# Physics Teasers : Part 1

1. Dec 20, 2004

### Gokul43201

Staff Emeritus
THE RULES :

1. Each person may attempt only one teaser, total, unless I announce otherwise.

2. If teasers remain unsolved after a significant length of time ( a couple of days), I will announce a 'free for all', at which time anyone (including those that have already answered) may attempt any number of the remaining questions.

3. Anyone may attempt the bonus question, even if they have already tried another question.

Note :

1. Some of these teasers require back-of-envelope type calculations, or 'order of magnitude' estimates.

2. All teasers are roughly aimed at the college/grad level. Some are accessible to high schoolers.

THE TEASERS :

1. When liquid nitrogen or dry ice evaporates/sublimes, you can see thick, white fumes. If nitrogen and carbon dioxide are really white gases, why is air colorless ?

2. A planet forms from a large, cold cloud of dust. Neglecting radiation, find the radius of the planet (order of mag.) if its temperature keeps the planetary material at about its melting point.

Show working/explanation.

3. Experimental particle physicists have elaborate experiments set up (some, deep down in modified mine shafts) to observe proton decay. Derive a lower limit (very rough number) on the lifetime of the proton on the based on the fact that you are alive.

Assume that a dosage of about 600 rad is lethal. Also assume that a proton always decays into a positron and a neutral pion. Use Google for decay info or for conversions.

4. Sound can not propagate through vacuum.

In a popular demonstration of this, an electric bell is set up inside a bell-jar, which is then evacuated using a vacuum pump. When the air pressure drops below about 1cm of Hg, the bell is no longer audible. Voila !

Show why this demonstration is spurious, and, if you can, provide the correct interpretation.

5. BONUS QUESTION :

The earth is not a sphere, but is, approximately, an oblate spheroid, with its equatorial diameter a about 13 miles (21 km) greater than its polar diameter b. Modeling this in 2 dimensions, as an ellipse, you have $x = a~cos \theta, ~~y=b~sin \theta$. Writing $a-b=\delta$ and neglecting squares of $\delta /a$, the equation yields $r = \sqrt{x^2 + y^2} = a - \delta sin^2 \theta$, where $\theta$ is the latitude. This gives $r_{30} - r_{50} = 4.6~$mi (about 7.4 km). So a point on the sea level at a latitude of 30N is about 4.6 miles higher than a similar point at 50N.

Now the Mississippi River flows from roughly 50N to about 30N, where it drains into the Gulf of Mexico. The altitude at the source is about 1.5 miles above mean sea-level, and the mouth is obviously at sea-level. Now, what this means is that the river has effectively climbed a height of about 3 miles in its journey to the sea.

But we all know that water can not flow upwards against gravity, from a smaller height to a greater one. So the Mississippi is a figment of our imagination.

That's all for now...more coming up after this.

Last edited: Dec 20, 2004
2. Dec 20, 2004

### DarkEternal

i am going to take a quick guess at the bonus question. would it have anything to do with the centrifugal fictional force caused by the earth's rotation?

3. Dec 20, 2004

### Gokul43201

Staff Emeritus
I won't say whether or not the bonus question has been answered correctly, till the end (say in a few days or so). I want everyone to give it a shot.

4. Dec 21, 2004

### NoTime

1-> Condensed water vapor

5-> Um. You pretty much give the answer in the question.
>So a point on the sea level at a latitude of 30N is about 4.6 miles higher than a similar point at 50N.
Sea level defines the gravitation equipotential surface.

5. Dec 21, 2004

### Gokul43201

Staff Emeritus
No Time, you're right on Q1. But now, you may not attempt any more until I throw it open to all.

No comment on Q5.

No takers for 2, 3, 4 ? Fraidy cats, eh ?

6. Dec 21, 2004

### hemmul

N4:

-- [select]
hmm...
a bell can't hang in the air freely. it needs some support (a basement, a rope) e.i. something connects it to a surface of the bell-jar. The sound propagates directly through this material, and reaches us (of course it is slightly disrtorted). Now the only way out i see is to "hang" the bell inside the jar with a magnet or something, avoidint the direct contact ;)

-- [deselect]

7. Dec 21, 2004

### hemmul

N5:

actually the heights that differ are calculated respectively to the geometrical center of the ellipse (and there is no surprise in that) If that was so for a sphere, it would be strange - because the potential phi~1/r. Now what about ellipse?
Consider an ellipse with axes a & b. now lets expand/shrink the coordinate system (e.i. b/a-factor scaling transformation) via OX axis so, that the ellipse transforms into a sphere of a radius b. everywhere on the surface of a sphere the potential~1/b. now "i don't think" that the fact of the surface "equipotentiality" will change after a backward transformation to the ellipse ("i don' think" - because i believe it can be prooved, making certain substitutions in the potential integral for an ellipse, but i have no back-of-envelope nearby right away :rofl:)
So, the potential everywhere on the surface is same (no wonder - the ocean water chooses least-potential states), and there is no potential-energy losses during Misissipi's voyage! However, the height relatively to the centre of the Earth does change.

maybe i have actually never seen it...

