(a) Assuming the earth to be a sphere of uniform density, calculate the value of acceleration due to gravity at a point (i) `1600 km` above the earth, (ii) `1600 km` below the earth, (b) Also find the rate of variation of acceleration due to gravity above and below the earth's surface. Radius of earth `=6400 km, g 9.8 m//s^(2)`.

Explanation

At depth `x`
`g_(x)=g((1-x)/R)=g(1-(1600km)/(6400km))=gxx3/4=7.35m//s^(2)`
At distance `r` from centre of earth above the earth's surface
`g=(GM)/(r^(2))`
`implies(dg)/(dr)=-2/(r^())3GM`
As `r=R+h, dr=dh, (dg)/(dh)=-=(2GM)/((R+h)^(3))`
ii. At distanc `x` below earth's surface
`g=g((1-x)/R)`
`(dg)/(dx)=-g/R=-((GM)/(R^(2)))1/R=-(GM)/(R^(3))`
`(G4/3piR^(2)rho)/(R^(3))=-4/3piGrho`