Question

An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of escape velocity from the earth. The height of satellite above the surface of earth is (where R = radius of earth)

• A. R
• B. 2R
• C. 3R
• D. 4R

Answer 1

Answer: (A)

R

Answer 2

$\frac{1}{2}$ $\sqrt{\frac{2GM}{R}}$ = $\sqrt{\frac{GM}{R + h}}$

$\frac{1}{2}$ × $\frac{2GM}{R}$ = $\frac{GM}{R + h}$

2R = R + h

or h = R