why value of g is 9.8 m/s2 on the surface of earth | knowledge of physics
The relation for acceleration due gravity (g) acting on a body at the surface
of the earth is given by
Where,
G = gravitational constant = `6.67408 \times 10^{-11} \frac{Nm^2}{kg^2}`Source:
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M = Mass of the earth = `5.972 \times10^{24} kg`,
R = radius of the earth = `6.371 \times 10^{6} m`
Source of data:
https://solarsystem.nasa.gov/planets/earth/by-the-numbers/ (visited on 12th
May 2021)
Using values of G, M and R in (1),
`g = \frac{GM}{R^2}`
`g= \frac{3.9858 \times 10^(14) Nm^2kg^(-1)}{(6.371X10^(2)m)^2}`
`g = 9.8196 ms^{-2}`
`g ≈9.8 ms^(-2)`
Thus, depending on the values of mass and radius of earth at it's surface, value of g becomes approximately 9.8 ms-2.
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