Question

(a) What is the escape speed on a spherical asteroid whose radius is 301 km and whose gravitational acceleration at the surface is 0.412 m/s2

Answer 1

`V.E=498.02m/s^2`

Step-by-step explanation:

From the question we are told that:

Radius
`r=301Km`

Gravitational acceleration
`g=0.412 m/s^2`

Generally the equation for Escape velocity is mathematically given by

`V.E^2=2gr`

`V.E^2=2*0.412m/s^2*301000`

`V.E^2=248024`

`V.E=√(248024)`

`V.E=498.02m/s^2`