Question 277: Gravitation

Question

For a satellite moving in an orbit around the earth at height ' $h$ ' the ratio of kinetic energy to potential energy is

Accepted Answer

Let $\mathrm{M}_{\mathrm{e}}=$ Mass of earth
$\mathrm{m}=$ Mass of satellite
$\mathrm{R}_{\mathrm{e}}=$ Radius of earth
$\mathrm{G}=$ Gravitational constant
$	herefore \quad$ Potential Energy,
$$\mathrm{U}=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{~m}}{\mathrm{R}_{\mathrm{e}}}$$
Kinetic Energy,
$$\begin{aligned}
& \mathrm{K}=\frac{1}{2} \frac{\mathrm{GM}_{\mathrm{e}} \mathrm{~m}}{\mathrm{R}_{\mathrm{e}}} \
& \frac{\mathrm{~K}}{\mathrm{U}}=\frac{\frac{1}{2} \frac{\mathrm{GM} \mathrm{M}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}{\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{~m}}{R_{\mathrm{e}}}} \
& \frac{\mathrm{~K}}{\mathrm{U}}=\frac{1}{2}
\end{aligned}$$

Answer

$2: 1$

Answer

$1: 2$

Answer

$1: \sqrt{2}$

Answer

$\sqrt{2}: 1$

Gravitation

For a satellite moving in an orbit around the earth at height ' h ' the ratio of kinetic energy to potential energy is

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