Question

If the escape velocity of a body from the surface of the earth is 11.2 km/s, then the orbital velocity of a satellite in an orbit which is at a height equal to the radius of the earth is

• A. 11.2 km/s
• B. 2.8 km/s
• C. 22.4 km/s
• D. 5.6 km/s

Hint:

Escape velocity: $v_e = \sqrt{\frac{2GM}{R}}$. Orbital velocity: $v_o = \sqrt{\frac{GM}{R+h}}$.

Answer 1

The Correct Option is D

Escape velocity $v_e = \sqrt{\frac{2GM}{R}}$, where $G$ is the gravitational constant, $M$ is the mass of the earth, and $R$ is the radius of the earth. Orbital velocity at a height $h$ is given by $v_o = \sqrt{\frac{GM}{R+h}}$. Given $h=R$, $v_o = \sqrt{\frac{GM}{2R}} = \frac{1}{\sqrt{2}}\sqrt{\frac{2GM}{R}} = \frac{v_e}{\sqrt{2}}$. Since $v_e = 11.2$ km/s, $v_o = \frac{11.2}{\sqrt{2}} = \frac{11.2}{1.414} \approx 5.6$ km/s. $v_0 = \sqrt{\frac{GM}{R+h}}$. Since $h=R$ then $v_0=\sqrt{\frac{GM}{2R}}=\frac{1}{\sqrt{2}}\sqrt{\frac{2GM}{R}} = \frac{v_e}{\sqrt{2}} = \frac{11.2}{1.414} \approx 7.92 \approx 8$, but the closest option given is 5.6. $v_e = \sqrt{2gR}$, and $v_0 = \sqrt{\frac{gR^2}{R+R}} = \sqrt{\frac{gR}{2}}$. $v_e = 11.2 = \sqrt{2gR}$ so $\sqrt{gR}=7.919$. so $v_0=\frac{7.919}{\sqrt{2}}=5.6$.