Question

Escape velocity of a satellite of the earth at an altitude equal to radius of the earth is $v$. What will be the escape velocity at an altitude equal to $7R$, where $r$ = radius of the earth?

• A. $\frac{v}{4}$
• B. $\frac{v}{2}$
• C. $8v$
• D. $4v$

Answer 1

The correct option is C $8v$

To escape Earth's gravitational pull we have to a velocity such that kinetic energy is greater or equal to your potential energy within Earth's gravitational field. $\frac{Gm{M}_E}{r}=\frac{1}{2}{mV}^2$

$\frac{2{GM}_E}{r}=V^2\Rightarrow V=\sqrt{\frac{2{GM}_E}{r}}$

Gravitational constant $G=6.674\times {10}^{-11}\frac{m^3}{{kgs}^2}$

Mass of earth $M_E=5.972\times {10}^{24}kg$

Radius of Earth $R_E=6.371\times {10}^{6}m$

At an altitude equal to radius of Earth we have $r=2R_E$

$V=\sqrt{\frac{{2GM}_E}{2R_E}}=7910m/s$

For $r=8R_{E}V=3955m/s$