If the radius of the earth were to shrink by 1% its mass remaining the same, the acceleration due to gravity on the earth's surface would

Options

A. decrease by 1%
B. increase by 1%
C. increase by 2%
D. decrease by 2%

Answer

C

Explanation

`g=(GM)/(R^(2))`
[`g prop (1)/(R^(2))` hence g will increase if R decreased]
`therefore log g = log G + log M - 2 log R`
Differentiating, we get
`(d_(g))/(g)=0+0-(2dR)/(R ) " "(because "G and M are constant.")`
`therefore ((d_(g))/(g))xx100=-2((dR)/(R ))xx100`
`=-2(-1%)=2%`
`therefore` g increases by 2%.