Assume that the earth is a solid sphere of uniform density and a...

The gravitational force acting on a particle of mass $m$ at a distance$r ( < R )$ from the centre of Earth is

$F = - G\dfrac{{M}^{'}m}{{r}^{2}}$ ....(1)

where ${M}^{'}$ is the mass of portion of Earth enclosed by a sphere of radius $r$ concentric with Earth. Here $-$ sign indicates radially inward direction of force.

Taking the density of Earth to be uniform

$\dfrac{{M}^{'}}{4{\pi r}^{3}} = \dfrac{M}{4{\pi R}^{3}}$

$⇒ {M}^{'} = \dfrac{M{r}^{3}}{{R}^{3}}$ ....(2)

Substituting ${M}^{'}$ from equation 2 into equation 1:

$F = - \left(\dfrac{GMm}{{R}^{3}}\right)r$

Which is similar to the equation of motion $F = - kx$ with $k = G\dfrac{Mm}{{R}^{3}} = \dfrac{mg}{R} ( ∵ g = G\dfrac{M}{{R}^{2}} )$

For such motion, the time period is given by:

$T = 2\pi\sqrt{\dfrac{m}{k}}$

Substituting k from above,

$T = 2\pi\sqrt{\dfrac{R}{g}}$

Putting the values, we get

$T = 2 \times 3 . 14\sqrt{\dfrac{6400 \times {10}^{3}}{10}} = 2 \times 3 . 14 \times 800 \approx 1hour24minutes$