# I What does it take to spaghettify?

1. Jan 4, 2017

### Andreas C

Let's say we have a long, straight, one-molecule-thick string of connected molecules under the influence of Earth's gravity, pointing towards the center of the Earth. This string will then be subject to tidal forces. Does the length of the string have anything to do with whether or not the bonds between the molecules will break at some point, or does it only have to do with the gravitational gradient of the point where it will break (meaning that it would break anyways, regardless of the length of the chain)?

2. Jan 4, 2017

### Bandersnatch

The longer the chain, the larger the difference in gravitational attraction experienced at its ends, and the larger the tension in the whole string.

3. Jan 4, 2017

### Andreas C

Are you sure about that? Is there a mathematical justification?

4. Jan 4, 2017

### jbriggs444

If we are imagining a strand in free fall, held together by tension alone then the acceleration of the center of mass will be given by the integral of the local linear mass density multiplied by the local acceleration of gravity divided by the total mass of the strand.

The tension at the ends is obviously zero. The tension at any other point can be computed as the integral of the local linear mass density multiplied by the local acceleration of gravity minus the acceleration of the center of gravity.

5. Jan 4, 2017

### Andreas C

Over what is the integral?

6. Jan 4, 2017

### jbriggs444

Over the strand.

7. Jan 4, 2017

### Andreas C

Over the length of the strand?

8. Jan 4, 2017

### jbriggs444

You can parameterize the strand any way you please. It would be convenient to parameterize it by "distance from the center of the earth" in meters and label that parameter r. Then the linear mass density could be expressed in kilograms per meter and you could integrate from 0 up to 6 million meters above center.

If you wanted to parameterize by depth below the surface in miles then you could do that too. The mass density might then be in pounds per mile and you could integrate from 0 down to 4000 miles below surface.

Or you could come up with some other wonky parameterization that suits your fancy. For instance, you might parameterize it by molecule number and integrate from 0 at one end up to umpteen bazillion at the other end. If the molecules are packed closer together at one end than the other, this might not be the same thing as a parameterization by length. But the same integral would still work.

9. Jan 4, 2017

### Andreas C

Wait, if I integrate the local mass density like that, and the strand is of uniform density, I'm just going to get the total mass. If I divide the total mass by the total mass, I get... 1. The assumption is that it's of uniform density. So I guess that's is irrelevant, and we can just say that the acceleration of the center of mass is equal to the local g.

10. Jan 4, 2017

### jbriggs444

That's why I said to multiply each mass element by the local acceleration of gravity.

The acceleration of the center of mass will NOT, in general, be equal to local g at the center of mass.

11. Jan 4, 2017

### Andreas C

Ok, so then if I had say a strand that is 10km long, with the center being say 100km from the center of the Earth, and I wanted to compute the tension at the center, wouldn't that be 0? Intuitively that doesn't seem quite right...

12. Jan 4, 2017

### jbriggs444

Start by computing the free fall rate for the center of mass. That rate will be larger than the acceleration of gravity at the top of the strand and lower than the acceleration of gravity at the bottom.

Continue by setting the tension at either end to 0 and integrating to find how each mass element going from that end toward the other either adds or subtracts from that tension.

Suppose, for instance, that you are integrating down from the top end. The first little bit of strand is falling faster than the local acceleration of gravity. It is being pulled downward by the rest of the strand. It adds to the tension in the strand -- tension is greater at its bottom end than its top. The farther you go down the strand, the greater the tension becomes. This falls out of the second integral I had you perform.

At some point along the strand, the local acceleration of gravity will be equal to the acceleration rate of the entire strand. At that point, tension is maximum. If you keep integrating, the tension will be decreasing. Each incremental bit of strand is being pulled upward by the rest of the strand rather than being pulled downward. The tension at the bottom of an incremental segment will be less than at its top. Note well that the transition point from increasing to decreasing tension will likely not be at the center of mass of the strand.

