Question

If 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 of the satellite above the surface of the earth is

• A. $2R$
• B. $\frac{R}{2}$
• C. $R$
• D. $\frac{R}{4}$

Answer 1

The correct option is C $R$

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

$\frac{1}{4}\times \frac{2GM}{R}=\frac{GM}{R+h}$

$\frac{1}{2}\times \frac{1}{R}=\frac{1}{R+h}$

$2R=R+h$

or $h=R$