Question 437: Gravitation

Question

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

Accepted Answer

From equation of acceleration due to gravity.
$${g_e} = {{G{M_e}} \over {R_e^2}} = {{G\left( {4/3} ight)\pi R_e^3} \over {R_e^2}}{ho _e}$$
$${g_e} \propto {R_e}{ho _e}$$
Acceleration due to gravity of planet
$${g_p} \propto {R_p}{ho _p}$$
$$ 	herefore $$ $${R_e}{ho _e} = {R_p}{ho _p} \Rightarrow {R_e}{ho _e} = {R_p}2{ho _e}$$
$$ \Rightarrow {R_e} = {1 \over 2}R$$   ($$ \because $$ R
e
= R)

Answer

2R

Answer

4R

Answer

$${1 \over 4}$$R

Answer

$${1 \over 2}$$R

Gravitation

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

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