If you keep integrating all the way to the other end, it should work out so that the final tension is again equal to zero -- if you've done the integration properly and if the prescription I've given you is correct.

13. Jan 4, 2017

### Andreas C

"Start by computing the free fall rate for the center of mass."

Where does that come into the rest of the procedure?

I'm a bit confused by it, let's try a specific example: say I have a 10km long uniform strand with the center being 100km from the Earth's center of gravity. Let's label the edge of the strand closest to the center of earth a, the center of the strand b, and the other edge c. How would find the tension at a point a) 2km away from point a, b) at point b, c) 2km from point c and d) the point where the tension is maximum. Don't worry, it's not a homework question, I just find things easier to understand if I try an example.

14. Jan 4, 2017

### jbriggs444

Have you done the first part yet? What's the fall rate of the strand as a whole?

For this you will need:

1. The mass density of the strand. Let's call it $\rho$ kg per km.
2. A formula for the local acceleration of gravity. Let's call it $kr$ for some constant k in Newtons per kilometer where r is the distance from the center of the earth in kilometers.
3. The integral of (local acceleration of gravity times the mass density) over the length of the strand -- from r=90km to r=110 km. This is the total gravitational force on the strand.
4. The total mass of the strand. $20 \rho$ kg.

Divide 3) by 4) and see what you get. (a = F/m)

15. Jan 5, 2017

### Andreas C

Oh I see now! It's more clear what you mean, I'll try it when I get home. Thanks, you've been very patient and helpful.

16. Jan 5, 2017

### Andreas C

Hmm, that's a bit odd... The dimensions of this are kg×m/s^2, not m/s^2, and k wouldn't be a constant...

17. Jan 5, 2017

### jbriggs444

You are 100 km from the center of the earth, deep into the inner core. By the spherical shell theorem, we can ignore the gravitational influence of the portion of the Earth above the point of interest on the strand.

The mass of the portion of the Earth below that point scales as the cube of the radius. The gravitational attraction from that mass scales as the inverse square of the radius. The net is that the gravitational acceleration scales directly proportionally with radius.

But you are correct that I missed a factor of inverse kilograms on k.

18. Jan 5, 2017

### Andreas C

Well, thanks for the reply, but through it all, when there was doubt, I ate it up and spit it out, I faced it all and I stood tall and did it my way, as Frank Sinatra would put it. Actually I don't think I did anything different, but anyways, here's what I did, stop me if I did something wrong:

Let's call r the distance from the center of the earth the end of the strand closest to earth, x the distance of a point of the strand from that point, and l the length of the strand. As such, the local g at any point will be GM/(r+x)^2
I first found dF, which is the force on a point with mass dm on the strand. dm/dx=ρ, therefore dm=ρ*dx, so we've got $dF=\frac{ρGM}{(r+x)^2}dx$. Now $∫dF=F$ so $F=ρGM\int_{0}^{l} \frac{1}{(r+x)^2} dx⇔F=ρGM\frac{l}{r^2+rl}$. The total mass of the strand is m=ρl, and if we divide the two we get the final result:

$a=\frac{GM}{r^2+rl}$.

As a sanity check, if we put l=0 we get the standard inverse square law for a point particle.

Is this solution right?

19. Jan 5, 2017

### jbriggs444

So you are hanging the strand somewhere outside the earth rather than buried deep in the inner core. And you are parameterizing the string by x with x=0 at the bottom end and x=l at the top end. That's all just fine.

That all looks good.

You've factored the $\rho G M$ out of the integral. That's good.
You've evaluated the definite integral, subtracted the bottom value from the top value and simplified. That looks correct.
That looks correct as well. It is independent of the density, as it should be and is dimensionally correct as it must be.

Perfect. I love it when folks sanity-check their work. Yes, that result seems correct.

Now we are ready to work the second part of the exercise. The strand is assumed to be in free fall -- not suspended from any point. The tension at each end must be zero. We assume for the moment that the strand falls together as a unit. The acceleration of every portion of the strand is identical. That is, it stays straight, does not stretch (much), does not break and, most importantly, does not bunch up anywhere. We can eventually sanity-check the bunch-up condition by making sure that tension is positive everywhere.

