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

Answer:

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

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

 $V.E^2=2gr$

 $V.E^2=2*0.412m/s^2*301000$

 $V.E^2=248024$

 $V.E=\sqrt{248024}$

 $V.E=498.02m/s^2$