Question:

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

Updated On: May 12, 2022
  • $1$
  • $1/2$
  • $2/1$
  • $1/4$
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The Correct Option is A

Solution and Explanation

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 $E / K=\frac{G M m}{2 R} \times \frac{2 R}{G M m}=1$
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Concepts Used:

Energy In Simple Harmonic Motion

We can note there involves a continuous interchange of potential and kinetic energy in a simple harmonic motion. The system that performs simple harmonic motion is called the harmonic oscillator.

Case 1: When the potential energy is zero, and the kinetic energy is a maximum at the equilibrium point where maximum displacement takes place.

Case 2: When the potential energy is maximum, and the kinetic energy is zero, at a maximum displacement point from the equilibrium point.

Case 3: The motion of the oscillating body has different values of potential and kinetic energy at other points.