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Concept of Gravity
In article , bobofficers@
127.0.0.7 says... On 12 Nov 2006 14:59:39 -0800, in alt.astronomy, "Doug" wrote: skddlbyp wrote: I don't really understand the astro-physics of it. I'm not sure if the sun spins, or, if it does, it exerts a gravitational force in the direction of the spin. I may be wrong, but if the sun spins, wouldn't centrigugal force cause Centripetal force... Not centrifugal force. Actually, in the situation described, it *is* centrifugal force! Centripetal forces are forces acting toward the center, in this case gravity, which acts towards the center of the Sun. Centrifugal force is a fictitious force acting away from the center, I.e. in the opposite direction or up. Just because it is a fictitious force doesn't mean it's not real! :-) Centrifugal force arises because the problem is being viewed from the perspective of someone on the surface of the Sun, which is a rotating frame of reference. When viewed from a "normal", inertial frame, there is no centrifugal force because the only forces involved are gravity and the upwards pressure on the bottom of the person's feet, which is what keeps him from falling to the center of the Sun. If the person is standing on the equator of a rotating sun, he isn't stationary, but is moving in a circle and therefor is being accelerated. For this to happen, the forces have to be out of balance. (If they were completely balanced, he wouldn't be moving, or would be moving in a straight line at constant velocity.) To get constant circular motion, gravity pulls at right angles to the direction of motion (I.e. toward the center of the sun, since the direction of motion at any instant is horizontal, toward the East.) The surface exerts an almost equal force upwards, which the person would feel as his weight, but this force is slightly less than the force of gravity. So the net force is inwards, toward the center of the Sun, at right angles to the direction of motion. (The result of a constant force at right angles to the direction of motion is circular motion, since the sideways (or in this case, downward) deflection in any small unit of time is constant.) Since the upward pressure is slightly less than the downward gravity, the person feels like he weighs less at the equator. Why does the upward force behave like this, allowing the slight difference from gravity that results in circular motion? Well, actually it is a given in the problem, since the problem states that the person is standing "stationary" on the surface of the Sun, neither rising nor falling nor flying off into space! Fortunately, it is perfectly possible to satisfy this condition, just by standing there and moving at the same horizontal speed as the surface of the Sun is rotating. All this discussion is from the view point of an inertial frame of reference (one that is stationary or moving at a constant velocity), where the math is usually easiest. However, when viewed from a rotating frame of reference (I.e. the frame of the person standing on the rotating Sun), the forces have to be transformed, which is where the fictitious centrifugal force arises. Normally the math is much harder in a non-inertial frame, but in the special case of uniform circular motion, the math actually gets simpler, you just have to add in the constant centrifugal force, which is the fictitious force necessary to keep a stationary object from flying off, since it is actually moving. All this applies to an object held by gravity to the surface of any rotating body, not necessarily the Sun or the Earth, but it applies equally well to the Moon or Jupiter or an asteroid or another star. Disclaimer: it's been 35 years since I took freshman physics, so this could be horribly garbled, but I think it is essentially correct. less gravitational force in the direction of the spin? I read that on The gravitational force is *not* in the "direction of the spin". Actually, "direction of the spin" can mean two different things. What I am taking you to mean is the direction that the surface (and the person standing on it) are moving at any given instant. (It is very important to note that this direction is constantly changing!) "Direction of spin" could also mean the direction of the spin vector. (Imagine looking down on the object from above one of its poles. The equator would appear as a circle at the outside edge of the object. From above one pole, the object would appear to be spinning clockwise, and from above the other pole, it would appear to be spinning counterclockwise. The spin vector points in the direction of the counterclockwise pole (IIRC) and its magnitude depends on how fast the object is spinning.) However, that's way beyond this discussion. The gravitational force is towards the center of the Sun, which is neither of these two directions. earth you weigh slightly less at the equator than at the poles due to centrifugal force. Would this be true of the sun also? Yes. However the Sun doesn't rotate very fast (about once a month), so the effect is fairly small. On Jupiter, which rotates in about 10 hours, the effect is strong enough to make the planet slightly but visibly oblate. In another thread in this forum, a discussion of black holes, it was implied that black holes have stronger gravity at their poles than at their equators due to their very rapid spin. Probably, but this could also be due to some general relativity effect. (I didn't get far enough in physics to study GR, but there are lots of strange and un-intuitive effects.) Doug -- John |
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