Johnny5

Yesterday, I was asked to figure out how my weight would be reduced, if I were standing upon the equator of the earth, rather than at one of its poles. (My first thought was that since you are farther away from the center of earth's gravity when you are at the equator than when you are at its poles there is nothing to think about. Just use Newton's gravity formula, and figure out the answer. I am no longer sure this is correct.) There was discussion on the centrifugal force, as being responsible for a decrease in my weight. I am trying to decide whether or not this is true. In classical mechanics, Newton's third law states that for every action, there is an equal an opposite reaction. One way to interpret this statement, is that in any inertial reference frame, you will never find a non-rocket which causes itself to self-accelerate. Things don't just change speeds on their own, an external force must be applied (F=ma) which is Newton's second law. I have been reading things about centripetal and centrifugal force a bit, in order to answer the question. One of things that I read, was that the centrifugal (center fleeing) force is the action/reaction pair force, for the centripetal (center seeking) force. Mathematical analysis of circular motion reveals that an object in circular motion is accelerating, even though it's speed is constant. The reason the object is accelerating, is because acceleration isn't change in speed, it is change in velocity, and velocity is speed times direction. In the case of circular motion, an object moving in a circle is changing direction in some frame. Since its direction of travel in the frame is changing, its velocity in the frame must be changing, hence the object is accelerating in the frame. Upon analysis, you will discover that the direction of the acceleration always points at the center of the circular path of the object, and has the following magnitude: [math] \frac{v^2}{R} [/math] Where v is the speed of the object in the reference frame, and R is the orbital radius of this object. The quantity above, has units of acceleration, and using polar coordinates we can write this "centripetal acceleration" as: [math] \vec a_c= -\frac{v^2 \hat r}{r} [/math] in a coordinate system whose origin is the center of the circular path. The symbol r^ is a unit vector which always points from the origin to the particle. So the minus sign indicates that the direction of this acceleration, is opposite to the direction of r^, and therefore this acceleration points to the point the particle is orbiting. If we now multiply the centripetal acceleration by the inertial mass m of the object, we will have the quantity known as the centripetal force. [math] \vec F_c = -\frac{mv^2 \hat r}{r} [/math] In the case of gravitation, one thinks of gravity as pulling us towards the center of the earth, and Newton's gravity formula is: [math] \vec F_G = -\frac{GMm\hat r}{r^2} [/math] Where m is the inertial mass of a body sitting on the surface of the earth, and M is the inertial mass of the earth, and r is the distance from the center of inertia of earth to the center of inertia of the body sitting on the surface of earth. So can we equate these two forces? I just read this in wikipedia:

Yes, but during the derivation, i want to put dots over things, to denote differentiation with respect to time, and I'm sure you can do it with Latex, but I just don't know how. [math] \dot\omega [/math] WOW! I just made a blind child's guess and it worked. Ok i answered my own question, with my very first guess. I'm sure that doesn't happen too often.

Is there any way to put a dot above something. I am going to use [math] \omega [/math] to denote angular velocity, and I want to put a dot over the omega symbol, to denote the angular acceleration. The notation [math] \frac{d\omega}{dt} [/math] is too cumbersome for what I want to do. Thank you

Doesn't only one side of the moon face us at all times? So to visualize the 27.3 day rotation' date=' I have to stand on the moon? Then, the "fixed stars" will orbit me once every 27.3 days, do I have that right? Assuming I do, you are saying that if I were at the equator of the moon, there is a slight "centrifugal force" reducing my weight? I have read that centrifugal force is a fictious force. Can a fictitous force reduce my weight? Is there an action reaction pair, for the centrifugal force? Without the rotation say my weight is W. I could measure W by standing upon a scale on the moon. Now, let the moon be spinning with respect to the fixed stars. In this case, the moon is buldging at the equator. Let my scale be located on the equator of the moon. So I am looking up at the stars, and they are orbiting me once every 27.3 days. But because the moon is spinning, my weight is now [i']reduced [/i] from W? How do you compute my new weight? I thought the centrifugal force wasn't a real force, how can it reduce my weight. If a force acts on me, there is an action reaction pair, what is the reaction force called? This is highly interesting.

I only know of one mathematical process which operates on two vectors, and outputs a vector... the cross product. The dot product also operates on two vectors, but it outputs a scalar.

That location is the center of gravity of the earth-moon system. I think he wants to know if that location is inside the earth, at the surface of the earth (possibly Pittsburg PA), or outside the surface of the earth. He wants to know where that location is, in relation to the CM of the earth, and the CM of the moon.

It appears that your astronomy book is correct, if indeed the radius of the moon is really 27% the radius of the earth. The earth is an oblate spheroid actually, and therefore doesn't have a single 'radius' to speak of. The real earth buldges at the equator, because it is spinning (as JC and swansont both pointed out).

