Question

Find the value of acceleration due to gravity near the surface of the earth.

Answer 1

We know that,
$mg=\frac{GMm}{r^2}$
Here, m = Mass of the body
g = Acceleration due to gravity
M = Mass of the Earth = $5.98\times {10}^{24}kg$
G = Universal gravitational constant = $6.673\times {10}^{-11}N{m}^{2}k{g}^{-2}$
r = Radius of the Earth = $6.38\times {10}^{6}m$
So,
$g=\frac{GM}{r^2}=\frac{6.673\times {10}^{-11}\times 5.98\times {10}^{24}}{(6.38\times {10}^6{)}^2}=9.8m/s^2$