Ever wanted to get from America to China with incredible speed? Well, ok, so you don't have to go to China. The fact is, you can go anywhere on the earth in under an hour. "But how?" you ask. Well, just sit back and let me explain.

Step one: dig a hole straight from point A (your current location) to point B (your destination.) How you accomplish this first step is up to you. I suggest hiring an army of gophers to do your dirty work. Oh yeah, you'll need to take all the air out of this new tunnel too. In other words, for the sake of simplicity, any kind of friction (including air resistance) will be left out of the solution.

Now on to the explanation. When you're on the surface of the earth, it's known that things fall at a rate of 9.8 meters per second per second. This is because the gravity of the earth below you is pulling down. But, what if you were at the center of the earth? All the earth's mass would be pulling on you from all sides, basically making the rate at which you fall zero meters per second per second because you're being pulled from every which way by the mass around you. Well, it turns out that anywhere inside a symmetrical hollow sphere, of any mass and radius, that the acceleration due to gravity is zero. As you fall through your new hole in the earth, this means you can divide the earth into two parts: the hollow sphere surrounding you, and the inner part that you're right "above." It is only this inner part that pulls on you (because of gravity,) and as you fall deeper and deeper into the earth this inner part of the earth gets smaller and smaller. That means the acceleration of gravity is also getting smaller. This rate of change of the acceleration of gravity is linear, as you approach the earth's core. What does that mean? At the surface of the earth, you fall at 9.8 m/s2. Halfway to the earth's core your acceleration due to gravity is 4.9 m/s2. At the core it's 0 m/s2.

The result is that acceleration of gravity is proportional to the distance from the center of the earth, at a linear rate. This is very similar to how a spring works. As you pull back on the spring the force pulling back on you increases at a linear rate. In fact, if you jumped in the hole, as soon as you reach the other side you'd begin to accelerate back to your original position (just like a spring,) because all the earth's mass is now on the other side of you. So, the force of gravity equals the distance from the earth's center times some constant k. Now, even if the hole does not go through the center of the earth, the acceleration of gravity will still be linear and it will still vary based on the distance from the center of the earth. The acceleration of gravity will simply "bottom out" when you are closest to the center of the earth. You would travel at a slower rate, but you would also have a shorter distance to travel. As a result, the time it takes to get there stays constant.

Because the earth is acting as a spring, other equations that apply to springs can also be used as well. The one that solves this particular problem is: T=2π√(m/k), where T is time of one period (of a spring), m is the mass of an object attached to the spring (you,) and k is the spring constant (which was solved for above.) (You'll just have to take my word for it that this equations works. I don't want to get off topic.) Now, T is one period, which means it takes T/2 to travel the diameter of the earth (or 1/2 a period.) Simply plug in known constants, and you get your answer.


See the more mathematical version of the proof below.

There are six equations that are needed to solve this problem:

F = ma
Where F is force, m is mass, and a is acceleration.

Fg = (GMm) / R2
Where Fg is the force of gravity, G is the gravitational constant, M is the mass of object one, m is mass of object two, and R is the distance between the center of both masses (the distance between you and the earth, in this case.)

M = ρV
Where M is the mass of the earth, ρ is the average density of the earth, and V is the volume of the earth. (Note: the average density of the earth will be used for simplicity's sake.)

V = (4/3)πR3
Where V is volume of the earth and R is the radius of the earth.

F = kx
Where F is force, k is a constant, and x is the distance from the center of the earth.

T = 2π√(m/k)
Where T is the time of one period in simple harmonic motion, m is the mass of an object attached to a spring (you, in this case,) and k is a constant.

G = 6.67259 × 10-11 Nm2/kg2
ρ = 5520 kg/m3


Fg = (GMm) / R2 = ma
...divide by m...

(GM) / R2 = agravity
...substitute ρV for M...

(GρV) / R2 = agravity
...substitute (4/3)π R3 for V...

(Gρ(4/3)πR3) / R2 = agravity
...R3 over R2 becomes R3-2, or just R...

So, we know that the acceleration of gravity at any distance from the center of earth for any object is Gρ(4/3)πR. Note: this equation is only true when measuring acceleration of gravity inside or at the surface of the earth.

For the next part of the problem it is necessary to recognize that the force exerted on an object is proportional to it's distance from the center of the earth. This means the equation F = kx can be used.

F = magravity = kx
...Replace agravity with (Gρ(4/3)πR)...

m(Gρ(4/3)πR) = kx
...R and x are the same thing, the distance from the center of the earth, so they cancel out...

Therefore, k = m(Gρ(4/3)π).

Because F = kx, the problem can be thought of as a "simple harmonic motion" problem. So, the equation T = 2π√(m/k) can be applied.

...replace k with Gρ(4/3)π...
T = 2π√( m/(mGρ(4/3)π) )
...mass cancels...

T = 2π√(Gρ(4/3)π)

So, the time it takes to travel to the other side of the earth is equal to ½T (one half a period.) The only variable that T is dependent is the density of the earth, which has already been discovered. Simply plug in the numbers and you'll get (approximately) 3382 seconds. That is equal to 56.37 minutes.

Log in or register to write something here or to contact authors.