# Average acceleration problem:

sonhouse
Posers and Puzzles 04 Oct '07 14:24
1. sonhouse
Fast and Curious
04 Oct '07 14:24
With our discussion of the idea of driving a tunnel through the center of the earth, evacuating the air for said tunnel, and dropping an object, maybe a train loaded with passengers and freight, it reaches some maximum velocity dead center of the earth and then is decellerated with the exact same value of force and it reaches the other end of the tunnel on the surface with zero velocity, so you apply the brakes, unload, reload, and you are good to go again. Ignoring the engineering problems in such an impossible venture, ignoring friction, since we specified an evacuated tunnel, what is the average acceleration value and thus what is the final velocity as it hits the center of the earth. I made one assumtion that allowed me to calculate a number but I don't know if my assumtion is correct:
I visualized a graph of the mass as you go down and you see that immediately you have mass above you and below you so the accel rate has to start going down, so I artificially pegged the average at 50% of surface acceleration numbers, 32 Ft/sec^2 or 9.8 M/sec^2 and divide those by two and you get 16 Ft/sec^2 or 4.9 M/sec^2 and simplifying the distance to the center of the earth to 4000 miles or 6400 Km, I came up with a velocity at the center of just under 8,000 meters per second or 26,000 F/sec or 17,000 miles per hour.
Is my assumtion of the average at 50% correct? It might be the square root of it or something, don't know. Any help? How would you calculate the total trip time, surface to surface?
2. AThousandYoung
A Hispanic Invention
04 Oct '07 14:27
Originally posted by sonhouse
With our discussion of the idea of driving a tunnel through the center of the earth, evacuating the air for said tunnel, and dropping an object, maybe a train loaded with passengers and freight, it reaches some maximum velocity dead center of the earth and then is decellerated with the exact same value of force and it reaches the other end of the tunnel on ...[text shortened]... ething, don't know. Any help? How would you calculate the total trip time, surface to surface?
Average acceleration is 0, since the speeding up and slowing down exactly cancel. Maximum velocity shouldn't be too hard to figure out but I don't have time right now.
3. 04 Oct '07 15:28
Average acceleration = (9.81t+-9.81t)/t = 0
Time to centre = Root(2(6,400,000)/a)= 1142seconds
Velocity at centre = 11203ms^-1
4. sonhouse
Fast and Curious
04 Oct '07 15:29
Originally posted by AThousandYoung
Average acceleration is 0, since the speeding up and slowing down exactly cancel. Maximum velocity shouldn't be too hard to figure out but I don't have time right now.
If the average accel is zero, how can it move? It starts out at max, 1G, going down till it reaches zero G in the center, I think you mean the average is zero for the entire trip, starting at one G, going down to zero G at the center then accel going negative to a max of -1G at the other end. I am talking about the average from the surface to the center, that should be somewhere around 0.5 G.
5. 04 Oct '07 15:47
You need to solve a differential equation:

d^2x/dt^2 = g(R-x)^3/R^3,

x = 0 at t = 0.
R = radius of Earth, g = acceleration due to gravity at surface.

Good luck 🙂
6. 04 Oct '07 18:441 edit
Hi,

Assuming the earth's density is constant, then the train's motion is a simple harmonic motion, with period equal to the time a satelite in VERY low orbit (height=0) takes to circle the earth. Actually, if you project the satellite onto thr tunnel, the projection of the setellite is the train's position.

This means the time it takes to reach the center of the earth is about 21 minutes and it's speed there is about 7.9 km/sec, so the average acceleration is about 6.2 m/sec^2, or (2/pi)g.
7. AThousandYoung
A Hispanic Invention
04 Oct '07 19:27
Originally posted by sonhouse
If the average accel is zero, how can it move? It starts out at max, 1G, going down till it reaches zero G in the center, I think you mean the average is zero for the entire trip, starting at one G, going down to zero G at the center then accel going negative to a max of -1G at the other end. I am talking about the average from the surface to the center, that should be somewhere around 0.5 G.
On average it does not move. It goes down, comes back up, and we're back to the beginning. Then again, and again, and again...always ending up at the same place.
8. AThousandYoung
A Hispanic Invention
04 Oct '07 19:28
Originally posted by sonhouse
If the average accel is zero, how can it move? It starts out at max, 1G, going down till it reaches zero G in the center, I think you mean the average is zero for the entire trip, starting at one G, going down to zero G at the center then accel going negative to a max of -1G at the other end. I am talking about the average from the surface to the center, that should be somewhere around 0.5 G.
Oh, ok, I understand now.
9. PBE6
Bananarama
05 Oct '07 14:31
Originally posted by David113
Hi,

Assuming the earth's density is constant, then the train's motion is a simple harmonic motion, with period equal to the time a satelite in VERY low orbit (height=0) takes to circle the earth. Actually, if you project the satellite onto thr tunnel, the projection of the setellite is the train's position.

This means the time it takes to reach the ce ...[text shortened]... eed there is about 7.9 km/sec, so the average acceleration is about 6.2 m/sec^2, or (2/pi)g.
That's a really interesting way of looking at things! I was trying to do an integral of the forces over each horizontal layer of the Earth above and below the train's position, but I kept getting indeterminate expressions in the answer. 🙁

How do you prove that the low orbit and the straight line path take the same amount of time to reach the other side? And can you apply this to other situations where you're trying to find the gravitational force at the centre of an object? I've always wondered what the gravitational strength is like at the centre of the Earth.
10. 05 Oct '07 15:122 edits
Originally posted by PBE6
That's a really interesting way of looking at things! I was trying to do an integral of the forces over each horizontal layer of the Earth above and below the train's position, but I kept getting indeterminate expressions in the answer. 🙁
There's a useful result (that Newton worked out) that the gravitational effect of a uniform shell is zero everywhere within the shell. That means that at any point you can ignore all the mass 'above' you, and just consider the effect of a smaller planet made up of the bits closer to the centre than you.

