# Integration of acceleration

1. May 13, 2015

### Owl2256

1. The problem statement, all variables and given/known data
Movies are always chock-full of scientific inaccuracies. As educated people, it’s our duty to ruin all the fun of it by taking the time to work out what should happen. For this problem, let’s look at the transport in the 2012 remake of “Total Recall.”

In the movie, there are only two places to live, the UK and Australia, due to toxic pollution everywhere else in the world. Since they’re literally on opposite sides of the Earth, people commute between the locations by a gravity elevator. The movie states that the travel time between the ends of the Earth is about 17 minutes.

Assuming a constant acceleration of 9.8m/s2 through the entire fall and starting from rest, how long would it take to fall the diameter of the Earth? Derive your result from integration of the acceleration.Assume object is dropped in a vacuum, the Earth is not rotating, and there’s enough space for that long of a drop.

2. Relevant equations
The radius of the Earth r0 = 6.37 × 106 meters.

3. The attempt at a solution
I took the integral and i got 4.9 m^2/s^2, but I'm not sure what that means or what comes next. it asks to derive the integral so i feel like that is a double negative and in the end you would still have 9.8 m/s^2 so I'm really confused.

2. May 13, 2015

### rude man

This is the old "hole-thru-the-earth" SHM problem. Solve the 2nd-order ODE to get T.

3. May 13, 2015

### Owl2256

my teacher has barely touched upon differential equations what do you mean by solve the second order ODE to get time? I have only done first order so far.

Last edited: May 13, 2015
4. May 13, 2015

### ehild

You integrated acceleration with respect to what?
The first question is to get the time assuming constant acceleration of 9.8 m/s2 Recall how acceleration is defined.

5. May 14, 2015

### SammyS

Staff Emeritus
Well, it's SHM if you assume uniform density of the Earth, but the problem here says to find the time of transport assuming a constant acceleration of 9.8m/s2 .

Taking those instructions literally will give a time that's much too small.

Your instructions say to do this by integration of acceleration.

You need to integrate acceleration, with respect to time to get velocity, (as a function of time).

Integrate that to get distance as a function of time. You just get the usual kinematic equations.

You need to know the distance through the center of the earth to get your answer for time.

Furthermore, it appears that in your attempt, you integrated acceleration with respect to distance, but acceleration is not a derivative with respect to distance.

Last edited: May 14, 2015
6. May 14, 2015

### SammyS

Staff Emeritus

7. May 14, 2015

### rude man

SHM ODE is m d2r/dt2 = -kr.
r is distance from earth's center. Initial conditions for you are r=R, dr/dt=0. R is radius of earth. You assumed the force is constant at -mg but that is incorrect. The closer you get to the center (r=0) the weaker the force.
You should be able to calculate what k is. It will be due to all of earth's mass inside r. (Mass outside r produces zero force.)
So this is just a mass-spring thing, aka SHM. You'll get something like r(t) = A cos(ωt) where A results from the initial conditions and ω is a function of k. The crucial thing is that k is indeed a constant for all r<R. (You don't care about r>R.) Your answer will be half the period of the oscillation of the elevator dropping from the U.K. to Australia and back.

8. May 14, 2015

### SammyS

Staff Emeritus
The motion is SHM only if uniform density is assumed for the earth.

The problem asks for a (very under-estimated) lower bound for the time in question, by assuming a constant acceleration of 9.8 m/s2 for the entire distance traveled.

9. May 14, 2015

### rude man

Obviously.
I hadn't noticed that. That assumption is so absurd it should never have been posited in the first place. But OK, that's what it posited.

10. May 14, 2015

### Ray Vickson

The assumption is especially absurd in a post that claims to be checking for scientific accuracy in movies! (Note: I am not blaming the OP, who just follows orders.)

11. May 14, 2015

### rude man

Agreed, 100%.

12. May 14, 2015

### SammyS

Staff Emeritus
Well, it does turn out that even with this wild assumption which should give an absurdly small time, that time is somewhat greater than 17 minutes. (Also, the velocity as the "elevator" emerges on the far side would be more than escape velocity for earth!)

A slightly more reasonable, but still very rough lower bound for the time would result from using constant acceleration of 9.8m/s2 to Earth's center, then -9.8m/s2 to the far surface. That way the velocity at the surface would be 0. That total time is more than twice the 17 minutes quoted in the movie.

13. May 14, 2015

### rude man

well, ya gotta step out quick and hang onto something solid!
I think that's what they had in mind. g always points to earth center.[/QUOTE]