Assuming there is no drag, no friction and no other objects affecting you. If you drop into the earth (through a tube). Your velocity will be 7900 m/s at the center of the earth according to http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/earthole.html

I assume that was the summation of all the velocity gained as gravitational force changes/decreases such as v= (9.8* first nano-meter traveled) + (9.7...*next nano-meter traveled)+...+(0*last nano-meters traveled)

I am sure there is a way to write a function using calculus (integration) that calculates your velocity at every point as you get closer. Lets approximate the earth's radius is 6,371 km and the acceleration due to gravity on the earth's surface is 9.807 m/s^2.

I will like a function that calculates your velocity at every given distance from the earth center as you fall.

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    $\begingroup$ The acceleration as a function of depth depends on the density of the material below. For a spherical distribution, per shell theorem the only mass that causes a net attraction is in the shells of material below the falling object, i.e. the gravity will always decrease. For a homogeneous body this decrease would lead to a linear force over radius law and a harmonic motion. In practice, however, the density of the material increases, which means that the attraction decreases less than linearly. The general case can, of course, be solved with a differential equation. $\endgroup$ – CuriousOne Jul 7 '16 at 20:54
  • $\begingroup$ Since gravitational force is proportional to mass and not volume. So I assume right at the center of the earth (center of mass of the earth), all the gravitational force of all molecules of the earth's balances themselves out, therefore a net gravitational force of zero, right? $\endgroup$ – readytolearn Jul 7 '16 at 21:05
  • $\begingroup$ That's correct. In the center of mass the gravitational attraction is zero. $\endgroup$ – CuriousOne Jul 7 '16 at 21:07

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...

  • $\begingroup$ Where G is the gravitational constant and R is distance from earth’s center ? $\endgroup$ – readytolearn Jul 7 '16 at 21:22
  • $\begingroup$ Sorry - yes. I will add... $\endgroup$ – Floris Jul 7 '16 at 21:23
  • $\begingroup$ I think there is a small typo: I believe the equation given is the acceleration, not force. $\endgroup$ – Paul T. Jul 7 '16 at 21:25
  • $\begingroup$ I thought gravitational force and acceleration due to gravity is equal. since N/kg = (kgm)/kgs^2 = m/s^2 $\endgroup$ – readytolearn Jul 7 '16 at 21:27
  • $\begingroup$ @PaulT. - you are right. I wrote "acceleration" but used 'g'. Fixed. $\endgroup$ – Floris Jul 7 '16 at 21:28

A curious result of the physics involved is that the dropped mass oscillating up and down through the hole (say from North Pole to South Pole and back) would be matched exactly by a mass in a polar circular orbit at ground level.

Note that the max velocity in the answer above is the same as the circular orbital velocity.

If the object were dropped at the North Pole end, just as the satellite passed by, the two objects would cross the equator plane together and reach the South Pole together, and so on...


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