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The radii of two planets are respectively ${R_1}$ and ${R_2}$, their densities are ${\rho _1}$ and ${\rho _2}$. The ratio of the accelerations due to the gravity at their surfaces is:
$(A)\dfrac{{{R_1}}}{{{R_2}}}.\dfrac{{{\rho _1}}}{{{\rho _2}}}$
\[(B)\dfrac{{{R_2}}}{{{R_1}}}.\dfrac{{\rho _1^2}}{{{\rho _2}}}\]
$(C)\dfrac{{{R_2}}}{{{R_1}}}.\dfrac{{{\rho _2}}}{{{\rho _1}}}$
$(D)1:1$

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Last updated date: 23rd May 2024
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Answer
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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 $9.8m/{s^2}$.