If the radius of the earth were to shrink by one percent its mass remaining the same, the acceleration due to greavity on the earth's surface would

Explanation

Consider the case of body of mass m placed on the earth's surface (mass of the earth M and radius R). If g is acceleration due to gravity, then we known that
`g_(s) = (GM_(e))/(R_(e)^(2))` …(1)
Now, when the radius is reduced by 1%, i.e. radius becomes 0.99 R, let acceleration due to gravity be g', then
`g' = (GM)/((0.99R)^(2))` ...(2)
From equation (1) and (2), we get
`(g')/(g) = (R^(2))/((0.99R)) = (1)/((0.99)^(2))`
or `g' = g xx ((1)/(0.99))^(2)`
or `g' = 1.02 g`
Thus, the value of g is increased by 2%.