Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# A planet has twice the density of earth but the acceleration due to gravity on its surface is exactly the same as on the surface of earth. Its radius in terms of radius of earth $R$ will beA. $\dfrac{R}{4}$ B. $\dfrac{R}{2}$ C. $\dfrac{R}{3}$ D. $\dfrac{R}{8}$

Last updated date: 19th Sep 2024
Total views: 421.5k
Views today: 4.21k
Verified
421.5k+ views
Hint: The acceleration due to gravity is equal to the product of gravitational constant and mass of the earth or planet divided by the square of the distance between the object and the planet or earth. The distance between the object and the planet or earth is taken as the radius of the earth or planet in this case and the mass is the product of volume and density.

The density of a given planet is twice the density of earth but the acceleration due to gravity on the surface of this planet is the same as that on the earth. According to Newton’s law of gravitation, the force of attraction (F) between the body and the earth is given by the formula F$F = \dfrac{{GMm}}{{{R^2}}}$ where G is known as Gravitational constant, M is mass of the earth, m is mass of the body and R is the radius of the earth but the force with which a body is attracted towards the earth gives us the weight of the body i.e., $F = mg$ where g is acceleration due to gravity on the surface of earth. Thus,
$mg = \dfrac{{GMm}}{{{R^2}}}$
$\Rightarrow g = \dfrac{{GM}}{{{R^2}}}$
$\Rightarrow g = \dfrac{{G\left( {\rho \times V} \right)}}{{{R^2}}}$ where ρ is the density of the earth, V is the volume of the earth and $mass = Volume \times density$ .
$g = \dfrac{{G\left( {\rho \times \dfrac{4}{3}\pi {R^3}} \right)}}{{{R^2}}}$ [we have assumed that the earth is a sphere]
$g = \dfrac{{4G\rho \pi R}}{3}$ [eqn. 1]
Now, let ${g_1}$ be the acceleration due to gravity on the surface of the given planet and ${R_1}$ be the radius of the planet and ρ1 be the density of the planet. Thus,
${g_1} = \dfrac{{4G\pi \left( {\rho_1} \right)\left( {R_1} \right)}}{3}$
$\Rightarrow g = \dfrac{{4G\pi \left( {2\rho } \right)\left( {R_1} \right)}}{3}$
As the acceleration due to gravity on the surface of the planet is the same as g and its density is twice the density of earth.
$g = \dfrac{{8G\pi \rho \left( {R_1} \right)}}{3}$ [eqn. 2]
Now, we will divide eqn.1 by eqn.2,
$\Rightarrow\dfrac{g}{g} = \dfrac{{\dfrac{{4G\pi \rho R}}{3}}}{{\dfrac{{8G\pi \rho (R_1)}}{3}}}$
$\Rightarrow 1 = \dfrac{R}{{2\left( {R_1} \right)}}$
$\therefore{R_1} = \dfrac{R}{2}$
Hence, the radius of the planet is half of the radius of the earth.

Therefore, option B is correct.

Note:Kindly remember that the value of gravitational constant is the same for all the places as it is a constant. Thus, it is the same for both the values of acceleration due to gravity i.e., in both the formulae of g, we’ll take the same value of G. Also, remember that the mass should be written as the product of density and volume as in the question the value of mass is not given. So, we have to change it to volume and density.