Answer
Verified
405k+ 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.
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
Why Are Noble Gases NonReactive class 11 chemistry CBSE
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
At which age domestication of animals started A Neolithic class 11 social science CBSE
Which are the Top 10 Largest Countries of the World?
Give 10 examples for herbs , shrubs , climbers , creepers
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Write a letter to the principal requesting him to grant class 10 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE