Question

The value of acceleration due to gravity ′g′, at the earth's surface, is $10m/s^2$. Its value at the center of the earth which is assumed to be a sphere of radius ′R′ and uniform mass density is :

Answer 1

Step 1: Given data:

The acceleration due to gravity at the surface of the Earth is $10m/s^2$.

Step 2: Formula Used:

The value of acceleration at a depth h from the Earth's Surface is given by $g^{\text{'}}=g(1-\frac{h}{R})$

where,

1. g^' is the gravity at a depth of h meters from the Earth's Surface,

2. g is the acceleration due to gravity at the surface of the Earth,

3. h is the depth below the Earth's Surface and

4. R is the radius of the Earth.

Calculating gravity at the center :

At the center of the Earth, h = R, So,

$$\begin{gathered} g^{\text{'}}=g(1-\frac{R}{R}) \\ {g}^{\text{'}}=g(1-1) \\ {g}^{\text{'}}=0 \end{gathered}$$

Hence, the value of acceleration due to gravity at the center of the Earth is zero.