Question

what will be the value of g at the centre of the earth?

Answer 1

Answer: Acceleration due to gravity varies directly proportional to distance from center ,i.r. r

but varies 1/r2 When r>R

expression is given by g=GM/r2 if r>R

and g =GMr/R 3 if r

So when r =0 then g will be zero

Explanation: Hope it helpful

Answer 2

Value of g at centre of earth is zero.

The general expression of gravitational acceleration beneath the earth surface is:

$g' = g \times \bigg(1 - \dfrac{d}{r} \bigg)$

• g is gravitational acceleration at the surface, d is the depth, r is the radius of the earth.

Now, at the centre of the earth, the value of depth will be equal to the radius of the earth

So, we can say:

$g' = g \times \bigg(1 - \dfrac{d}{r} \bigg)$

$\implies g' = g \times \bigg(1 - \dfrac{r}{r} \bigg)$

$\implies g' = g \times \bigg(1 - 1 \bigg)$

$\implies g' = g \times 0$

$\implies g' = 0$

So, gravitational acceleration at the centre of the earth will be zero.