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

For a satellite moving in an orbit around the earth at height ' $h$ ' the ratio of kinetic energy to potential energy is
  1. $2: 1$
  2. $1: 2$
  3. $1: \sqrt{2}$
  4. $\sqrt{2}: 1$

Solution

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

Asked in: MHT CET 2024 (11 May Shift 2)

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