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 Rₑ):

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

Hint:

Always remember: • Orbital velocity depends on the radius of orbit. • Escape velocity depends only on the radius of the planet. • Height of satellite = orbital radius - earth’s radius.

Answer 1

The Correct Option is B

Step 1: Velocity of a satellite in a circular orbit of radius r is v = √((GM)/(r)) Escape velocity from the earth’s surface is vₑ = √((2GM)/(Rₑ))
Step 2: Given that v = (vₑ)/(2) √((GM)/(r)) = (1)/(2)√((2GM)/(Rₑ))
Step 3: Squaring both sides, (GM)/(r) = (GM)/(2Rₑ) ⟹ r = 2Rₑ
Step 4: Height of the satellite above earth’s surface is h = r - Rₑ = 2Rₑ - Rₑ = Rₑ