Question

An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of the escape velocity from the earth. The height \(h\) of the satellite above the earth’s surface is (Take radius of earth as \(R_e\)):

• A. \(h = R_e^2\)
• B. \(h = R_e\)
• C. \(h = 2R_e\)
• D. \(h = 4R_e\)

Hint:

For circular orbits, remember: \[ v_{\text{orbital}} = \sqrt{\frac{GM}{r}}, \quad v_{\text{escape}} = \sqrt{\frac{2GM}{R}} \] Always distinguish between orbital radius and height above the surface.

Answer 1

The Correct Option is B

Step 1: Escape velocity from the surface of the earth is \[ v_e = \sqrt{\frac{2GM}{R_e}} \]
Step 2: Given speed of the satellite is half of escape velocity: \[ v = \frac{1}{2}v_e = \frac{1}{2}\sqrt{\frac{2GM}{R_e}} = \sqrt{\frac{GM}{2R_e}} \]
Step 3: Orbital speed of a satellite in a circular orbit of radius \(r\) is: \[ v = \sqrt{\frac{GM}{r}} \]
Step 4: Equating the two expressions for velocity: \[ \sqrt{\frac{GM}{r}} = \sqrt{\frac{GM}{2R_e}} \] \[ \Rightarrow r = 2R_e \]
Step 5: Height of the satellite above earth’s surface: \[ h = r - R_e = 2R_e - R_e = R_e \]