1=\sqrt{1.5}{v}_{orb}=\sqrt{1.5gR}\text{and}(r_1=R)$

$\because g=\frac{GM}{R^2}\Rightarrow GM=g{R}^2$

$\therefore$ conservation of angular momentum gives,

$m{v}_1{r}_1=m{v}_2{r}_2$...(1)

$\sqrt{1.5gR}R=v_2{r}_2$

$\therefore v_2=\frac{\sqrt{1.5gR}R}{r_2}$
Using conservation energy,

$\frac{1}{2}m{v}_1^2-\frac{GMm}{r_1}=\frac{1}{2}m{v}_2^2-\frac{GMm}{r_2}...\left(2\right)$

Substituting the values of $r_1,v_1,v_2$ and $GM$ in eq (2) and solving we get,

$r_2^2-4r_{2}R+3R^2=0$ .......(3)

Solving eq (3) we get,

$r_2=R$ and $r_2=3R$

So maximum distance of satellite from the center of earth is $3R$

$\therefore$ Maximum distance of the satellite from the surface of earth will be,

$d=3R-R=2R$

$\therefore n=2$