What will be the percentage increases or decrease in the value of acceleration due to gravity on the surface of the earth , if the radius of the earth decreases by 1% and the mass of the earth remains unchanged ?

Explanation

Acceleration due to gravity on the surface of the earth, for the present value of the earth.s radius R, is g=`(GM)/(R^2)`.
Let `R^.` =radius of the earth when it decreases by 1% ,i.e., `R^.=0.99R`.
At that time, the value of the acceleration due to gravity on the earth.s surface is `g^.=(GM)/(R^(.2))`.
`therefore (g^.)/(g)=(R^2)/(R^(.2))=(R^2)/((0.99R)^2)=1/((0.99)^2)=1.02`
Hence, `g^. =1.02g=g+0.02g`
Hence in this case, the value of acceleration due to gravity on the earth.s surface increases by 2%
ALTERNATIVE METHOD :
We know, `g=(GM)/(R^2)`
`or, (dg)/(dR)=-GM2/(R^3)=-2(GM)/(R^2)*1/R=-(2g)/R`
`therefore (dg)/(g) =-2(dR)/R`
Radius decreased by 1% implies that `(dR)/Rxx100=-1`
`therefore` Percentage change in `g=(dg)/gxx100=-2(dR)/Rxx100=2`
Hence, the value of g increases by 2% .