If we consider an infinitesimal section of the strand of length dx then this section must be accelerating at the same rate as the rest of the strand. That is, at a rate equal to $\frac{GM}{r^2+rl}$. The section is subject to a downward pull of gravity based on its incremental mass ($\rho dx$) multiplied by the local acceleration of gravity $\frac{GM}{(r+x)^2}$

By inspection it is clear that the downward pull of gravity on the top portion of the strand is too weak to account for the calculated acceleration. Similarly, on the bottom portion of the string, the downward pull of gravity is too strong. The discrepancy is made up by a gradient in the tension along the strand. The tension must change along the length of the strand.

Let us work from the bottom up. The tension on the top of an incremental section in this region will be greater than the tension at its bottom. Call the tension T. Can you see how to express $\frac{dT}{dx}$ in terms of the data you have available or have already calculated?

Can you see where to go next?

20. Jan 5, 2017

### Andreas C

I thought about it this way: let's say we have an infinitesimal segment of the strand situated at a point O. I will perform a similar process as before but now instead for two pieces of the strand from r to O and from O to l. Now we know how those two pieces would accelerate if they were not connected with the segment. We find the absolute value of the difference of that acceleration and the acceleration of the connected strand, we multiply those differences by the respective masses of the pieces, we add them, lo and behold, we got the tension on that segment. At the ends of the strand, it's obviously 0. Is this correct?

21. Jan 5, 2017

### jbriggs444

That approach is viable. But there is a catch. What is the tension on the rope in a tug-of-war when both sides are pulling with 1000 Newtons of force?

22. Jan 5, 2017

### Andreas C

2000 Newtons.

23. Jan 5, 2017

### Staff: Mentor

Nope..... That's everyone's instinctive first answer but it's not right.

The little bit of rope next to each person is not accelerating, so the net force on it zero. The tension of the rope is acting on that bit in one direction, the force the person is exerting is acting on that bit in the other direction, and the two forces are cancelling.

The best way of understanding this problem is to draw a complete free-body diagram including the forces between each person and the ground. The earth isn't accelerating because the two people are using their feet to push it in opposite directions with equal force; the people aren't accelerating because the tension of the rope on them is balanced by the force of the earth against their feet; every segment of the rope isn't accelerating because the force pulling it to the left is equal to the force pulling it to the right; and that's true even of the segment at the ends where the tension is pulling in one direction and a person is pulling in the other.

It's a good exercise to also draw a free body diagram of a person pulling on one end of the rope with a force of 1000 newtons when the other end of the rope is fastened to a solid wall. Compare the two diagrams (which end up yielding the same tension in the rope) and it will be clear what's going on here.

Last edited: Jan 5, 2017
24. Jan 5, 2017

### Staff: Mentor

There's an easy way of understanding this problem. Look at the little bit of the string at the upper end. Calculate the gravitational force on it, using $F=Gm_1m_2/r^2$. Do the same for the little bit at the bottom end. You'll get different forces because $r$ is different, so $F=ma$ tells us that unless some other force is involved, they will have different accelerations so will tend to move apart. But they're part of one string so they're constrained to move together, which is to say that they don't have different accelerations so some other force must be involved. That force is obviously the tension in the string, and to make the accelerations come out the same that force must exactly cancel out the difference between the gravitational force on the two ends.

So we conclude that the tension must be equal to and opposite to the difference between the forces on the two ends, to exactly cancel out that difference. The greater the distance between the two ends, the greater the difference between the forces we calculated using $F=Gm_1m_2/r^2$, and therefore the greater the tension in the string. So the greater the distance, the greater the force that is trying to break the string.

25. Jan 6, 2017

### Andreas C

Oh wait, there'
Ah ok, that makes sense. So I don't have to add them, I can just evaluate the force with which that segment is pulling the other upwards or downwards.