Question

A planet having the same mass as that of the 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) 4g
b) 2g
c) $\dfrac{g}{2}$
d) $\dfrac{g}{4}$

Answer 1

The universal law of gravitation holds everywhere. It is due to the gravitational force of the earth which holds all the bodies. If a body is allowed to fall freely from a certain height than it falls under the influence of gravity and the acceleration of the body is g whose value is 9.8 $m/{{s}^{2}}$
The value of acceleration due to gravity changes with the height and the depth.

Complete step by step answer:
Let the density of the planet be $\rho$and let the density of the earth be $\varsigma$. Let the mass of both the planet and earth be M
Given, $\rho =8\varsigma$
Writing the formula for density and equating it we get,

& \dfrac{M}{\dfrac{4\pi R_{p}^{3}}{3}}=8\dfrac{M}{\dfrac{4\pi R_{e}^{3}}{3}} \\\ & \Rightarrow \dfrac{M}{\dfrac{4\pi R_{p}^{3}}{3}}=8\dfrac{M}{\dfrac{4\pi R_{e}^{3}}{3}} \\\ & \Rightarrow 2{{R}_{p}}={{R}_{E}} \\\ \end{aligned}$$ Now to find the acceleration due to gravity on the planet, we write the formula for acceleration due to gravity, We know on earth, $${{g}_{e}}=\dfrac{GM}{R_{E}^{2}}$$, so, $\implies$ $$\dfrac{{{g}_{e}}}{{{g}_{p}}}=\dfrac{\dfrac{GM}{R_{E}^{2}}}{\dfrac{GM}{R_{P}^{2}}}=\dfrac{\dfrac{GM}{4R_{P}^{2}}}{\dfrac{GM}{R_{P}^{2}}}=\dfrac{1}{4}$$ $\therefore$ $${{g}_{P}}=4{{g}_{E}}$$ **Thus, the acceleration due to gravity on the planet’s surface is 4g. So, the correct option is (a)** **Note-** The g, is the acceleration due to gravity and this is not a constant. At the earth’s surface its value is 9.8 $$m/{{s}^{2}}$$and as one goes above the earth surface its value changes and if one goes inside the earth crust its value also changes. The value also varies from pole to the equator. The angular velocity of rotation of earth affects the value of g.