How Long Would it Take to Drop an Object Through a Tunnel on the Moon?

[clow]

I have always been curious about what would happen to the gravity at the center of the earth. After some googling I found some neat stuff. So IF it was possible to endure the heat and pressure to get there (impossible) it would feel weightless. So that answered my question but then I read it would only take 42 minutes to drop something and have it come out the other side, then of course fall back.

Now the other part comes when I hear since the moon is cold and frozen so a tunnel would be more possible than Earth (still almost impossible). So the question I had was since the moons diameter is smaller than earth, but the gravity is less how long would it take for an object to make to the other opening of the tunnel before falling back.

Gravity: 1.63 m/s^2

Diameter: 3476 KM

I have had very little physics but and even less calculus in my background. Any help on where to start with this?


Thanks,

Clow

 

[Bill_K]

Clow,

Actually, the answer is a very interesting result. This is going to take a little bit of physics, but no calculus. The force of gravity acting on the object will be F = GMm/r² where G is the gravitational constant, M is the Earth's (or moon's) mass, and r is the object's distance from the center.

However the effective value for M is not the entire Earth. It's really M(r). Only that part of the Earth which is below you counts. In other words, once you've fallen 1000 miles, say, the outermost 1000 miles of the Earth is a hollow sphere over your head, and its gravitational pull on you cancels itself out. Only the innermost 3000 miles is still pulling you.

So using the volume of a sphere, M(r) = 4/3 πρr³ where ρ is the Earth's density. (I'm going to assume that the Earth has a uniform density all the way down, although it really does not.) Newton's law is F = ma, and combining things we have a 1/r² in F and r³ in M, which together make F ~ r:

a = - (4/3 πGρ) r

(There was an m on each side, which I canceled.) Notice from this equation that as we already guessed, the force decreases as you go down. By the time you get to the center r = 0 the force is zero. Now why is this equation interesting? Because it's a familiar one, and writing down the solution is easy! It's what's called a simple harmonic oscillator, and it works just like the motion of an oscillating spring. It's as if the falling object was on the end of a spring attached to the Earth's center, with spring constant k = 4/3 πGρ. The object will oscillate back and forth, up and down the tunnel, with frequency given by ω = 1/sqrt(k) = 1/sqrt(4/3 πGρ). The period of the motion is T = 2π/ω. Of course the time it takes to fall through the Earth all the way through once is half of that. Well you can work out the value and check that you get 42 minutes.

Ok, now here's what I think is really interesting about this problem. The answer does not depend on the diameter of the planet, whether it's Earth you're talking about or the moon. All it depends on is the density! I looked it up, and ρ = 5.51 g/cm³ for the Earth and 3.34 g/cm³ for the moon. So if you say it takes 42 minutes to fall through the Earth, I can tell you it must take 42 x sqrt(3.34/5.51) = 32.7 minutes to fall through the moon.

 

[Lsos]

Another interesting thing is that it will take the same time to pass through ANY straight tunnel through the Earth (with gravity being the propelling force, and no friction). The tunnel need not go through the center, it just needs to be straight. It can even exit only a few miles away from the entrance.

 

[JaredJames]

I have always been curious about what would happen to the gravity at the center of the earth. After some googling I found some neat stuff. So IF it was possible to endure the heat and pressure to get there (impossible) it would feel weightless. So that answered my question but then I read it would only take 42 minutes to drop something and have it come out the other side, then of course fall back.

Just a note, it's actually ~37 minutes. We had this in a thread not so long ago and produced some basic calcs that gave us a much more accurate result.

I'll try to find it for you.

 

[clow]

Oh, that is very interesting. The equations are coming back to me. Maybe not how to use them all at once but in general.

Thanks for the input!

 

[Bill_K]

Just a note, it's actually ~37 minutes. We had this in a thread not so long ago and produced some basic calcs that gave us a much more accurate result.

I agree with clow's original value. From the results above, the tunnel time is π sqrt(R/g), where R is the Earth's radius and g is the acceleration due to gravity. Using R = 6.378 x 10⁸ cm and g = 978 cm/s² (equatorial values from Wikipedia) it comes out to be 42.28 minutes.

 

[JaredJames]

I agree with clow's original value. From the results above, the tunnel time is π sqrt(R/g), where R is the Earth's radius and g is the acceleration due to gravity. Using R = 6.378 x 10⁸ cm and g = 978 cm/s² (equatorial values from Wikipedia) it comes out to be 42.28 minutes.

Yes, but our results compensated for the varying g value as you pass through the earth. g isn't constant!

EDIT: Original thread: https://www.physicsforums.com/showthread.php?t=471524

I get 38.19 minutes using a 1000 step leapfrog integrator and using piecewise linear interpolation to determine gravitational acceleration at some point inside the Earth.

 

[Bill_K]

If you read any of the above, you know that we have already taken the variation of gravity into account. The final result is expressed in terms of the surface value g for convenience, but that doesn't mean we assumed it was the same all the way down.

 

[JaredJames]

If you read any of the above, you know that we have already taken the variation of gravity into account. The final result is expressed in terms of the surface value g for convenience, but that doesn't mean we assumed it was the same all the way down.

Please see the updated link I provided above. It is not 42 minutes.