28-06-2003 08:26									#61
	Suffering from ads?

quote: Originally posted by Rogan Josh Of course the Earth's gravity affects the ISS - I didn't say it didn't. But anything inside the ISS is in the ISS's rest frame and doesn't feel gravity (just like being weightless in a freely falling elevator).

That's different. You're weightless in a free falling elevator because it does not exert a force on you. A pendulum does not require such a force, it merely requires gravitational pull.

quote: Originally posted by Rogan Josh The moon is orbiting the sun too. If fact both the Earth and the moon are orbiting the sun. The orbit of the moon around the Earth is not affected (much) by the sun because both the Earth and moon are in freefall, so in their rest-frame they don't feel the gravitational effects of the sun.

They are rotating around the sun, that means they are both accelerating (changing direction) all the time. That's not freefall.

	28-06-2003 08:28									#62
	Suffering from ads?

quote: Originally posted by Big Crunch What many people are forgetting is that if the moon and Earth are rotating around the sun. In effect, if you look at the Earth-Moon system as being stationary you can ignore the sun's attractive force as it is cancelled out by the centripetal force of a rotating frame of reference.

But you can say the same thing about the Earth-Moon system.

	28-06-2003 08:34									#63
	Support Apolyton, buy Civilization: The Boardgame

quote: Originally posted by Adalbertus I wouldn't pose the ISS question for testing, unless the students made their first step into general relativity (or it is a really messy calculation in accelerated frames of reference! In classical physics, the ISS is not an inertial system, in GR, it is).

Angular acceleration doesn't count in GR?

	28-06-2003 11:27									#64
	Got spare money?

ixnay -

quote: Actually, he's right. The forces between you and the earth are the same when you jump in the air. But since force = mass * acceleration, and the earth has so much more mass than you, the acceleration of the earth is negligible compared to your acceleration.

Does that mean if we could slow down my return to the Earth after I jump to the same speed as the Earth approaching me, both objects would meet in the middle?

Rogan -

quote: Since there is no gravity to change the rotational speed it doesn't oscilate. It just goes round in a circle until air resistance stops it (or if constrained, to bounces against the walls until it dissipates its energy against the walls and stops, probably on the 1st bounce since wood on a grandfather clock isn't very springy).

Assuming a free swinging pendulum, if it was here on Earth, it's not so much the gravity causing the scribed circle, but the Earth's spin, true? Gravity would only cause it to swing, not oscilate. Therefore, up in space the Earth's spin becomes irrelevant so the pendulum would still swing back and forth with the gravity focus being Earth, albeit further away. So the period - the length of the swing - would shorten due to less gravitational force pulling on the weight.

	28-06-2003 11:30									#65
	Got spare money?

On second thought, a pendulum up on a space station orbiting the Earth would still have a spin - the orbital velocity of the station. Hmm...

	28-06-2003 16:05									#66
	Inflate your Upload Space

quote: Originally posted by Urban Ranger quote: What many people are forgetting is that if the moon and Earth are rotating around the sun. In effect, if you look at the Earth-Moon system as being stationary you can ignore the sun's attractive force as it is cancelled out by the centripetal force of a rotating frame of reference. But you can say the same thing about the Earth-Moon system.

Why would you want to, and why would the average person look at it that way? It just makes things much more complicated.

	28-06-2003 19:41									#67
	Support Apolyton buy from Amazon

Ignoring air resistance, there will be no difference in the behaviour of a pendulum in orbit whether or not it has a space station around it. So, consider a pendulum orbiting the Earth, with a certain angular velocity. Will that angular velocity change?

At first glance it might appear that it will not, that the pendulum will keep spinning at its initial rate forever. However, the gravitational acceleration of the pendulum varies from one end to the other - if the bob is closer to the Earth, it will be experiencing a greater gravitationl force per unit mass than the stick[1].

When the pendulum is perfectly radial to the Earth, it will not experience any torque from this tidal effect. When it is in any other orientation, it will experience a force restoring it to that state. If the initial angular velocity is sufficiently small, it will oscillate back and forth across the radial orientation, much like it would on Earth - save that the axis of rotation would pass through the centre of mass rather than the end of the stick. Due to the comparitive weakness of the tidal effect (which I can't be bothered calculating just now), it would have a much longer period than it would on Earth.

If it were initially rotating much faster, it would continue to do so, but not at a constant angular velocity - it would speed up slightly twice each rotation, as it approached the radial position, and slow down by an equal amount as it departed it. Finally, there are two unstable equilibria, where the pendulum is oriented tangentially to its orbit.

