Question

Two satellites P and Q of same mass are revolving near the earth surface in the equatorial plane. The satellite P moves in the direction of rotation of earth whereas Q moves in the opposite direction. The ratio of their kinetic energies with respect to the frame attached to earth will be

• A. ${\left(\frac{8363}{7437}\right)}^2$
• B. ${\left(\frac{7437}{8363}\right)}^2$
• C. $\frac{8363}{7437}$
• D. $\frac{7437}{8363}$

Answer 1

The correct option is A ${\left(\frac{8363}{7437}\right)}^2$

Force balance ,

$\frac{GmM}{r^2}=\frac{m{v}^2}{r}$

this gives velocity of satellite near earth's surface,

V = $(gR{)}^{0.5}=(9.8\times 6.4\times {10}^6{)}^{0.5}=7.92\times {10}^{3}m/s$

${\omega }_e=\frac{2\pi }{24\times 60\times 60}$

$V_e={\omega }_e\times r$

=> $V_e=\frac{2\pi 6.4\times {10}^6}{24\times 60\times 60}=.465\times {10}^3$
Relative velocities as seen from earth , $V_1=(7.92+.465)\times {10}^3=8.385\times {10}^3,V_2=(7.92-.465)\times {10}^3=7.455\times {10}^3$

Ratio of kinetic energies =$(\frac{V_1}{V_2}{)}^2=(\frac{8.385\times {10}^3}{7.455\times {10}^3}{)}^2=(\frac{8385}{7455}{)}^2$