High School physics questions

[KrazyHorse]

Originally posted by Cruddy
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?

No. Coriolis force interferes. The object retains its orbital momentum about the earth as it gets closer since the earth acts as a central force. Since L=VXR (X the vector cross product) = w*R^2 (w the angular velocity) w must increase as R decreases. This means that from the point of view of someone standing on the ground immeditately "beneath" a geostationary ISS (the ISS isn't even close to high enough for this, IIRC) the object would appear to pick up velocity in the direction of orbit of the ISS as it fell. If you didn't throw it hard enough the object would miss the earth entirely and go into its own elliptical orbit.

[KrazyHorse]

Originally posted by Urban Ranger


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.





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

That's exactly what freefall is. As long as there is no force other than gravity accelerating you you are in freefall. In GR terms all freely falling frames are physically equivalent, whether you're alone in the universe or following a complicated orbit in space.

[KrazyHorse]

Originally posted by MichaeltheGreat
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.

Not exponentially. Varies with the square of the distance. Exponentially would mean of form f propotional to e^(-d). Minor nitpick. Good point about tides not being accurate predictors of total gravitational field, though. Tides are caused by the gradient of G (first derivative of G), which is dependent on r^-3, whereas G is only dependent on r^-2. This means that tides are weighted much more heavily for closer objects. So much so that the moon's tidal effect on earth is ~twice as strong as the sun's tidal effect on earth (IIRC)

[KrazyHorse]

Originally posted by Berzerker


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

Because on the boat it was displacing its own mass. In the water it displaces its own volume.

[Rogan Josh]

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

Yes, you're right. That was rather sloppy of me. apologies. (but it is a hell of a long time since I had to remember stuff like that!)

b) You've forgotten that the w is squared.

Ahh, no I didn't! Notice the little * beside the word 'twelfth' and my Edit note (I felt it would be unfair to edit directly, even although I edited it about 3 seconds after I posted).

The forces are on the same order of magnitude (only different by a factor of 2.5, if you see above)

You didn't read the whole thread, did you

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.

I agree. After putting in the numbers I was surprised at how close they are.

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...

What is the variation in the orbits of the moon and Earth? It must be pretty small not to have a bigger effect. If so, why are the orbits so circular?