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Derive an Expression for Acceleration Due to Gravity at Depth ‘D’ Below the Earth’S Surface. - Physics

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Derivation

Derive an expression for acceleration due to gravity at depth ‘d’ below the earth’s surface.

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Solution

 
 

Let M be mass of the earth, R be the radius of the earth

gd be gravitational acceleration at depth 'd ' from the earth surface

g be gravitational acceleration on the earth surfaces.

ρ be the density of the earth.

‘P’ be the point inside the earth at depth 'd ' from earth surfaces.

∴ CS-CP=d, ∴ CP=R-d .............(1)   (since CS=R)

`g=(GM)/R^2`,

`therefore g=(G4/3piR^3rho)/R^2`

`therefore g=(4GpiRrho)/3`  ....(2)

gd = acceleration due to gravity at depth 'd '

`g_d=(Gxx"Mass of the sphere with radius CP")/(CP^2)`

`thereforeg_d=(G4/3piCP^3rho)/(CP^2)`

`thereforeg_d=(4GpiCPrho)/3` .... (3)

Dividing eq. (3) by eq. (2)

`g_d/g=(CP)/R=(R-d)/R`

`therefore g_d=g(1-d/R)`

c

 

 
Concept: Acceleration Due to Gravity and Its Variation with Altitude and Depth
  Is there an error in this question or solution?

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