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Rogan Josh
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OK, let me explain.
1. Maybe the wording of this is a little confusing. Of course, I wasn't really thinking about the pendulum being constrained by the box, but it doesn't really matter. 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).
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.
Hmm... from the responses this thread has had I think these problems are probably too hard for school kids. 
UR: Could you rephrase your question - I have not sure what our problem is. Whether he can beat his record depends on his starting velocity. Clearly if he starts from rest then he won't since 2*0=0. The minimum speed he would have to start with is (approx.) (1000m)/(2^14s)=0.062ms^-1
Is this what you mean?
Edit: *...err, a 144th obviousy. duh!
Last edited by Rogan Josh on 27-06-2003 at 20:11
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Urban Ranger
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Apolyton Duke of Off-Topic
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quote:
Originally posted by Rogan Josh
1. Maybe the wording of this is a little confusing. Of course, I wasn't really thinking about the pendulum being constrained by the box, but it doesn't really matter. 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). |
Not really, because if earth's gravity doesn't affect the ISS, it would not rotate around the earth, it would have gone into deep space. Hence, this question depends on the initial velocity of the pendulum.
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. |
That doesn't make sense at all. If the Sun's gravational pull is greater, the moon would have stop orbiting the earth and start orbiting around the sun. Come to think of it, it is incorrect to assume the angular velocity of the moon around the earth is 12 times greater than that of the sun, because the orbit around the sun is much greater than that of around the earth.
Last edited by Urban Ranger on 27-06-2003 at 20:31
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Rogan Josh
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quote:
Originally posted by Urban Ranger
Not really, because if earth's gravity doesn't affect the ISS, it would not rotate around the earth, it would have gone into deep space. Hence, this question depends on the initial velocity of the pendulum. |
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).
quote:
That doesn't make sense at all. If the Sun's gravational pull is greater, the moon would have stop orbiting the earth and start orbiting around the sun. |
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.
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.
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Neutrino
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Austin, Texas, USA
Jun 2002
time: 23:29
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From http://hypertextbook.com/facts/2002/AdaLi.shtml :
Mass of moon = 7.34 x 10^22 kg
Mass of Earth = 5.98 x 10^24 kg
Mass of Sol = 1.99 x 10^30 kg
Gravitation force = G[(m1*m2)/d^2)], where G is the gravitational constant, m1 and m2 are the masses of the 2 objects, and d is the distance between their centers of mass.
From http://scienceworld.wolfram.com/phy...alConstant.html ,
G = 6.672 x 10^-11 N m^2 / kg^2
From http://www.freemars.org/jeff/planets/Luna/Luna.htm ,
Distance from Earth to Moon:
Perigee 363,300 km
Mean 384,400 km
Apogee 405,500 km
From http://neo.jpl.nasa.gov/glossary/au.html ,
(mean) Distance from Earth to Sun:
149,597,870.691 km
Approx min and max distance from Sun to Moon (using mean dist of Earth-moon):
149,597,870.691 +- 384,400 km
So,
Earth-moon gravitational force :
1.98e26
Min Sun-moon gravitational force (assuming moon is on far side of earth to maximize distance):
4.33e26
Note that the distances are probably not distances between centers of mass, but I don't think that would change the results by that much. Feel free to check my results, I'm at work so I didn't have time to doublecheck.
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Rogan Josh
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Thanks Neutrino, I just worked it out with numbers and got the same thing.
The ratio of gravitational moon-sun force to moon-earth force is:
(Ms/Me)(Re/Rs)^2 = (2*10^30/6*10^24)*(4*10^5/15*10^7)^2 =(1/3)*10^6 * ((4/15)*10^-2)^2=16/(3*225) *100=2.37
So the moon-sun force is a little over twice the earth-moon force, confirming neutrino's answer.
Edit: this is actually quite a bit less than I was expecting...
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Adalbertus
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Cologne, Germany
Feb 2001
time: 06:29
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quote:
Edit: this is actually quite a bit less than I was expecting... |
Rogan,
This is because you think in terms of rotational energy and neutrino calculated forces. In the case of energy, you've got m(sun)^2/m(earth)^2*d(earth-moon)/d(sun-moon), if I calculated correctly. This should turn out to bigger.
The problem with this question is that "effect" is not a precise term in physics. You should state what you mean in physically well-defined quantities (or accept a lot of different but well-explained answers). It's a mistake teachers make quite often (and then insist on their private interpretation).
That's a nice question, but not for "thinking" only, but doing the calculation - the problem here is that we look from earth and tend to neglect the influence of the sun on the moon entirely, because its the same as that on the earth.
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.
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). I think it works more as a sort of guesswork to introduce general relativity.
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raghar
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I wish somewhere else.
Oct 2002
time: 05:29
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Suppose you have an alien ship in L1. what is affecting that alien ship more earth, sun or moon? And what if we add jupiter? Could he answer it?
Could you post all his questions?
If he will try to get work as an teacher this questions could work, it could impress director of that institution. Well more or less. If he will try them on bored irritated students that have a lots of other problems, for example how we would pay for school, It probably wont be as big - how to say it nicely. Then again students in GB are different that from rest of europe, not too much however. Last time, I listened "...in prison is better than in school..."
Lets hope he would get experience rather quickly and don't forget to remind him about his ideals after few years. Or he could turn into some kind of maniac like most of teachers.
Lots of begining teachers are forgeting that teaching isn't just about stupid questions, but is also about common sense from teacher and what he could give to students. Like tolerance and lots of unpaid work.
As an ex astronomer amateur that had rather nasty experience with less capable teachers, I could tell the real difference between a teacher and a fool.
Last edited by raghar on 28-06-2003 at 05:14
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Cruddy
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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.
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. |
Yeah, but surely the highest gravity effect win?
If the sun had a greater gravitational effect on the moon than the Earth, why doesn't the moon get further away from the Earth and closer to the sun?
Of course the sun has a bigger gravity well than the Earth - but as the Earth's gravity well is stronger on the moon, in the moon's current position, that makes the Earth have a bigger gravitational effect?
Maybe the wording doesn't come over in the translation...
Anyway, here's an alternative question -
Assuming the ISS is in geo-sync orbit - it stays over the same spot of Earth all the time - and an astronaut throws an object at the earth - assuming it survived re-entry - would it hit near that local spot, or much further away?
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Cruddy
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quote:
Originally posted by Big Crunch
You'd have to throw the object pretty fast, it would more likely end up in an elliptical orbit than fall to Earth. |
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...
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Cruddy
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You forgot the "at standard air pressure" disclaimer Pekka.
You try that trick in a vacuum and you are in for a BIG surprise.
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27-06-2003 17:45
(I will explain why later but I have to dash of to a meeting now...)
(well I remember it somewhat from last year, heh)
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