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 gravity at the surface 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. `1/2R`
B. 2R
C. 4R
D. `1/4R`

Answer

A

Explanation

`implies` Acceleration of gravitation `g = (GM)/R^2`
where mass `m = d xx (4)/3 piR^3` , where d = density
`:.g = G4/3 dpiR`
From Formula,
For earth : `g_e = G(d) piR_e g G, 4/3 pi` is constant
For planet : `g_p = G (2d) pi R_p:.d R = ` constant
Now , `g_e =g_p implies dR_e = 2dR_p`
`:. d_eR_e = d_p R_p`
`:. R_p = R_e / 2` But `R_2 = R and dp = 2de `
`:. R_p = R/2`