Answer

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.

**Complete step by step answer:**

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.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Mark and label the given geoinformation on the outline class 11 social science CBSE

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Trending doubts

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Which are the Top 10 Largest Countries of the World?

Give 10 examples for herbs , shrubs , climbers , creepers

10 examples of evaporation in daily life with explanations

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Why is there a time difference of about 5 hours between class 10 social science CBSE