The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is `R`, the radius of the planet would be

Options

A. `2R`
B. `4R`
C. `1/4R`
D. `1/2R`

Answer

D

Explanation

The acceleration due to gravity on an object of mass `m`
`g=F/m`
but from Newton's law of gravitation
`F=(GMm)/(R^(2))`
where `M` is the mass of the earth and `R` the radius of earth.
`:. G=(GMm//R^(2))/m=(GM)/(R^(2))`
Given: `rho_("planet")=g_("earth")`
`(GM)/(R_(p)^(2))=(GM_(e))/(R_(e)^(2))`
or `(Gxx4/3piR_(p)^(3)rho_(p))/(R_(p)^(2))=(Gxx4/3piR_(e)^(3)rho_(e))/(R_(e)^(2))`
or `R_(p)rho_(p)=R_(e)rho_(e)`
or `R_(p)xx2rho_(p)=R_(e)rho_(e)`
or `R_(p)=(R_(e))/2=R/2`
Alternative: `g=4/3pirhoGR`
`implies (R_(p))/(R_(e))=((g_(p))/(g_(e)))((rho_(e))/(rho_(p)))=(1)xx(1/2)`