|
Post by moonglow on Feb 3, 2006 18:50:42 GMT -4
Did people come up with the moons gravity to be 1/6th of earths because the moon is 1/6th the size of earth? If they did there is no accounting for the centrifical force that the moon has about thirty times less of.
A lot of the stuff discussed here we would have to go to space or go to the moon to prove either side. Centrifical force is a law I can prove here on earth myself. So if the earth really is spinning around a thousand mph centrifical force would have to be a factor wouldn't it? And if it is shouldn't the moon have more gravity than 1/6th that of the earth since it spins about thirty times less?
Also, if the earth really is spinning that fast how come people don't weight more at the poles where the centrifical force would be a lot less. In the center of either of the earths pole's the C force wouldn't even exist I would think, and the gravity would be vary strong there compared to the equator wouldn't it?
If the moon is bigger than 1/6th that of the earth then that would explain it. But I would still wonder why we don't weigh more at the pole's if the earth really is spinning that fast.
|
|
|
Post by Stout Cortez on Feb 3, 2006 18:54:40 GMT -4
A person does weigh more at the poles than at the equator, but not for the reason you think.
|
|
Bob B.
Bob the Excel Guru?
Posts: 3,072
|
Post by Bob B. on Feb 3, 2006 19:06:57 GMT -4
Did people come up with the moons gravity to be 1/6th of earths because the moon is 1/6th the size of earth? No, this is not how surface gravity is determined nor is the Moon 1/6th the size of Earth. Surface gravity is proportional to mass divided by the square of the radius. Earth is 81.3 times more massive than the Moon and 3.67 times larger in radius. Thus Earth's gravity is 81.3/3.67^2 = 6.04 times more than the Moon.
|
|
|
Post by grashtel on Feb 3, 2006 19:21:30 GMT -4
Also, if the earth really is spinning that fast how come people don't weight more at the poles where the centrifical force would be a lot less. In the center of either of the earths pole's the C force wouldn't even exist I would think, and the gravity would be vary strong there compared to the equator wouldn't it? If the moon is bigger than 1/6th that of the earth then that would explain it. But I would still wonder why we don't weigh more at the pole's if the earth really is spinning that fast. You're partly right, gravity at the Earth's poles is higher than at the equator, but only by 0.5% and a large portion of that isn't directly related to centrifugal force. The difference is mostly due to being closer to the center of the Earth as Earth isn't perfectly spherical, it bulges out at the equator due to centripetal force. Though even if the difference was entirely due to centripetal force the fact that it is so small on Earth means that the effect on the Moon would be non-existently small.
|
|
|
Post by moonglow on Feb 3, 2006 22:21:24 GMT -4
Thanks for the replys. I was really curious about that when I thought of it. One thing I don't get is why the difference is mostly due to being closer to the center of the Earth as earth isn't perfectly spherical, why would that matter? And is there a difference in meaning between centrifugal force and centripetal force?
|
|
|
Post by JayUtah on Feb 3, 2006 23:02:07 GMT -4
First question. Every object has a center of gravity. Even though gravity acts on every particle in that object, we can do the math pretending that it acts only at that point, and the answer comes out the same. The center of gravity for a sphere or a spheroid of uniformly dense (or appropriately symmetrical) matter is the geometric center. The strength of the gravity force between two bodies is inversely proportional to the square of the distance between their centers of gravity. The Earth is squashed at the poles, so if you're standing at the pole you're actually a bit closer to the Earth's center of mass and so the attractive force between you and the Earth is a wee bit stronger.
Second question. Centrifugal and centripetal forces are basically opposites. If we're talking about orbits, the centripetal force is that which draws a moving body into a circular path -- in this case, gravity. In the case of the classic ball on a string swung overhead, centripetal force is the force of the string pulling against the ball. It's the force -- whatever it is -- that pulls the body toward the center of the circle.
The centrifugal force actually derives from a fairly esoteric discussion of inertial reference frames, which is about as exciting as dirt if you just want a quick answer. It's best explained with the classic example: You're in a roller-coaster car and it takes a sharp corner. The car exerts a force on you to coerce you into a circular path. That's the centripetal force. But pretend that you're in a completely sealed car and you can't see what's going on outside. So from your point of view, you're simply mashed up against the side of the car. You exert a force outward. That force derives from inertia and is relatively straightforward. But in this way of thinking (the sealed car), it has the special name "centrifugal" force.
|
|
|
Post by moonglow on Feb 4, 2006 3:36:21 GMT -4
Makes sense, thank you!
|
|
|
Post by PhantomWolf on Feb 4, 2006 5:00:26 GMT -4
centrifugal force is really a mathematical construct, rather than a real force.
|
|
|
Post by Joe Durnavich on Feb 4, 2006 10:42:23 GMT -4
centrifugal force is really a mathematical construct, rather than a real force.