8. Dec 21, 2004

### Gokul43201

Staff Emeritus
hemmul,

You've made an interesting point for Q4, but that would only be a reason for you to hear the bell ringing, not the opposite.

9. Dec 21, 2004

### NateTG

The paradox makes the tacit assumption that surfaces of gravitational equipotential on the surface of the earth are spherical while explictly modeling the earth as an elipsoid, but elipsoids generally do not have spherical surfaces of equal gravitational potential.

Moreover, there is an additional force that is not being accounted for: the earth is a rotating reference frame, and the Mississippi is traveling with the action of the centrifugal force.

10. Dec 22, 2004

### Bartholomew

4.)I'm going to go out on a limb and say that the demonstration is valid. I was thinking that the speed of sound might have been decreased by the lowered air pressure and that this speed change could cause the sound to be reflected internally within the bell but wikipedia tells me that air pressure has no effect on the speed of sound. With that idea discarded, I can't think of any reason why there would be silence except that as the air becomes vacuumlike there are just fewer particles to transmit the sound.

11. Dec 23, 2004

### DaveC426913

4. Well, I don't know about 'spurious', it certainly demonstrates the point in layman's terms (you said "...popular demonstration..."), but it is not rigorous science, no.

For one, it is not in vacuum, it is only in partial vacuum. You cannot make more than a good partial vacuum on Earth using pumps.

12. Dec 23, 2004

### DaveC426913

4. Waitaminnit, when offered the "sound wuld be transmitted through the support" explanation, you responded with "...only be a reason for you to hear the bell ringing, not the opposite..."

So, you're looking for an answer that shows that:
- you would not be able to hear the bell ringing, AND
- that is not proof that sound doesn't travel through a vacuum
?

Which means you propose that the bell is not heard for some other reason.

Is this correct?

13. Dec 23, 2004

### Gokul43201

Staff Emeritus
Yes, that's right.

Bart came close, but I don't have time to explain now. Got a flight to catch.

14. Dec 23, 2004

### Rogerio

4.
In some situations, the sound transmitted through the air could be cancelled by the sound transmitted through the solid material of the bell-jar. This effect depends on some parameters (phase and transmission coeficients) but it could be possible.

Last edited: Dec 23, 2004
15. Dec 23, 2004

### NoTime

Well, since Gokul is flying off somewhere I guess he won't mind me saying that's not it for him.

16. Jan 12, 2005

### Rogerio

Hi Gokul,

17. Jan 12, 2005

### Gokul43201

Staff Emeritus

18. Jan 12, 2005

### NoTime

Does that mean I can answer them now

4) Bell frequency change.

3) Opinions to the contrary, protons do not decay.

I think the first time I saw this idea I think the estimate was 10^17 years.
Not sure what its up to now, ?10^133 or something?
Perhaps we would have acquired some of the properties of radiodurans.

2)This one I originally thought there was not enough info for.
Now I'm thinking that you might have the idea of using something like the gas law, then cross this with the mass required to provide the appropriate pressure.

19. Jan 12, 2005

### Gokul43201

Staff Emeritus
This one has been answered correctly by NoTime.

The white fumes are from the condensation of water vapor due to the local cooling of the air below its dew point.

That was just to get you warmed up. This is the only question among 1,2,3,4 that has satisfactorily been answered.

Since I completely forgot about this thread and I promised a "free for all" after a certain period of inectivity, this is that free-for-all. Anyone may attempt any number of questions among 2, 3, 4, 5.

I'll acknowledge correct solutions, or else post my solutions around this time tomorrow. Remind me, if I forget.

Last edited: Jan 12, 2005
20. Jan 16, 2005

### Gokul43201

Staff Emeritus
The gravitational potential energy lost by the dust cloud upon coming together must go into thermal energy gained by the planet thus formed.

The gravitational PE of a sphere of mass M, radius R, is $\frac{3GM^2}{5R}$. Since the initial cloud was considered large, its PE ~ 0.

If you didn't know this you could have either derived it or gone with the rough value $\frac{GM^2}{R}$.

$$\frac{3GM^2}{5R} = MCT_m~, ~~M = \frac{4\pi R^3 \rho}{3}$$

$$=> R^2 = \frac{5CT_m}{4\pi G \rho }$$

Now plugging in typical (rough) values

$$C = 10^3~ J/K \cdot kg~,~~ T_m = 2*10^3~ K~, ~~\rho = 5*10^3~ kg/m^3$$

Gives : $$R = (about)~2*10^6 ~m~or~2000~km$$

Notice that this is smaller than the radii of most planets in the solar systems (Radius of earth ~ 6000 km), suggesting that they were all mostly molten when they formed, but have eventually cooled off through radiation. Indeed the inner core of the earth is still molten.