Answer:
[tex]g_K=22.0m/s^{2}[/tex]
Explanation:
From the Newton's Law of Universal Gravitation, we have that the acceleration due to gravitational force near an object of mass M, is given by:
[tex]g=\frac{GM}{R^{2}}[/tex]
Where g is the acceleration due to gravity, G is the gravitational constant, and R is the distance between the objects.
In this case, we have the given values for planet Krypton:
[tex]g_K=\frac{(6.6726*10^{-11}\frac{Nm^{2}}{kg^{2}})(4*10^{23}kg)}{(1.1*10^{6}m)^{2}}=22.0m/s^{2}[/tex]
This means that the acceleration of an object in free fall near the surface of Krypton is of 22.0m/s².