Is centripetal force a "real force" then? Using Jay's example of a roller coaster car, presumably on a level stretch of track, that takes a sharp corner. Is there a "real force" here that tends the car towards the center of the arc?
The wonderful explanations above came at a perfect time for me. I have been reading several books on Newton and it was said that a breakthrough in his thinking was to consider orbiting bodies in terms of centripetal force instead of centrifugal force. As I think about this too much, I seem to lose sight of the difference. (And no, I'm not man enough yet to survive the Principia except for bits and pieces.)
|
|
Bob B.
Bob the Excel Guru?
Posts: 3,072
|
Post by Bob B. on Feb 4, 2006 11:07:37 GMT -4
Is centripetal force a "real force" then? Using Jay's example of a roller coaster car, presumably on a level stretch of track, that takes a sharp corner. Is there a "real force" here that tends the car towards the center of the arc? Yes, centripetal force is real. We know that velocity is a vector, that is, it has both magnitude and direction. If an object deviates a straight path then its velocity has changed. A change in velocity (whether direction or magnitude) is acceleration, and acceleration can only occur when a force is applied. Therefore, if an object is moving in a curved path there must be a force causing it to do so. In the cause of a planet or satellite in orbit, the centripetal force comes from gravity. For the roller coaster car the force is applied to the wheels by the track. For the passenger the force is applied to his/her body by the car.
|
|
|
Post by JayUtah on Feb 4, 2006 11:54:35 GMT -4
Centripetal force is real. Centrifugal force is often characterized as fictional because it arises simply out of a reaction that is considered in a particular reference frame; it is only a "force" when considered according to the reference frame of the moving body.
|
|
|
Post by Joe Durnavich on Feb 4, 2006 12:17:49 GMT -4
First, let me apologize ahead of time for trying the group's patience. I am just trying to get clear on what sense centripetal force is "real" and centrifugal force is not.
A change in velocity (whether direction or magnitude) is acceleration, and acceleration can only occur when a force is applied.
Does this statement merely define what force is, or does it establish its "reality".
For the roller coaster car the force is applied to the wheels by the track.
What I was struggling with was getting clear on what sense the track "applies" a force to the wheels. Would it be correct to say that the section of curved track is applying a force even when no roller coaster is present? Or does force appear on the scene only when we consider the track in reference to the motion of the roller coaster?
|
|
|
Post by JayUtah on Feb 4, 2006 14:11:22 GMT -4
Centrifugal force is just a special application of inertia. There is no agent or phenomenon that actually generates the force; it's merely a reaction to the action of a moving body's linear path being changed. Centripetal force has to have some agent or phenomenon to exert the force: gravity, the roller-coaster track. the string on the swung ball.
In the case of the roller-coaster the track doesn't actually apply a force until the car moves over it. It's like the ball-on-a-string where the whole thing is just lying on a table; the string isn't actually applying any force on the ball until you pick it up and start swinging.
You're not trying anyone's patience. The original question on the table was a simple one, and often the formalities get in the way of comprehension when a simple answer is wanted. But at some point you have to know why it's formulated the way it is. In engineering especially, getting the reference frame right means the difference between success and failure.
If you have a reference frame that moves with the roller-coaster car, the turning action isn't apparent. And so the inertial response that pushes you to the outside of the turn seems like a force that arises out of nowhere in that frame. But it has no cause aside from inertia. And inertia is just a fancy word for the Newtonian status quo. So basically if your reference frame doesn't include the caused forces, the reactive forces appear as phantom forces -- but only in that frame. Generally only engineers and physicists and similar people need to worry about reference frames. So when you have these esoteric, philosophical definitions that arise out of that worry, as soon as you're in for a penny you're in for a pound.
These can be insidious. As we discussed, Roger McCarthy -- an engineer's engineer -- got wrong on national television the reference frames for rocket flight. If you get the reference frame wrong, you omit forces that occur and you conversely and wastefully consider forces that don't arise.
In the reference frame of the track, the car applies a lateral force to the track as the track begins to curve and the track exerts an oppositely lateral reactive force against the car wheel. The choice of reference frame depends on wether you're designing the track or the car.