I wouldn't expect an average high school student to be able to work that out, but they should at least be able to identify that, ignoring tidal effects, the pendulum will simply keep spinning. As for the other question, anyone knowing the formula F = G * M1 * M2 / R^2 should be able to work it out if they have access to a few astronomical figures for mass and distance. The calculations earlier in the thread are a perfect demonstration of this.

For an interesting physics question that doesn't require too much background knowledge, try this:

There is a boat floating in a tank of water, with a brick in the boat. The brick is heaved overboard, and sinks to the bottom of the tank. Does the water level rise, sink, or remain the same?

[1] The ends of the pendulum will henceforth be referred to as the 'bob' and the 'stick', assuming a rigid pendulum.

	29-06-2003 00:58									#68
	About the tidal effect Tired of ads?

quote: Originally posted by Adalbertus As for those who thought of tidal forces: They don't depend on the force itself but on the rate of change of the force - or better graviational acceleration. On earth, the gravitational field of the earth-moon system changes faster than that of the earth-sun system, and therefore the tidal effect of the moon is stronger than that of the sun.

The words "the rate of change of force" and "faster" are a bit confusing in this context, since one could understand that you meant how fast they change with time (btw, did you?). But in reality, what is important here is how fast they (the gravitational fields of the moon and the sun, respectively) vary in space, i.e. the degree of their nonuniformity in the earth's vicinity. In other words, the quantity determining the tidal effect is the difference between the gravitational field at the front and the rear sides of the earth, whereas the magnitude of the gravitational field itself plays no role. Using the data provided in Neutrino's post, I estimated that the difference in question is 2.3 times stronger for the moon than for the sun ( I would have expected the number to be larger than that), whilst the absolute value of the gravitational pull is much greater for the sun, of course.

	29-06-2003 02:06									#69
	Support Apolyton, buy GURPS/ Alpha Centauri

quote: Originally posted by CerberusIV Think a bit more laterally. The Sun and Moon both cause tides in the Earth's oceans but the Moon's effect is greater (that is why the spring and neap tides occur on a 28 day cycle). If the Moon has a greater gravitational effect on the Earth than the Sun does then it is pretty safe to infer that the Earth (which has a larger mass than the Moon) has a greater gravitational effect on the Moon than that of the Sun on the Moon.

That's a tidal gravitational effect, due to the relative difference in vectors because neither the earth nor the moon are point objects. Due to the distance between the two, and the diameter, there's a measurable difference in gravitational effect on the parts of the earth closest and farthest away (and every distance in between).

Since gravity varies with directly with the masses, but exponentially with distance, tidal effects caused by the closer object can't be used to conclude that the closer object exerts more gravitational effect.

quote: Going back to the original question, these questions are probably at about the right level or maybe a little difficult, not because of their technical demands but because they require the student to think about the question. That is the hardest skill to teach, getting the student to analyse the question and answer what is actually being asked.

I think they're a good type and level of question - at the high school level, IMO, it's more appropriate for students to be able to think through and analyze problems, rather than focus on specific knowledge or techniques. Problem solving and critical thinking skills they'll use in any field.

	29-06-2003 02:55									#70
	Help yourself to an AD-FREE life

quote: Originally posted by FrustratedPoet And skywalker is (IIRC) one of the high-school kids we have around here. So I guess that proves they aren't too hard. This thread also proves that some people tend to overthink problems.

Yes, but I'm not in physics yet

However, I am not exactly your average high-school student

quote: Originally posted by FrustratedPoet I think they're a good type and level of question - at the high school level, IMO, it's more appropriate for students to be able to think through and analyze problems, rather than focus on specific knowledge or techniques. Problem solving and critical thinking skills they'll use in any field.

Very true. My precalculus teacher this year would always teach us a concept, then immediately give us several "quizzes". The way they worked, each was just a single normal math problem (concerning the concept we had just learned) which we had as many tries to answer as we wanted, until he decided we had spent enough time. We just used printer paper. The good thing was, we figured out how to apply what we'd been taught to the problem.

	30-06-2003 00:56									#71
	Re: About the tidal effect Lose 30 kilos (of popups)

quote: Originally posted by The Vagabond The words "the rate of change of force" and "faster" are a bit confusing in this context, since one could understand that you meant how fast they change with time (btw, did you?).

I thought of space, but I tried to avoid the word "gradient" which is appropriate but probably not known to most non-scientists.

	30-06-2003 03:22									#72
	Support Apolyton

quote: Originally posted by Cruddy Correct. Very unlikely to hit the Earth at all - because it's already in orbit, all you would get is a changed orbit. Bit of a trick question really...

If it wouldn't rebound from athmosphere it would hit the ground. If of course it wont burn.
BTW not the spot directly under space station.

