If the radius of the earth decreases by 1% without changing its mass, will the acceleration due to gravity at the surface of the earth increase or decrease? If so, by what percent?

Explanation

We know that, `g = (GM)/(R^(2))` or `g prop (1)/(R^(2))`
So `(g_(2))/(g_(1)) = ((R_(1))/(R_(2)))^(2)`
If `R_(1) = R`, then `R_(2) = R +- n% R = (1 +- n%)R`
`:. (g_(2))/(g_(1)) = [(R)/((1 +- n%)R)]^(2) = (1)/((1 +- n%)^(2))`
`= (1 +- n%)^(-2)`
`%` increase or decrease in the value of `g` is
`= ((g_(2) - g_(1))/(g_(1))) xx 100 = ((g_(2))/(g_(1)) - 1) xx 100`
`= [(1 +- n%)^(-2) - 1] xx 100`