Question

two satellites are going round the earth one at a height 3000 km and the other at 9000 km.The ratio of their angular speed is.

• A. 4/3
• B. 16/9
• C. $\frac{8}{3\sqrt{3}}$
• D. none of these

Answer 1

The correct option is D none of these

The orbital speed $v_o$ of a satellite in a circular orbit of radius $r$ around a mass $M$ is given by

$v_o=\sqrt{\frac{GM}{r}}$

Hence, angular speed ${\omega }_o=v_o/r$ is given by

${\omega }_o=\frac{1}{r}\sqrt{\frac{GM}{r}}$

$\therefore {\omega }_o=\sqrt{\frac{GM}{r^3}}$

Taking the radius of earth to be $R=6400$ km, we can write

$r=R+h$

where $h$ is the height of the orbit.

for the first satellite, $h=3000$ km $⟹r=6400+3000=9400$ km

$\therefore {\omega }_o^1=\sqrt{\frac{GM}{(9400{)}^3}}$

for the second satellite, $h=9000$ km $⟹r=6400+9000=15400$ km

$\therefore {\omega }_o^2=\sqrt{\frac{GM}{(15400{)}^3}}$

$\therefore \frac{{\omega }_0^1}{{\omega }_0^2}=\frac{\sqrt{\frac{GM}{(9400{)}^3}}}{\sqrt{\frac{GM}{(15400{)}^3}}}$

$\therefore \frac{{\omega }_0^1}{{\omega }_0^2}={\left(\frac{15400}{9400}\right)}^{3/2}={\left(\frac{77}{47}\right)}^{3/2}$