Question

The value of the acceleration due to gravity is $g_1$at a height $h=\frac{R}{2}$ ( $R$= radius of the earth) from the surface of the earth. It is again equal to $g_1$ at a depth $d$ below the surface of the earth. The ratio $\frac{d}{R}$ equals :

• A. $\frac{4}{9}$
• B. $\frac{1}{3}$
• C. $\frac{5}{9}$
• D. $\frac{7}{9}$

Answer 1

The correct option is C

$\frac{5}{9}$

Step 1: Given data

Acceleration due to gravity at height $h$ is $g_1$.

Here, $h=\frac{R}{2}$

Acceleration due to gravity at depth $d$ is also $g_1$.

Here, $h$ is the height above the earth's surface and $d$ is the depth from the earth's surface.

Step 2: Formula used:

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

Where, $g$ is the acceleration due to gravity,

$G$ is gravitational constant,

$M$ is the mass of Earth and $R$ is the radius of the earth.

Density is defined as mass per unit volume and is given by,

$Density\left(\rho \right)=\frac{Mass\left(M\right)}{Volume\left(V\right)}$

The volume of a sphere is given as, $V=\frac{4}{3}{\mathrm{\pi R}}^3$

Step 3: Calculating the acceleration due to gravity at a height h

As we know that $g=\frac{GM}{{\left({\displaystyle R}\right)}^2}$

Therefore, acceleration due to gravity at height h will be,

$g_h=\frac{GM}{{(R+h)}^2}$

Substitute the value for height as $h=\frac{R}{2}$ in the above formula,

$$\begin{gathered} g_h=\frac{GM}{{(R+{\displaystyle \frac{R}{2}})}^2} \\ {g}_h=\frac{GM}{{\left({\displaystyle \frac{3R}{2}}\right)}^2} \\ {g}_h=\frac{GM}{{\displaystyle \frac{9R^2}{4}}} \\ {g}_h=\frac{4GM}{{\displaystyle 9R^2}}.......eq\left(1\right) \end{gathered}$$

Step 4: Calculating the mass of the earth at a depth d.

While calculating the acceleration due to gravity at a depth "d" we need to consider the mass of the earth only up depth "d" from the centre of the earth because the mass that is participating in the acceleration due to gravity is only up to a depth “d” as shown by the red color in the figure.

Now we need to calculate the mass of the earth at a depth d.

The density of the earth can be calculated by,

$Density\left(\rho \right)=\frac{Mass\left(M\right)}{Volume\left(V\right)}$

Where volume $V=\frac{4}{3}{\mathrm{\pi R}}^3$

Putting the value of volume we get,

$$\begin{gathered} \rho =\frac{M}{{\displaystyle \frac{4}{3}}{\mathrm{\pi R}}^3} \\ \rho =\frac{3M}{{\displaystyle 4{\mathrm{\pi R}}^3}} \end{gathered}$$

Now, since the density will remain the same even at a depth so we can calculate the mass of the earth at depth "d",

$$\begin{gathered} Mass=Density\times Volume \\ Mass\left(M^{\text{'}}\right)=\left(\frac{3M}{4{\mathrm{\pi R}}^3}\right)\times \left(\frac{4}{3}\mathrm{\pi }{\left(\mathrm{R}-\mathrm{d}\right)}^3\right) \\ Mass\left(M^{\text{'}}\right)=\frac{M{\left(R-d\right)}^3}{R^3} \end{gathered}$$

Where, $M^{\text{'}}$is the mass of earth at a depth “d”.

Step 5: Calculating the acceleration due to gravity at a depth "d".

$g_d=\frac{G{M}^{\text{'}}}{{(R-d)}^2}$

Putting the value of $M^{\text{'}}$ in the above equation, we get,

$$\begin{gathered} g_d=\frac{G\left({\displaystyle \frac{M{\left(R-d\right)}^3}{R^3}}\right)}{{\left(R-d\right)}^2} \\ {g}_d=\frac{GM\left(R-d\right)}{R^3}.........eq\left(2\right) \end{gathered}$$

Step 6: Calculating the ratio of $dandR$

Now in the question, it is given that the value of acceleration due to gravity is the same both at height “h” and depth “d”.

Therefore, $g_h=g_d=g^{\text{'}}$ (given in question).

Now on dividing equations (1) and (2), we get,

$$\begin{gathered} \frac{g_h}{g_d}=\frac{\frac{4GM}{9R^2}}{\frac{GM\left(R-d\right)}{R^3}} \\ \frac{g^{\text{'}}}{g^{\text{'}}}=\frac{4R}{9\left(R-d\right)} \\ 9R-9d=4R \\ 9R-4R=9d \\ 5R=9d \\ \frac{d}{R}=\frac{5}{9} \end{gathered}$$

Therefore the correct answer is option (C).