Salvatore Boone

2022-01-16

What does it mean to say that the gravity of the Earth is 9.8 m/s2?

servidopolisxv

Expert

As per Law of Universal Gravitation the force of attraction between two bodies is directly proportional to the product of masses of the two bodies. it is also inversely proportional to the square of the distance between the two. That is the force of gravity follows inverse square law.
Mathematically
${F}_{G}\propto {M}_{1}\cdot {M}_{2}$
Also ${F}_{G}\propto \frac{1}{{r}^{2}}$
Combining the two we obtain the proportionality expression
${F}_{G}\propto \frac{{M}_{1}\cdot {M}_{2}}{{r}^{2}}$
Follows that
${F}_{G}=G\frac{{M}_{1}\cdot {M}_{2}}{{r}^{2}}$
Where G is the proportionality constant.
It has the value $6.67408×{10}^{-11}{m}^{3}k{g}^{-1}{s}^{-2}$
r is the mean radius of earth and taken as
Mass of earth is $5.972×{10}^{24}kg$
If one of the body is earth the equation becomes
${F}_{G}=\left(G\frac{{M}_{e}}{{r}^{2}}\right)\cdot m$
See this has reduced to $F=mg$
Were $g=G\frac{{M}_{e}}{{r}^{2}}$
Inserting the values
$g=6.67408×{10}^{-11}\frac{5.972×{10}^{24}}{{\left(6.371×{10}^{6}\right)}^{2}}$
Simplifying we obtain
$g\approx 9.8\frac{m}{{s}^{2}}$
In other words if an object is dropped from a height h above the earth's surface, the object will fall towards earth with constant acceleration of $g=9.8\frac{m}{{s}^{2}}$

RizerMix

Expert

Gravity is a force, and according to Newtons

alenahelenash

Expert