Question

The total energy of an artificial satellite of mass \(m\) revolving in a circular orbit around the earth with a speed \(v\) is

• A. \(\frac{1}{2}mv^2\)
• B. \(\frac{1}{4}mv^2\)
• C. \(-\frac{1}{4}mv^2\)
• D. \(-mv^2\)
• E. \(-\frac{1}{2}mv^2\)

Hint:

In satellite dynamics, remember the "1:2:1" ratio rule for magnitudes: \(|Total Energy| = Kinetic Energy = \frac{1}{2}|Potential Energy|\). The Total Energy is always negative for a bound system!

Answer 1

Answer: (E)

Concept: For a satellite in a stable circular orbit, the total energy (\(E\)) is the sum of its Kinetic Energy (\(K\)) and Potential Energy (\(V\)).
• Kinetic Energy: \(K = \frac{1}{2}mv^2\)
• Potential Energy: \(V = -\frac{GMm}{r}\)
• Orbital Velocity: \(v = \sqrt{\frac{GM}{r}} \Rightarrow v^2 = \frac{GM}{r}\)

Step 1: Relate Potential Energy to Kinetic Energy.
From the orbital velocity formula, we know \(GMm/r = mv^2\). Substituting this into the Potential Energy formula: \[ V = -mv^2 \]

Step 2: Calculate Total Energy.
\[ E = K + V \] \[ E = \frac{1}{2}mv^2 + (-mv^2) \] \[ E = -\frac{1}{2}mv^2 \]