Question

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

• A. $1$
• B. $1/2$
• C. $2/1$
• D. $1/4$

Answer 1

Answer: (A)

$1$

Answer 2

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