Question

A satellite of mass $m$ is in an elliptical orbit around the earth. The speed of the satellite at its nearest position is $\left(6GM\right)/\left(5r\right)$ where $r$ is the perigee (nearest point) distance from the centreof the earth. It is desired to transfer the satellite to the circular orbit of radius equal to its apogee (farthest point) distance from the centre of the earth. The change in orbital speed required for this purpose is

• A. $0.35\sqrt{\frac{G{M}_e}{r}}$
• B. $0.075\sqrt{\frac{G{M}_e}{r}}$
• C. $\sqrt{\frac{2G{M}_e}{r}}$
• D. zero

Answer 1

The correct option is B $0.075\sqrt{\frac{G{M}_e}{r}}$

The speed has to be square root of the given value

Given the Speed at the perigee of $ellipticalorbit$ = $\sqrt{\frac{6GM}{5r}}$

Radius at perigee = $r$

Let radius at apogee be $R$

By using conservation of angular momentum we get velocity at apogee to be $\sqrt{\frac{6GM}{5R}}$

Velocity of the new circular orbit would be $\sqrt{\frac{GM}{R}}$

Now we need to find radius $R$

Total energy of the elliptical orbit at apogee is $\frac{-2GMm}{5R}$ which should be equal to$\frac{-GMm}{2a}$

where $2a$ = $r$ + $R$

From the above equation $R$ = $\frac{3r}{2}$

$\Delta v$ = $\sqrt{\frac{6GM}{5R}}$ - $\sqrt{\frac{GM}{R}}$

So, $\Delta v$ = $0.095\sqrt{\frac{GM}{R}}$
Now putting $R$ $=\frac{3r}{2}$

We get $\Delta v$ $=$ $0.095\sqrt{\frac{2GM}{3r}}$ $=0.075\sqrt{\frac{GM}{r}}$
So, the correct option is $\left(B\right)$