Tunnel across the Earth

The acceleration of a body of mass $m$ inside Earth at a depth $x$ is given by

$a = -\dfrac {GM_e}{R_e^3} (x) \Rightarrow F = ma = \dfrac {GM_em}{R_e^3} (x)$

Comparing $F$ with general equation of SHM $F_{SHM} = k (x)$

$k = \dfrac {GM_em}{R_e^3}$

And Time Period is given by, $T = 2π \sqrt {\dfrac mk} = 2π \sqrt {\dfrac {R_e^3}{GM_e}}$

We know that, acceleration due to gravity $g = \dfrac {GM_e}{R_e^2}$

$T = 2π\sqrt {\dfrac Rg} = 2π \sqrt {\dfrac {6.4 \times 10^6}{9.8}} = \boxed{5075.00638383} seconds$

Another interesting way of getting here is to use the UCM (uniform circular motion) whose projection gives this SHM. Both are governed by the Earth's gravity. The UCM gives a satellite grazing the Earth's surface. Since one is the projection of the other, they will have the same period. Hence we can use the satellite period formula -

$T = 2 \pi \sqrt{\frac{R^3}{GM}} = 2 \pi \sqrt{\frac{R}{g}} = 2(3.14)\sqrt{\frac{6.4 \times 10^6}{9.8}}=5075.006$

Time Period= 2pi (root L/g)

Substituting values, we get T= 5075