For a satellite in an orbit around the earth, the ratio of kinetic energy to potential energy is:

Question

Aicharya · Accepted Answer

$-\frac{\mathrm{GMm}}{R^2}+m \omega^2 R=0$ $\begin{aligned} 	herefore \quad \frac{\mathrm{GMm}}{R^2} & =m \omega^2 R \ 	ext { K.E. }=\frac{1}{2} I \omega^2 & =\frac{1}{2} m R^2 \omega^2 \ & =\frac{\mathrm{GMm}}{2 R} \end{aligned}$ $\begin{aligned} & P . E .=-\frac{\mathrm{GMm}}{R} 	herefore K . E=\frac{|P \cdot E|}{2} \ & 	herefore \quad \frac{K \cdot E}{P \cdot E}=\frac{1}{2} \end{aligned}$ .

For a satellite in an orbit around the earth, the ratio of kinetic energy to potential energy is:

Solution