Question

The orbital velocity $v_o$ of a satellite at height R from the surface in terms of escape velocity $v_e$ from the earth is:

• A. $\frac{v_{e}}{2}$
• B. $\frac{v_{e}}{4}$
• C. $\frac{v_{e}}{\sqrt{2}}$
• D. $v_{e}$
• E. $\sqrt{2}v_{e}$

Hint:

Remember to always measure distance from the center of the earth (Radius + Height) for orbital mechanics.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We need to find a relationship between the orbital velocity of a satellite at a specific altitude and the escape velocity from the surface of the Earth.
Step 2: Key Formula or Approach:
1. Escape Velocity (\( V_e \)) from the surface of the Earth (radius R, mass M): The minimum velocity needed to escape Earth's gravity. \[ V_e = \sqrt{\frac{2GM}{R}} \] 2. Orbital Velocity (\( V_o \)) of a satellite in a circular orbit at a radius r from the center of the Earth: \[ V_o = \sqrt{\frac{GM}{r}} \] Step 3: Detailed Explanation:
The satellite is at a height H=R from the surface. The orbital radius 'r' is the distance from the center of the Earth. \[ r = R + H = R + R = 2R \] Now, let's write the expression for the orbital velocity at this radius: \[ V_o = \sqrt{\frac{GM}{2R}} \] We want to express this in terms of \( V_e = \sqrt{\frac{2GM}{R}} \). From the escape velocity formula, we can write \( GM = \frac{V_e^2 R}{2} \). Let's substitute this into the orbital velocity formula. \[ V_o = \sqrt{\frac{1}{2R} \left( \frac{V_e^2 R}{2} \right)} \] \[ V_o = \sqrt{\frac{V_e^2 R}{4R}} = \sqrt{\frac{V_e^2}{4}} \] \[ V_o = \frac{V_e}{2} \] This result matches the provided correct answer. My initial thought about \(V_e/\sqrt{2}\) was for an orbit close to the surface (r=R). Step 4: Final Answer:
The orbital velocity is \( \frac{V_e}{2} \).