Question

The ratio of kinetic energy to potential energy for a satellite moving in an orbit around the earth is.

• A. $\frac{1}{\sqrt{2}}$
• B. $2$
• C. $\frac{1}{2}$
• D. $\sqrt{2}$

Answer 1

The correct option is C

$\frac{1}{2}$

Step 1: We know that potential energy,

$U=-\frac{G{M}_{e}m}{R_e}$

Here, $M_e$ is the mass of the earth.

$m$ is the mass of the satellite.

$R_e$is the radius of the earth.

$G$ is the gravitational constant.

Step 2: Calculating the ratio of kinetic energy to potential energy for a satellite moving in an orbit around the earth.

$\left|U\right|=\frac{G{M}_{e}m}{R_e}$

Now, kinetic energy

$K=\frac{1}{2}\frac{G{M}_{e}m}{R_e}$

Then,

$\frac{K}{\left|U\right|}=\frac{1}{2}\frac{G{M}_{e}m}{R_e}\times \frac{R_e}{G{M}_{e}m}$

Step 3: Final answer

$\frac{K}{\left|U\right|}=\frac{1}{2}$

Hence, the ratio of the kinetic energy to potential energy is $\frac{1}{2}$

Therefore, the correct option is (c).