Answer:
Acceleration, [tex]a=9.39\ m/s^2[/tex]
Explanation:
Given that,
Mass of the planet Krypton, [tex]m=8.8\times 10^{23}\ kg[/tex]
Radius of the planet Krypton, [tex]r=2.5\times 10^{6}\ m[/tex]
Value of gravitational constant, [tex]G=6.6726\times 10^{-11}\ Nm^2/kg^2[/tex]
To find,
The acceleration of an object in free fall near the surface of Krypton.
Solution,
The relation for the acceleration of the object is given by the below formula as :
[tex]a=\dfrac{Gm}{r^2}[/tex]
[tex]a=\dfrac{6.6726\times 10^{-11}\times 8.8\times 10^{23}}{(2.5\times 10^{6})^2}[/tex]
[tex]a=9.39\ m/s^2[/tex]
So, the value of acceleration of an object in free fall near the surface of Krypton is [tex]9.39\ m/s^2[/tex]
