Question

What is the minimum velocity needed for a spaceship to escape the Earth's gravitational force?

• A. $5.4{\text{km s}}^{-1}$
• B. $8.6{\text{km s}}^{-1}$
• C. $11.2{\text{km s}}^{-1}$
• D. $9.8{\text{km s}}^{-1}$

Answer 1

The correct option is C $11.2{\text{km s}}^{-1}$

The escape velocity needed for a object to leave the earth's surface overcoming the gravitational pull, is called escape velocity.
It is given by: ${\text{v}}_{\text{esc}}=\sqrt{\frac{2\text{GM}}{\text{R}}}$, where:

• $\text{G}=6.67\times {10}^{-11}{\text{Nm}}^2{\text{kg}}^{-2}$ is the universal gravitational constant
• $\text{M}=6\times {10}^{24}\text{kg}$ is the mass of the Earth
• $\text{R}\approx 6400\text{km}$ is the radius of the Earth

$⟹{\text{v}}_{\text{esc}}=\sqrt{\frac{2\times 6.67\times {10}^{-11}\times 6\times {10}^{24}}{6400\times {10}^3}}$
$⟹11.2{\text{km s}}^{-1}$