Question

assume the earths mass is 34.102 kg, and the radius is 56.867 x 10^3 miles, what would be the gravitational acceleration on such an planet in unit of m/s^2?

Answer 1

Answer:

The gravitational acceleration of that planet is $2.718\times 10^{-25}$ meters per square second.

Explanation:

Under the assumption that Earth is a sphere and with an uniformly distributed density, the gravitational acceleration at the surface of the planet is given by the following expression:

$g = G\cdot \frac{M}{r^{2}}$ (Eq. 1)

Where:

$G$ - Gravitation constant, measured in newton-square meters per square kilogram.

$r$ - Radius of the planet, measured in meters.

$M$ - Mass of the planet, measured in kilograms.

If we know that $M = 34.102\,kg$, $r = 91.5\times 10^{6}\,m$ and $G = 6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$, then the acceleration on the planet is:

$g = \left(6.672\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot \left[\frac{34.102\,kg}{(91.5\times 10^{6}\,m)^{2}} \right]$

$g = 2.718\times 10^{-25}\,\frac{m}{s^{2}}$

The gravitational acceleration of that planet is $2.718\times 10^{-25}$ meters per square second.