Question

The Earth has a mass of 5.98X10^24 kg, and its& radius is 6.37X10^6 m. Using these values, calculate the acceleration due to gravity at the surface of the earth.

Answer 1

Q. The Earth has a mass of $5.98\times 10^{24}$ kg, and its radius is $6.37\times 10^{6}$ m. Using these values, calculate the acceleration due to gravity at the surface of the earth.

A. According to Newton’s law of universal gravitation, the force $F$ between Earth’s mass $M$ and arbitrary mass $m$ is

$F=G\,\frac{Mm}{r^{2}}$

where

$G$ is the gravitational constant ($G\approx 6.674\times 10^{-11}$ N m^{2} kg^{-2}) and

$r$ is the distance from Earth’s center.

In other hand, the force equals the mass multiplied by its acceleration (Newton’s 2nd law):

$F=ma$

Therefore we can rewrite first expression:

$ma=G\,\frac{Mm}{r^{2}}\qquad\Rightarrow\qquad a=G\,\frac{M}{r^{2}}$

and, finally,

$a=\frac{GM}{r^{2}}=\frac{6.674\times 10^{-11}\cdot 5.98\times 10^{24}}{(6.37\times 10^{6})^{2}}=9.83$ m/s^{2}