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

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

Step-by-step explanation:

Given that,

Mass of the planet Krypton,
`m=8.8* 10^(23)\ kg`

Radius of the planet Krypton,
`r=2.5* 10^(6)\ m`

Value of gravitational constant,
`G=6.6726* 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=(Gm)/(r^2)`

`a=(6.6726* 10^(-11)* 8.8* 10^(23))/((2.5* 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`