Question

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

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

Answer 1

We know that,

Potential energy,

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

Where,

$M_e=$Mass of the earth

$m=$Mass of satellite

$R_e=$ Radius of the earth

$G=$ Gravitational constant

Now,

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

Now, kinetic energy

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

Now,

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

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

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