Question

The planet Krypton has a mass of 8.8 × 1023 kg and radius of 2.5 × 106 m. What is the acceleration of an object in free fall near the surface of Krypton? The gravitational constant is 6.6726 × 10−11 N · m2 /kg2 . Answer in units of m/s 2 .

Answer 1

Answer:

Acceleration, $a=9.39\ m/s^2$

Explanation:

Given that,

Mass of the planet Krypton, $m=8.8\times 10^{23}\ kg$

Radius of the planet Krypton, $r=2.5\times 10^{6}\ m$

Value of gravitational constant, $G=6.6726\times 10^{-11}\ Nm^2/kg^2$

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 :

$a=\dfrac{Gm}{r^2}$

$a=\dfrac{6.6726\times 10^{-11}\times 8.8\times 10^{23}}{(2.5\times 10^{6})^2}$

$a=9.39\ m/s^2$

So, the value of acceleration of an object in free fall near the surface of Krypton is $9.39\ m/s^2$