High School physics questions

[Urban Ranger]

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.

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.

[Urban Ranger]

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.

[Urban Ranger]

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?

[Berzerker]

ixnay -

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 -

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.

[Berzerker]

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...

[Dauphin]

Originally posted by Urban Ranger

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.

[JellyBean]

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.

[The Vagabond]

About the tidal effect

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.

[MichaeltheGreat]

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.

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.

[Kuciwalker]

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

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.

[Adalbertus]

Re: About the tidal effect

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.

[raghar]

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.

[Dauphin]

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.

[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?

But would it move at all?

[Ecthy]

no