Question

If the density of the earth is doubled keeping the radius constant, find the new acceleration due to gravity? (take, $g=9.8m{s}^{-2}$)

• A. $9.8m{s}^{-2}$
• B. $19.6m{s}^{-2}$
• C. $4.9m{s}^{-2}$
• D. $39.2m{s}^{-2}$

Answer 1

The correct option is B

$19.6m{s}^{-2}$

Step 1: State assumptions and known data

Let the density of earth be $\rho$.

Let $R$ be the radius of Earth.

We know that the acceleration due to gravity on Earth's surface $g=9.8m{s}^{-2}$.

We need to find the value of $g$ when density is $2\rho$.

Step 2: Formulas used

Acceleration due to gravity is given as,

$g=\frac{GM}{R^2}$ $...\left(1\right)$

where $G$ is the gravitational constant, $M$ is the mass of the planet and $R$ is the radius of the planet.

Density of an object, $\rho =\frac{M}{V}$, where $V$ is its volume

Step 3: Derive expression for acceleration due to gravity in term of density of the earth

From the equation relating density and mass,
$M=\rho V$

The shape of a planet can be approximated to be that of a sphere.

The volume of a sphere is $V=\frac{4}{3}\mathrm{\pi }{r}^3$, where $r$ is its radius.

Thus, the mass of the sphere is, $M=\frac{4}{3}\mathrm{\pi }{r}^3\rho$

Substituting the above expression in $\left(1\right)$,

$\begin{array}{rcl}g& =& \frac{4}{3}\frac{G\mathrm{\pi }{r}^3\rho }{r^2}\\ \mathit{\therefore }g& =& \frac{4}{3}G\mathrm{\pi r}\rho \end{array}$

Step 4: Calculate the acceleration due to gravity when density is doubled

The acceleration due to gravity on Earth is,
$g=\frac{4}{3}G\mathrm{\pi }R\rho$

When density is doubled,
$\begin{array}{cc}& g\text{'}=\frac{4}{3}G\pi R2\rho \\ \Rightarrow & g\text{'}=2\times \frac{4}{3}G\pi R\rho \\ \Rightarrow & g\text{'}=2g\\ \Rightarrow & g\text{'}=2\times 9.8\\ \therefore & g\text{'}=19.6m{s}^{-2}\end{array}$

Therefore, if the density of the earth is doubled keeping the radius constant, the new acceleration due to gravity will be $19.62m{s}^{-2}$.

Hence, option B is correct.