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

`g =\frac{GM}{R^2}`  ................... (1)

 

Where,

G = gravitational constant = `6.67408 \times 10^{-11} \frac{Nm^2}{kg^2}`

 

Source:  Visit Now

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) 

                               Visit Now

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.

For more details watch video.

 

Interested to learn more? Then visit the Home page for more updates.