why value of g is 9.8 m/s2 on the surface of earth | knowledge of physics

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The relation for acceleration due gravity (g) acting on a body at the surface of the earth is given by

`g =\frac{GM}{R^2}`  ................... (1)
 
Where,
G = gravitational constant = `6.67408 \times 10^{-11} \frac{Nm^2}{kg^2}`

 
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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) 

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You can learn technique of full derivation of relation (1) from  Here. 



Using values of G, M and R in (1), 

`g = \frac{GM}{R^2}`

`g = \frac{6.67408 \times 10^{-11} Nm^2kg^{-2}X5.972X10^{24}kg}{(6.371X10^{6}m)^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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