Which is where my differential equation came from, except I missed something out.

If you're a distance r from the centre of the planet, the mass of this smaller 'planet' proportional to r^3. But you're closer to the centre (the bit that I missed), and the acceleration due to gravity is proportional to the 'mass'/r^2.

So the force on you is proportional to r. Or, the acceleration due to gravity equal to gr/R, where g is the acceleration due to gravity at the surface of the Earth, and R is the radius.

So d^x/dx^2 = g(R - x)/R,

which is an equation for simple harmonic motion, with the solution:

x = R[1 - cos(t.sqrt(g/R))]

So the time taken to reach the centre would be PI/2*sqrt(R/g).

Radius of the Earth is something like 6,300km, so the time taken is around 21 minutes. Which is what David said. Which means I probably got it right this time 🙂
11. sonhouse
Fast and Curious
05 Oct '07 15:20
Originally posted by David113
Hi,

Assuming the earth's density is constant, then the train's motion is a simple harmonic motion, with period equal to the time a satelite in VERY low orbit (height=0) takes to circle the earth. Actually, if you project the satellite onto thr tunnel, the projection of the setellite is the train's position.

This means the time it takes to reach the ce ...[text shortened]... eed there is about 7.9 km/sec, so the average acceleration is about 6.2 m/sec^2, or (2/pi)g.
So it must be going slower than the orbit of a low satellite, because the satellite would be going around the circumferance and the tunnel object would be going the diameter, if they do that in the same amount of time, wouldn't the tunnel object be going slower because it is covering less distance than the 90 degrees of travel of the satellite?
Simplifying the diameter of the earth to 8,000 miles or 12,800 Km times PI, makes the circumferance 25,000 miles and change or 40,000 Km. So 1/4th of that=90 degrees of travel=6250 miles or 10,000 Km covered in the same time the tunnel object goes only the radius or 4,000 miles or 6400 Km. So a satellite in low earth orbit is going about 5 M/S or about 18,000 MPH or 28,800 Km/hour which is about 83 minutes per orbit, divide by 4= 21 minutes like you said, and if the tunnel object goes 6250Km in 21 minutes then it would work out to 17,800 Km/hr average velocity or 11,100 MPH. I think that agrees with your numbers then. Interesting, it shows a satellite is really continuously falling and just happens to be taking a curve to do it where an object dropped in a tunnel takes the same time to make the radius trip than the satellite quarter orbit. Nice work!
12. 05 Oct '07 15:26
Originally posted by sonhouse
So it must be going slower than the orbit of a low satellite, because the satellite would be going around the circumferance and the tunnel object would be going the diameter, if they do that in the same amount of time, wouldn't the tunnel object be going slower because it is covering less distance than the 90 degrees of travel of the satellite?
Moving slower, but it's got the same velocity in the direction of the tunnel.

If you work out the force in that direction of a body on the surface, it equals the force of a body in the equivalent point in the tunnel, which confirms the analogy works.
13. sonhouse
Fast and Curious
05 Oct '07 19:25
Originally posted by mtthw
Moving slower, but it's got the same velocity in the direction of the tunnel.

If you work out the force in that direction of a body on the surface, it equals the force of a body in the equivalent point in the tunnel, which confirms the analogy works.
The only thing about that analogy that bothers me is this:
The satellite in its motion undergoes a constant acceleration, the vector never changes, so the accel is equal anywhere around the orbit. The thing about the force on the falling body is, unlike when you are say 1000 miles high and drop, you are going to a higher accel rate the lower you go but in the tunnel you start out at 1G and it changes all the way down till at the center you are at 0G but with the accumulated velocity. So the gist is, in spite of this variable accel force, if you had a magic marker that could follow the movement of the object in the tunnel and make a pretend laser beam tangent to the surface, 90 degrees from the fall, an object in orbit would remain in the beam of the laser all the way round the quarter circle? You see what I mean, the beam is stuck to the falling object and shining 90 degrees, at the surface it would be tangential to the surface of the globe and contacting a satellite, so it would track that satellite all the way to the 90 degree point? And by analogy, keep tracking it as the object in the tunnel passes the center and is now decellerating, our magic laser beam still tracks the satellite and would still shine on it as it gets to the surface on the other side of the planet?
14. AThousandYoung
A Hispanic Invention
07 Oct '07 23:12
Here's a site that helps explain the relationship between simple harmonic motion and orbit.

http://www.cbu.edu/~jvarrian/applets/sho1/harmo_g.htm
15. sonhouse
Fast and Curious
09 Oct '07 09:55
Originally posted by AThousandYoung
Here's a site that helps explain the relationship between simple harmonic motion and orbit.

http://www.cbu.edu/~jvarrian/applets/sho1/harmo_g.htm
Nice visual but how can you slow it down? I hit the mouse when I am in the circle but all it does is stop and then start up again when I let go. I would like to see it just go about ten times slower. Good visual anyway.