Answer
64.8k+ views
Hint: The acceleration gained by an object due to the force of gravitation is known as acceleration due to gravity. The acceleration due to gravity is represented by the letter $g$. Using the formula that relates radius, the density of the earth, and the gravitational constant.
Formula used:
$ \Rightarrow g = \dfrac{4}{3}\pi \rho GR$
Where,
$g$ is the acceleration due to gravity,
$\rho $ is the density of the earth,
$G$ is the universal gravitational constant,
$R$ is the radius of the earth.
Complete step by step answer:
The radius of the two planets is given as ${R_1}$ and ${R_2}$, their densities are given as ${\rho _1}$and ${\rho _2}$.
The acceleration due to gravity is the acceleration gained by an object because of the gravitational force. The value of $g$ can be calculated as,
$ \Rightarrow g = \dfrac{{mG}}{{{R^2}}}$
Where,
$g$ is the acceleration due to gravity,
$m$ is the mass of the earth
$G$ is the universal gravitational constant,
$R$ is the radius of the earth.
The mass of the earth can be determined by the formula,
$ \Rightarrow m = \dfrac{4}{3}\pi {R^3}\rho $
Where,
$\rho $ is the density of the earth.
Substituting the value of mass in the acceleration due to gravity, we get,
$ \Rightarrow g = \dfrac{{\dfrac{4}{3}\pi {R^3}\rho G}}{{{R^2}}}$
Canceling out the common term $R$ we get,
$ \Rightarrow g = \dfrac{4}{3}\pi \rho GR$
The above formula is used to calculate the acceleration due to gravity. There are two planets so there will be acceleration due to gravity for each planet.
Let ${g_1}$ be the acceleration due to gravity of the second planet.
$ \Rightarrow {g_1} = \dfrac{4}{3}\pi {\rho _1}G{R_1}$
Let \[{g_2}\] be the acceleration due to gravity of the second planet.
$ \Rightarrow {g_2} = \dfrac{4}{3}\pi {\rho _2}G{R_2}$
Universal gravitational constant $G$ will not change because it has a constant value.
Let us divide ${g_1}$ and \[{g_2}\].
$ \Rightarrow \dfrac{{{g_1}}}{{{g_2}}} = \dfrac{{\dfrac{4}{3}\pi {\rho _1}G{R_1}}}{{\dfrac{4}{3}\pi {\rho _2}G{R_2}}}$
Canceling out the common terms we get,
$ \Rightarrow \dfrac{{{g_1}}}{{{g_2}}} = \dfrac{{{\rho _1}{R_1}}}{{{\rho _2}{R_2}}}$
Separating the common terms, we get,
$ \Rightarrow \dfrac{{{g_1}}}{{{g_2}}} = \dfrac{{{R_1}}}{{{R_2}}}.\dfrac{{{\rho _1}}}{{{\rho _2}}}$
The Ratio of the accelerations due to the gravity at their surfaces is found as,
$ \Rightarrow \dfrac{{{g_1}}}{{{g_2}}} = \dfrac{{{R_1}}}{{{R_2}}}.\dfrac{{{\rho _1}}}{{{\rho _2}}}$
Therefore option \[\left( A \right)\] is the correct answer.
Note: The dimensional analysis for the acceleration due to gravity is ${M^0}{L^1}{T^{ - 2}}$. $g$ has both the direction and magnitude and hence is the vector quantity. The S.I unit of the $g$ is $m/{s^2}$. At the sea level, the standard value $g$ is approximately equal to $9.8m/{s^2}$.
Formula used:
$ \Rightarrow g = \dfrac{4}{3}\pi \rho GR$
Where,
$g$ is the acceleration due to gravity,
$\rho $ is the density of the earth,
$G$ is the universal gravitational constant,
$R$ is the radius of the earth.
Complete step by step answer:
The radius of the two planets is given as ${R_1}$ and ${R_2}$, their densities are given as ${\rho _1}$and ${\rho _2}$.
The acceleration due to gravity is the acceleration gained by an object because of the gravitational force. The value of $g$ can be calculated as,
$ \Rightarrow g = \dfrac{{mG}}{{{R^2}}}$
Where,
$g$ is the acceleration due to gravity,
$m$ is the mass of the earth
$G$ is the universal gravitational constant,
$R$ is the radius of the earth.
The mass of the earth can be determined by the formula,
$ \Rightarrow m = \dfrac{4}{3}\pi {R^3}\rho $
Where,
$\rho $ is the density of the earth.
Substituting the value of mass in the acceleration due to gravity, we get,
$ \Rightarrow g = \dfrac{{\dfrac{4}{3}\pi {R^3}\rho G}}{{{R^2}}}$
Canceling out the common term $R$ we get,
$ \Rightarrow g = \dfrac{4}{3}\pi \rho GR$
The above formula is used to calculate the acceleration due to gravity. There are two planets so there will be acceleration due to gravity for each planet.
Let ${g_1}$ be the acceleration due to gravity of the second planet.
$ \Rightarrow {g_1} = \dfrac{4}{3}\pi {\rho _1}G{R_1}$
Let \[{g_2}\] be the acceleration due to gravity of the second planet.
$ \Rightarrow {g_2} = \dfrac{4}{3}\pi {\rho _2}G{R_2}$
Universal gravitational constant $G$ will not change because it has a constant value.
Let us divide ${g_1}$ and \[{g_2}\].
$ \Rightarrow \dfrac{{{g_1}}}{{{g_2}}} = \dfrac{{\dfrac{4}{3}\pi {\rho _1}G{R_1}}}{{\dfrac{4}{3}\pi {\rho _2}G{R_2}}}$
Canceling out the common terms we get,
$ \Rightarrow \dfrac{{{g_1}}}{{{g_2}}} = \dfrac{{{\rho _1}{R_1}}}{{{\rho _2}{R_2}}}$
Separating the common terms, we get,
$ \Rightarrow \dfrac{{{g_1}}}{{{g_2}}} = \dfrac{{{R_1}}}{{{R_2}}}.\dfrac{{{\rho _1}}}{{{\rho _2}}}$
The Ratio of the accelerations due to the gravity at their surfaces is found as,
$ \Rightarrow \dfrac{{{g_1}}}{{{g_2}}} = \dfrac{{{R_1}}}{{{R_2}}}.\dfrac{{{\rho _1}}}{{{\rho _2}}}$
Therefore option \[\left( A \right)\] is the correct answer.
Note: The dimensional analysis for the acceleration due to gravity is ${M^0}{L^1}{T^{ - 2}}$. $g$ has both the direction and magnitude and hence is the vector quantity. The S.I unit of the $g$ is $m/{s^2}$. At the sea level, the standard value $g$ is approximately equal to $9.8m/{s^2}$.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)