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This is a surprisingly simple thing to calculate.

It is a well known result that a consequence of the inverse square law is that there is no force inside a symmetrical hollow shell. This means that as the object falls into the hole, it will appear to be attracted by a sphere of decreasing radius - the mass outside "doesn't count."

The acceleration of gravity at the surface of a sphere of radius R (assuming uniform density $\rho$) is given by

$$\begin{align} a &= \frac{GM}{R^2} \\ &= \frac{4G\pi R^3\rho}{3R^2} \\ &= \frac43\pi \rho G R\\ \end{align}$$

Where $G$ is the gravitational constant, and $R$ is the distance to the center of the earth. In other words - the acceleration is proportional to the distance to the center. The corollary is that an object dropped into a hole through the center of the earth will exhibit simple harmonic motion.

Do you think you can now solve the problem yourself?

This is a surprisingly simple thing to calculate.

It is a well known result that a consequence of the inverse square law is that there is no force inside a symmetrical hollow shell. This means that as the object falls into the hole, it will appear to be attracted by a sphere of decreasing radius - the mass outside "doesn't count."

The acceleration of gravity at the surface of a sphere of radius R (assuming uniform density $\rho$) is given by

$$\begin{align} a &= \frac{GM}{R^2} \\ &= \frac{4G\pi R^3\rho}{3R^2} \\ &= \frac43\pi \rho G R\\ \end{align}$$

Where $G$ is the gravitational constant, and $R$ is the distance to the center of the earth. In other words - the acceleration is proportional to the distance to the center. The corollary is that an object dropped into a hole through the center of the earth will exhibit simple harmonic motion.

Let's do the math in more detail. Put the distance from the center as $r$, then the acceleration $\frac{d^2r}{dt^2}$ is given by

$$\frac{d^2r}{dt^2}=-\frac43 \pi \rho G r$$

(since the acceleration is pointing towards the center). This looks like the differential equation for simple harmonic motion:

$$\frac{d^2x}{dt^2} = -\omega^2 x$$

for which the solution (if velocity is zero at t=0, and amplitude is $x_0$) is

$$x(t) = x_0 \cos\omega t$$

and the velocity is

$$v(t) = -\omega x_0 \sin\omega t$$

It follows that we can write the expression for the velocity as a function of position by eliminating time:

$$\begin{align}\\ v(x) &= -\omega x_0 \sin \cos^{-1}\left(\frac{x}{x_0}\right)\\ &= -\omega x_0\sqrt{1-\left(\frac{x}{x_0}\right)^2}\end{align}$$

If we now substitute $\omega^2 = \frac43 \pi \rho G = \frac{g}{R}$ where $g$ is the gravitational acceleration at the surface of the earth, we get

$$\begin{align}\\ v(r) &= -\sqrt{\frac{g}{R}} R \sqrt{1-\left(\frac{r}{R}\right)^2}\\ &=\sqrt{\frac{g}{R}\left(R^2-r^2\right)} \end{align}$$

If we substitute $r=R$, we get $v=0$ as expected; and when we put in $r=0$, we get $v = \sqrt{gR}$ = 7904 m/s using your values for $R$ and $g$. I think that's pretty close...

This is a surprisingly simple thing to calculate.

It is a well known result that a consequence of the inverse square law is that there is no force inside a symmetrical hollow shell. This means that as the object falls into the hole, it will appear to be attracted by a sphere of decreasing radius - the mass outside "doesn't count."

The acceleration of gravity at the surface of a sphere of radius R (assuming uniform density $\rho$) is given by

$$\begin{align} F &= \frac{GM}{R^2} \\ &= \frac{4G\pi R^3\rho}{3R^2} \\ &= \frac43\pi \rho G R\\ \end{align}$$

Where $G$ is the gravitational constant, and $R$ is the distance to the center of the earth. In other words - the acceleration is proportional to the distance to the center. The corollary is that an object dropped into a hole through the center of the earth will exhibit simple harmonic motion.

Do you think you can now solve the problem yourself?

This is a surprisingly simple thing to calculate.

It is a well known result that a consequence of the inverse square law is that there is no force inside a symmetrical hollow shell. This means that as the object falls into the hole, it will appear to be attracted by a sphere of decreasing radius - the mass outside "doesn't count."

The acceleration of gravity at the surface of a sphere of radius R (assuming uniform density $\rho$) is given by

$$\begin{align} a &= \frac{GM}{R^2} \\ &= \frac{4G\pi R^3\rho}{3R^2} \\ &= \frac43\pi \rho G R\\ \end{align}$$

Where $G$ is the gravitational constant, and $R$ is the distance to the center of the earth. In other words - the acceleration is proportional to the distance to the center. The corollary is that an object dropped into a hole through the center of the earth will exhibit simple harmonic motion.

Do you think you can now solve the problem yourself?

This is a surprisingly simple thing to calculate.

It is a well known result that a consequence of the inverse square law is that there is no force inside a symmetrical hollow shell. This means that as the object falls into the hole, it will appear to be attracted by a sphere of decreasing radius - the mass outside "doesn't count."

The acceleration of gravity at the surface of a sphere of radius R (assuming uniform density $\rho$) is given by

$$\begin{align} F &= \frac{GM}{R^2} \\ &= \frac{4G\pi R^3\rho}{3R^2} \\ &= \frac43\pi \rho G R\\ \end{align}$$

In other words - the acceleration is proportional to the distance to the earth. The corollary is that an object dropped into a hole through the center of the earth will exhibit simple harmonic motion.

Do you think you can now solve the problem yourself?

This is a surprisingly simple thing to calculate.

It is a well known result that a consequence of the inverse square law is that there is no force inside a symmetrical hollow shell. This means that as the object falls into the hole, it will appear to be attracted by a sphere of decreasing radius - the mass outside "doesn't count."

The acceleration of gravity at the surface of a sphere of radius R (assuming uniform density $\rho$) is given by

$$\begin{align} F &= \frac{GM}{R^2} \\ &= \frac{4G\pi R^3\rho}{3R^2} \\ &= \frac43\pi \rho G R\\ \end{align}$$

Where $G$ is the gravitational constant, and $R$ is the distance to the center of the earth. In other words - the acceleration is proportional to the distance to the center. The corollary is that an object dropped into a hole through the center of the earth will exhibit simple harmonic motion.

Do you think you can now solve the problem yourself?