Celestial mechanics is a bit different because centripetal force varies independently according to tensor field mechanics -- i.e., gravity just is what it is. When you swing the ball on a string in a circular path, you generate greater momentum by swinging faster. The ball's quantitative tendency to stay on its straight path increases. But the centripetal force increases and tension on the string rises. Whether you consider that the ball is pulling on the string or the string is pulling on the ball depends on your reference frame. In an orbit, if momentum increases, centripetal force stays the same (i.e., follows gravity's rules) and the path changes.
|
|
|
Post by Halcyon Dayz, FCD on Feb 4, 2006 17:05:26 GMT -4
In an orbit, if momentum increases, centripetal force stays the same (i.e., follows gravity's rules) and the path changes. No strings attached, hey. ;D
|
|
|
Post by Joe Durnavich on Feb 5, 2006 0:15:49 GMT -4
Let's say we are at the Apollo Hoax Testing and Demonstration Grounds and scanning across the grounds, we find it vibrant with activity. Count Zero arranges assortments of sun shadows; John Keller fires a rocket engine into the dirt; Phantom Wolf irradiates doubting conspiracy theorists; Kiwi dozes with camera in hand waiting for an evening of short-exposure stellar photography; and Jay and Bob are shooting me down a railroad track at high velocity...
So basically if your reference frame doesn't include the caused forces, the reactive forces appear as phantom forces -- but only in that frame.
Your explanations in terms of reference frames do the best in clearing my self-induced mist. What have been sources of puzzlement for me are terms like "real," "fictional," and "phantom" that are interspersed with the explanations. I couldn't help but wonder that if centrifugal force was not real, but centripetal force was, what constituted the "reality" of centripetal force?
Back to the track...Is the following an equivalent way to state your roller coaster example? Let's say the car I am in is a large, sealed train car--as large as the imagination can bear. I am wearing roller skates or suitable devices on my feet with negligible friction. You and Bob are up in the observation tower watching me fly down the straight section of track. You can see me standing in the car through a one-way-mirrored roof.
You have radioed me to expect a force. Inside the car, I am just standing there motionless waiting and wondering when you guys are going to get things moving. The train car charges onto the curved section of the track. Inside the car, I see myself start to move on the roller skates towards a wall. I radio up, "Guys, I am staring to move. My velocity has changed. Hence, there must be a real force acting on me."
You laugh and radio back, "No, Joe. From up here, we can see that you are simply continuing on the straight line that you were on. What is happening is the car is now moving in a different direction. The wall has started to move towards you; you have not started to move towards the wall. What you think is a force is just a fiction, a phantom."
I finally grasp the point and, then, promptly, slam into the wall.
First, if I have that imagined right, would it be proper to call my motion on the skates in the car centrifugal force? Or is centrifugal force reserved for the time when I am plastered spread-eagled against the wall fighting for breath?
If that is centrifugal force, then I can see that from my reference frame, I change from a position of rest to one of moving towards the wall. From your and Bob's reference frame, up in the observation tower, however, you see the car turn, but you see me maintain the same linear motion I had before. For your viewpoint, no force has acted on me.
Now, where do notions of "real" and "fictional" fit in here? Is your and Bob's reference frame privileged such that the movements you see entail "real" forces and the movements I see become relegated to a lower status?
You said, "Centripetal force has to have some agent or phenomenon to exert the force..." So, perhaps the non-reality of the force I experience in the car resides in the fact that I cannot observe a cause. But if so, then I might pester you guys about why gravity is considered a real force and not a phantom one, when all we see are apples, say, falling to the ground with no agent or phenomenon apparent as a cause.
One other question about centripetal force: The car is said to exert a force on the track or vice-versa. The car moves along under its own inertia. The track, anchored to the Earth, is at its own intertial rest. Does the centripetal force arise from the car and the earth-anchored-track resisting each other's inertia? (I'm also thinking of two billiard balls floating through the International Space Station that bump into each other and change direction. If we say there is a force here, are we simply describing the motion of the balls, or is the force constituted by something else?)
Centripetal force, as I understand it, is a tendency towards a center. But there is nothing special, right, about the patch of real-estate that comprises the center of the arc of our track? I wouldn't think that you could wave a tricorder over it and detect some sort of centripetal rays emanating from it. Do we, then, simply call any motion around a center "an object being acted on by a centripetal force"?
|
|