Question

A planet has a miss of eight times the mass of earth and density is also equal to eight times the average density of the earth. If g be the acceleration due to earth's gravity on its surface, then acceleration due to gravity on planet's surface will be:
(A) $2g$
(B)$4g$
(C)$8g$
(D)$16g$

Answer 1

The density of any fluid provides information about how much closer the molecules are in the object. Simply It is the ratio of the mass of the object and the volume of the object, as the mass increases the density of the object also increases, and if the volume of the object increases, the value of the density decreases.

Complete step by step answer:
Given,
Let the mass of the planet is $m$ and the mass of the earth is $M$.
Let the density of the planet is $d$ and the density of the earth is $D$.
Let the radius of the planet is $r$ and the radius of the earth is $R$ .
It is given that mass of the planet is eight times the mass of earth, so we write this relation,
$m = 8M\;...........{\rm{(1)}}$
It is also given that the density of the planet is eight times of the density of the earth, so we can write this relation as,
$d = 8D\;...........{\rm{(2)}}$
As we know that the formula of the density of an object is,
${\rm{density}}\left( d \right) = \dfrac{{{\rm{mass(}}m{\rm{)}}}}{{{\rm{volume}}\left( v \right)}}$
Taking the planet and the earth as spherical objects.
The expression for the volume of the planet and volume of the earth is given as follows,
$\Rightarrow V = \dfrac{4}{3}\pi {r^3}\\\ \Rightarrow V = \dfrac{4}{3}\pi {R^3}$
Here, $v$ is the volume of the planet and $V$ is the volume of the earth,
Now we substitute the values in equation (2) and solve further,

$$\Rightarrow\dfrac{m}{v} = 8\dfrac{M}{V}\\\ \Rightarrow\dfrac{{8M}}{{\dfrac{4}{3}\pi {r^3}}} = 8\dfrac{M}{{\dfrac{4}{3}\pi {R^3}}}\\\ \Rightarrow R = r$$

The expression for the acceleration due to gravity on planet's surface is given as follows,
${g_p} = \dfrac{{Gm}}{{{r^2}}}\;............{\rm{(3)}}$
Here, $G$ universal constant of gravitation.
The expression for the acceleration due to gravity on the earth's surface is given as follows,
$g = \dfrac{{GM}}{{{r^2}}}\;............{\rm{(4)}}$
Divide equation (3) by equation (4) and solve further,

$$\Rightarrow\dfrac{{{g_p}}}{g} = \dfrac{{\dfrac{{Gm}}{{{r^2}}}}}{{\dfrac{{GM}}{{{R^2}}}}}\\\ \Rightarrow\dfrac{{{g_p}}}{g} = \dfrac{m}{M}\\\ \Rightarrow\dfrac{{{g_p}}}{g} = \dfrac{{8M}}{M}\\\ \therefore{g_p} = 8g$$

Therefore, the acceleration due to gravity on the planet's surface is $8g$, and the correct answer is option (C).

Note: The acceleration due to the gravity of the earth is calculated by using the universal law of gravitation given by Newton, this acceleration depends on the location of the object from the centre of the earth, that means it varies with altitudes.