Question

A planet with a radius of 6.00 × 107 m has a gravitational field of magnitude 46.0 m/s2 at the surface. What is the escape speed from the planet?

Answer 1

In this case, if you equal the centripetal force to the force due to the magnitude of the gravitational field, you obtain:

$\frac{1}{2}m\frac{v^2}{R}=mg^{\prime}$

By solving for v into the previous equation, you get:

$v=\sqrt[]{2Rg^{\prime}}$

where,

R: radius of the planet = 6.00*10^7 m

g; gravitational acceleration constant = 46.0 m/s^2

v: escape speed.

Replace the values of the parameters into the formula for v:

$\begin{gathered} v=\sqrt[]{2(6.00\cdot10^7m)(46.0\frac{m}{s^2})} \\ v\approx74,296.7\frac{m}{s} \end{gathered}$

Hence, the escape speed is approximately 74,296.7 m/s