Question

The magnitude of binding energy of the satellite is $E$ and kinetic energy is $K$ . The ratio $E/K$ is

• A. $2$
• B. $\frac{1}{4}$
• C. $1$
• D. $\frac{1}{2}$

Answer 1

Answer: (C)

Kinetic energy of satellite $(E)$
$KE\left(k\right)=\frac{1}{2}m{v}^2$
$=\frac{1}{2}m{\left(\sqrt{\frac{GM}{R}}\right)}^2$ $\left(\because v=\sqrt{\frac{GM}{R}}\right)$
[ $m=$ mass of satellite]
$=\frac{1}{2}m\frac{GM}{R}$
[Symbols have usual meanings]
Potential energy of satellite,
$U=-m\frac{GM}{R}$
$\therefore$ Binding energy of the satellite,
$E_b=KE+U$
$=\frac{1}{2}m\frac{GM}{R}+\left(-m\frac{GM}{R}\right)$
$=\frac{1}{2}m\frac{GM}{R}-m\frac{GM}{R}$
$=-\frac{GMm}{2R}$
$\therefore$ Magnitude of the binding energy $E=\left|E_b\right|=\frac{GMm}{2R}$
The ratio of $\frac{E}{K}=\frac{GMm}{2R}\times \frac{2R}{GMm}=1$