I think it's worth talking about direction a bit. Suppose I am somewhere in space. I can rotate my head, and look in a multiplicity of directions. So given any location in space, we can define a theoretically infinite number of directions, simply by picking another location in space. So, correct me if I am wrong, but to define a direction in space, requires that two points be given to us. Now, it seems to me that the distance between the two points is immaterial. Once we have those two points, they uniquely define TWO different directions. Call one point X, and the other point Y. There is the direction pointing from X to Y, and there is the direction pointing from Y to X. Direction doesn't have length. It has nothing to do with length whatsoever. Now, since two arbitrary points don't uniquely define one direction, we need more information. We need to specify which point is the tail, and which point is the head of a "unit vector." But really, as I said, a direction doesn't have a length. And direction vectors are stipulated to have a length of 1 unit. I'm not sure about what's going on right now, I've never questioned this type of thing before. But my point is, that we need to be given two points in space, one of which is tail, the other head, in order to define direction. My point is only that since we are given two points, I don't see why we cannot use one of them, in order to define the inverse of the multiplication we are talking about. We don't know what that multiplication is in the first place, as you said. But, if we allow it to resemble the dot product, and additionally stipulate that the multiplicative inverse must have the same tail as A, then like I previously explained, given any single vector A (other than the zero vector), there is one and only dot product multiplicative inverse. I shouldn't call it dot product multiplicative inverse at all, since we can dot vectors which don't have the same tail, but I don't know what else to call it.

I just needed a number' date=' I pulled it out of thin air. I'm not really talking about the real moon, and the real earth, but it helps to visualize them though. As was correctly pointed out, my weight on the moon depends upon the mass of the moon, and the density (through R). So that really I should have written this: [math'] W_{moon} = \frac{1}{3} W_{earth} [/math] That is the fact that I would discover when I did my handstand. If the acceleration of gravity at the surface of the moon, were identical to the acceleration of gravity at the surface of the earth, then what I wrote would have been correct. But the acceleration of the moon's gravity is 1/6 that of the earth. At any rate, I wasn't talking about the real moon, and real earth, the question centers more around... what am I pushing against, when I do the handstand. What I was really interested in was, does the vacuum offer resistance or not.

I guess it would. Are you saying that when I do my handstand on earth, that I am also pushing slightly on our atmosphere (on the other side of the world? Ok suppose that formula is correct. [math] \vec F = \frac{d\vec P}{dt} = 1/c \frac{dE}{dt} [/math] Pressure is force per unit area. So when I do my handstand on the moon, I push the moon away slightly from me. So the area in question is half of the surface area of the moon?

Yes of course, R matters as well as M. If the moon was squeezed into a volume the size of say a football stadium , the inertial mass M would have remained constant, but the density of the moon would have increased dramatically. Then when I went to do my handstand, the force upon me would be much greater, because of gravity. The gravitational force of an object A on an object B is inversely proportional to the distance object B is from the center of mass of object A. So by squeezing the moon into the volume of a football stadium, I can get much closer to the center of gravity, (R decreases a lot, so F increases a lot), and M stayed the same (i think).

Ok I read a little further and I see very clearly where you wrote this

I read about Kaluza-Klein string theory years and years ago, I was just a boy. So far as I am aware, space is three dimensional. That pretty much puts a damper on that.

Suppose I want to measure the mass of the moon in the following way. I want to do a handstand on the moon. Suppose that the moon is a perfect sphere, and that it has a uniform density. Will it matter where on the moon I do the handstand? Let us suppose that I have an excellent "muscle memory" and that I remember how hard it is to do a handstand on earth. Define the earth to have one inertial mass unit. I go to the moon, do my handstand, and go wow, that was 1/3 easier therefore [math] M_{moon} = \frac{1}{3} M_{earth} [/math] So that the moon has a mass of 1/3 imu The question really has to do with the vacuum, not the moon, nor my method of measurment. Is inertial mass a direction dependent quantity? That's really what I'm asking. Suppose the following diagram illustrates where the center of mass of the universe is in relation to the moon. CMU . _______________________________( L-moon-R ) I can either choose to do the handstand on the left side of the moon, which is closer to the CMU, or the right side of the moon, which is further away from the CMU. Will which side I'm on affect my measurement? And if the location of the center of mass of the universe doesn't affect the measurement, then let me ask this instead... Does the vacuum have any effect on the measurement? And if it does, at what level does this effect become appreciable? Another way to ask this kind of question is this... Is there friction in the vacuum? if yes, how much does this vary, as a function of position? To a good approximation, the vacuum does not impede the foward motion of objects. But this cannot be literally true, since there are hydrogen atoms scattered throughout the universe, and also the vacuum is filled with EM radiation. Does the EM radiation impede our foward motion at all? Thank you

Is there a way to use latex to make a matrix? And for that matter a determinant? Thank you