	30-06-2003 03:39									#73
	Inflate your Upload Space

quote: Originally posted by JellyBean For an interesting physics question that doesn't require too much background knowledge, try this: There is a boat floating in a tank of water, with a brick in the boat. The brick is heaved overboard, and sinks to the bottom of the tank. Does the water level rise, sink, or remain the same?

The boat has a weight of X, the brick of B, and define water to have a unitary density.

In order to float the boat and brick displace water weighing X + B. Once the brick is thrown overboard the boat only displaces water weighing X, and the brick displaces a volume equal to its own volume b. Given that the brick sinks it has a higher density than water. So B > b. Therefore X + B is more than X + b and so therfore the water level sinks as less of it is being displaced.

	30-06-2003 05:25									#74
	Support Apolyton buy from Amazon

Isn't a pendulum's T (time of erm.. period] equal : 2*pi*sqrt(length / gravitational pull) ?

If so, then when the gravitational pull is nearing 0 (in orbit) the period lengt would strive to infinity? Meaning it would never return?

But would it move at all?

	30-06-2003 05:27									#75
	Support Apolyton buy from Amazon

no

	30-06-2003 05:40									#76
	Support Apolyton buy from Amazon

quote: therfore the water level sinks as less of it is being displaced.

But doesn't the brick continue displacing water regardless where it is? Why does the brick displace less water when it's submerged?

	30-06-2003 05:46									#77
	Support Apolyton, buy Galactic Civilizations

quote: Originally posted by raghar If it wouldn't rebound from athmosphere it would hit the ground. If of course it wont burn. BTW not the spot directly under space station.

Eventually the orbit will decay and it will enter the atmosphere... but if you consider the difference of the orbital velocity (1000s of MPH) with the tiny vector from an astronaut throwing it, you'll see the orbit isn't hugely affected at first.

Give it a couple of 100s of orbits at least to hit the earth... maybe more, I'm not sure how high the ISS is.

	30-06-2003 06:38									#78
	Support Apolyton, buy Civilization III: Complete

Big Crunch is right. Berzerker, the brick displaces less water when it's submerged because then it's only displacing its own volume. When it was in the boat, it was causing the boat to displace an additional quantity of water with the same mass as the brick.

If you throw something towards the Earth from a space station, the orbit it ends up in won't even be entirely inside the one you started in. If you want it to hit the Earth, you're better off throwing it in the opposite direction to your velocity, back along your orbit. If you throw it at your orbital speed, it will simply drop from your altitude. If you throw it more slowly, then at least it'll end up in an elliptical orbit that's entirely inside yours, it's greatest distance from the Earth occurring at the point where you threw it.

	30-06-2003 07:04									#79
	Suffering from ads?

quote: Are these too easy, too hard or what?

These questions are perfect. If all of us are formulating ideas and spending this much time reasoning them then they are good questions. For High School it is probably more important to have a discussion about different ideas then to actually get the right answer. I would say these questions will do the trick.

	30-06-2003 08:18									#80
	Support Apolyton, buy Civilization III: Complete

quote: Big Crunch is right. Berzerker, the brick displaces less water when it's submerged because then it's only displacing its own volume. When it was in the boat, it was causing the boat to displace an additional quantity of water with the same mass as the brick.

When the brick is tossed overboard, the boat no longer displaces the same amount of water, but the difference is now under water. I'm not getting this, oh well...

	30-06-2003 14:32									#81
	Full PM-box? Change here!

Thanks for your help guys

	30-06-2003 15:57									#82
	Help yourself to an AD-FREE life

quote: Originally posted by Sirotnikov Isn't a pendulum's T (time of erm.. period] equal : 2*pi*sqrt(length / gravitational pull) ? If so, then when the gravitational pull is nearing 0 (in orbit) the period lengt would strive to infinity? Meaning it would never return?

Actually, the period would disappear, which is pretty much the same thing.

quote: But would it move at all?

Yes, it would spin forever.

	01-07-2003 07:44									#83
	Increase Your PM Length

If they're too tough for the first years (read: College Freshmen) I tutor, then they're too tough. They couldn't get this.

These are too tough for your average "high school" student.

	01-07-2003 08:26									#84
	Support Apolyton, buy Galactic Civilizations

I didn't realize that the pendulum one was sort of a trick question until today when I was telling my dad about this thread (he started laughing when I told him about that one, and I asked why). He said that students who had just been studying relativity would immediately start working out the time dialation on the ISS. I didn't even realize that was an issue

	01-07-2003 09:24									#85
	Lose 30 kilos (of popups)

quote: Originally posted by skywalker Actually no. It just has a slightly higher velocity than necessary to stay in orbit, thus it spirals away (slowly).

