Question

A planet with a radius of 6.00 × 107 m has a gravitational field of magnitude 42.9 m/s2 at the surface. What is the escape speed from the planet?

Answer 1

We are asked to determine the escape velocity of a planet of radius 6x10^7 m.

to do that we will use the fact that the escape velocity of a planet is given by the following formula:

`v_e=\sqrt[]{2gR}`

Where:

`\begin{gathered} v_e=\text{ escape velocity, }\lbrack(m)/(s)\rbrack \\ g=\text{ magnitude of gravitational field,}\lbrack(m)/(s^2)\rbrack \\ R=\text{ radius of the planet, }\lbrack m\rbrack \end{gathered}`

Now, we substitute the values:

`v_e=\sqrt[]{2(42.9(m)/(s^2))(6*10^7m)}`

Solving the operations:

`v_e=71749.6(m)/(s)`

Therefore, the escape velocity is 71749.6 m/s.