Question

Calculate the depth below the surface of the earth where acceleration due to gravity becomes half of its value at surface of the earth given radius of the earth is 6400 km

Answer 1

Let a be the gravity at depth
and of course g be the acc.due to gravity at surface
$a = g(1 - \frac{d}{r} ) \\ given \: a = \frac{g}{2} \\ \frac{g}{2} = g(1 - \frac{d}{r} ) \\ 1 = 2 - 2 \frac{d}{r} \\ 1 = 2 \frac{d}{r} \\ d = \frac{r}{2} \\ d = \frac{6400}{2} \\ d = 3200$

Answer 2

The depth will be 3200 km.

To calculate the depth, formula used is -

a = g(1 - d/r)

Since, acceleration due to ravuty is half of that at the surface, hence

a = g/2

Now, solving the equation -

g/2 = g (1-d/r)

(1-d/r) = 1/2

r = 6400 km

r-d = r/2

d = 6400 - 3200

d = 3200 km

Hence, depth at which acceleration due to gravity is half will be - 3200 km.