Uh uh. Two ideal spherical bodies interacting will orbit their centre of mass in a conic section. If the net energy of the system is negative the conic section will be an ellipse. If zero it will be a parabola. If positive it will be a hyperbola. This is known as the two-body problem and is easily solvable (every physics student sees it solved at least once).

The spiralling outward effect is a result of angular momentum transfer from the earth to moon as a result of the tides. As time goes on, the stretching of the earth caused by the gravitational gradient (~10 meters, IIRC) tends to transfer angular momentum from the earth's rotation to the moon and earth's revolution about each other. The net effect is that in a couple of billion years the earth and moon ought to become tidally locked so that each only faces one side to the other (i.e. the period of revolution of the two bodies will become equal to the periods of revolution of each). The moon has already become tidally locked to the earth, which is why we only ever see one face. IIRC, the final system will have a period of ~50 days, as opposed to today's 28.

I'm not sure what the effect on the final solution will be if you include the sun as a perturbing force. I would guess that the end solution would look like a permanent solar eclipse, although I'm guessing it would take trillions rather than billions of years to manifest itself.

	01-07-2003 09:35									#86
	Avatar Enlargement: We've got the solution

quote: Originally posted by Big Crunch 2 is easy. Just use common sense. You don't even need numbers.

Yes you do.

Put the moon twice as close to the earth as it is and see what the answer is...

The way I figured it out I used the period of rotation of the moon about the earth and about the sun, as well as the mean radius of rotation of each body. Then I applied a=w^2*r

The moon has an angular velocity about the earth ~13 times what it has around the sun (since it rotates about the earth in 28 days and the sun in 365). It also has a radius of rotation about the sun of ~400 times that of its distance from the earth.

400/13^2 = 2.5 or so.

Not too much different so far as I can see.

	01-07-2003 09:38									#87
	Lose 30 kilos (of popups)

quote: Originally posted by Berzerker dannubis - If you jump in the air, does the earth move in your direction as much as you move toward the earth when landing? If you were right, we'd be floating along the surface waiting to hit a bump and bounce off into space.

His statement has nothing to do with the question (he seems to have misread it) but it is quite true. Force and acceleration are not the same thing.

	01-07-2003 09:40									#88
	Suffering from ads?

quote: Originally posted by Starchild In the space station, there is no gravity drawing the pendulum downwards. So there is no potential energy, just the kinetic energy. So when a pendulum is started off, it will swing around and around.

Bastard. Stop confusing people. The ISS is in almost as strong a gravitational field as we are (~95% or so). The difference is that they're a freely-falling frame and we're not...

	01-07-2003 09:47									#89
	Suffering from ads?

quote: Originally posted by Rogan Josh 2. The sun has more effect on the moon. This one is probably easiest to think of in terms of the energies involved. The energy of a rotating body goes like w^2*r where w is the angular velocity and r is the radius. w is 12 times larger for the moon round the Earth than for the moon round the sun (a year^-1 compared to a month^-1), but the Earth-moon distance is much less than a twelfth* of the moon-sun distance. So w^2*r is greater for the moon-sun and they must therefore have more gravitational energy.

a) You're mixing terms, rogan. Gravitational energy and gravitational force are quite different. Getting sloppy...

b) You've forgotten that the w is squared. The forces are on the same order of magnitude (only different by a factor of 2.5, if you see above) Unless the kids are supposed to remember the relative orbital disatances of the earth-moon and earth-moon-sun systems fairly accurately then the question is impossible. BC's statement that the answer is self-evident from application of common sense is completely untrue. The answer is rather counterintuitive, actually. Most natural satellites in the solar system are held in a stronger gravitational field by their primary than by the sun. Although, to be fair, I'm pretty sure all natural satellites in the solar system are given a greater gravitational potential by the sun than by their primary...

	01-07-2003 09:53									#90
	Full PM-box? Change here!

quote: Originally posted by Rogan Josh The reason for the tides being caused by the moon is because the orbit of the moon around the Earth is not circular. The same effect in the Earth around the Sun is what causes the seasons.

This is untrue. The tides of the moon on the earth are caused by the moon's gravitational gradient across the earth, not by the moon's orbital eccentricity. In other words the moon pulls harder at the points on the earth nearest it. As the earth rotates underneath it, this point changes, causing periodic stretching of the earth's shape.

The seasons are caused by the earth's tilt, not by its eccentricity. As a matter of fact, the earth's perihelion occurs at around Jan. 20, in the depths of the Northern hemisphere's winter...