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:

`(1)/(2)m(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(m)/(s^2))} \\ v\approx74,296.7(m)/(s) \end{gathered}`

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