Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads
Derivation
Derive an expression for acceleration due to gravity at depth ‘d’ below the earth’s surface.
Advertisement Remove all ads
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)`
Concept: Acceleration Due to Gravity and Its Variation with Altitude and Depth
Is there an error in this question or solution?