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No. The Hubble doesn't have the resolution. From http://science.nasa.gov/headlines/y2005/11jul_lroc.htm

The latter requirement, that X is on the line defined by S and P is simply [math]\vec x = \vec s + \kappa(\vec p - \vec s)[/math] where [math]\kappa[/math] is some free parameter. Let kappa range from -infinity to infinity and voila! you have the line defined by S and P. Restricting kappa to non-negative values results in your desired ray. Rather than work this all the way through, I'll leave the rest up to you for now. Holler if you need more help.

Sure. You could do exactly what a computer does, by hand. The word computer is old. Computers were people. Gauss employed a whole army of computers to do numerical computations for him. The function at hand obviously has a zero somewhere between x=0 and x=1 as log(x) tends becomes unbounded negative as x->0 and all of the other terms are positive over (0,1). It has exactly one zero between 0 and 1 because the four varying terms are all monotonically increasing over 0 and 1. Finally, this is the only zero. For x>1, log(x), x2, x, and 1 are all positive, so the only places the function can be zero is where sin(x) is negative. As sin(x) >= -1, 1+sin(x) >= 0. As log(x), x, and x2 are all positive for x>1, there are no zeros for this function beyond x=1. Bottom line: The function has exactly one zero located somewhere between 0 and 1, with no number crunching whatsoever.

Yet another way to look at this problem is from the perspective of potential. Since gravitation is a conservative force, for any given mass distribution exists a scalar potential function [math]\phi(\boldsymbol x)[/math] such that the gravitational force one an object of mass M located at a point x is the additive inverse of the produce of the mass M and the gradient of the gravitational potential at that point: [math]\boldsymbol F = -M\,\nabla \phi(\boldsymbol x)[/math] What is this potential function? The gravitational potential at a point x due to a point mass of mass m located at a point ξ is [math]\phi_{\text{pt. mass}}(\boldsymbol x) = -\,\frac{Gm}{||\boldsymbol x - \boldsymbol \xi||} = -\,\frac{Gm}{l(\boldsymbol x,\boldsymbol \xi)} [/math] where [math]l(\boldsymbol x,\boldsymbol \xi)\equiv ||\boldsymbol x - \boldsymbol \xi||[/math] is the distance between the points x and ξ. The superposition principle applied to potential functions. Therefore, the gravitational potential for a mass distribution characterized by a density function [math]\rho(\boldsymbol x)[/math] over some volume V is [math]\phi(\boldsymbol x) = -\,\int_V \frac {G \rho(\boldsymbol \xi)}{||\boldsymbol x - \boldsymbol \xi||}\,d \boldsymbol \xi = -\,\int_V \frac {G \rho}{l}\,d \boldsymbol \xi = -\,\int_V \frac {G dm}{l} [/math] The problem at hand is to develop the gravitational potential for an infinitesimally thin spherical shell of mass, with the mass uniformly distributed over the shell. Denote the density (units = mass/length2) of the shell as ρ. Denote the radius of the shell as r and the distance between the shell and the point in question as R. We're working with a sphere, so using describing the points on the sphere in terms of spherical coordinates will take advantage of the spherical symmetry of the problem. Define [math]\hat{\boldsymbol z}[/math] as the unit vector from the center of the sphere toward the point at which the potential is to be determined, [math]\hat{\boldsymbol x}[/math] as a unit vector normal to [math]\hat{\boldsymbol z}[/math], [math]\hat{\boldsymbol y}[/math] = [math]\hat{\boldsymbol z}\times \hat{\boldsymbol x}[/math], [math]\theta[/math] as the angle between the [math]\hat{\boldsymbol z}[/math] axis and a point on the sphere, and [math]\phi[/math] as the angle between the [math]\hat{\boldsymbol x}[/math] axis and the project of a point on the sphere onto the x-y plane. With this notation, the Cartesian coordinates (x,y,z) of the point on the sphere with angular coordinates (θ,φ) are [math]\aligned x &= r\cos\theta\cos\phi \\ y &= r\cos\theta\sin\phi \\ z &= r\sin\theta \endaligned[/math] The distance between this point on the sphere and the point in question is, by the law of cosines, [math]l = \sqrt{R^2+r^2-2Rr\cos\theta}[/math] Note that the angle φ is not involved in this distance function. The gravitational potential at the point in question is [math]\aligned \phi &= -\,\int_S \frac {G dm}{l} \\ &= - \int_0^{\pi} \int_0^{2\pi} \frac{G \rho}{\sqrt{R^2+r^2-2Rr\cos\theta}} r^2\sin\theta\,d\phi d\theta \\ & = -G 2\pi \rho r \int_0^{\pi} \frac {r\sin\theta d\theta}{\sqrt{R^2+r^2-2Rr\cos\theta}} \endaligned[/math] To evaluate the above, make the substitution [math]l^2 = R^2+r^2-2Rr\cos\theta[/math]. With this substitution, [math]Rr\sin\theta\,d\theta = l\, dl[/math]. The numerator in the [math]\phi = -G 2\pi \rho r \int_{l(\theta=0)}^{l(\theta=\pi)} \frac {l\,dl}{lR} = -\,\frac{G 2\pi \rho r}{R}(l|_{\theta=\pi}-l|_{\theta=0})[/math] The integration limits, [math]l|_{\theta=0}[/math] and [math]l_{\theta=\pi}[/math], are simply [math]|R-r|[/math] and [math]|R+r|=R+r[/math], respectively. For points outside the sphere, R>r and thus [math]|R-r|=R-r[/math]. For points inside the sphere, R<r, and thus [math]|R-r|=r-R[/math]. Thus for points outside the sphere, [math]\phi = -\,\frac{G 2\pi \rho r}{R}((R+r)-(R-r)) = -\,\frac{G 4\pi \rho r^2}{R} = -\frac {G M_s}{R}[/math] where [math]M_s=4\pi \rho r^2[/math] is the mass of the sphere. This is exactly the same as the potential due to a point mass, and thus a spherical shell looks exactly like a point mass for points outside the shell. For points inside the shell, [math]\phi = -\,\frac{G 2\pi \rho r}{R}((R+r)-(r-R)) = -\,\frac{G 4\pi \rho r}{r} = -\frac {G M_s}{r}[/math] This is a constant. As the gradient of a constant function is zero, a spherical shell results in zero gravitational acceleration for all points inside the shell. This proves the shell theorem. I could end with that. However, geistkie (geistkiesel) has asked about hemispheres as well. Using the same coordinate system as before, let S be the half of the spherical shell with positive z coordinates. Without going through the math, the gravitational potential for this hemisphere for points located on the z axis is [math]\phi(0,0,z) = -\,\frac{G 2\pi \rho r}{|z|}(\sqrt{z^2+r^2}-|z-r|))[/math] Note that this expression applies only to points on the z axis. I'll leave it up to you, geistkie, to take the gradient of this expression (Hint: compute [math]\partial \phi/\partial z[/math]). I made a claim in an earlier post that for any point mass located inside the spherical shell, the gravitational forces resulting from any partitioning of the shell into two measurable parts will be equal and opposite. Suppose you partition the shell into two complementary sets, [math]S[/math] and [math]S_C[/math]. After Herculean series of calculations, you manage to find [math]\phi_S(x,y,z)[/math]. Since the sets are complementary, the potential function for the complement of [math]S[/math] must be, by virtue of the definition of the Lebesgue integral, [math]\phi_{S_C}(x,y,z) = \phi_{\text{shell}} - \phi_S(x,y,z)[/math] For points inside the sphere, the potential for the entire sphere is just a constant. Thus for points inside the sphere, [math]\phi_{S_C}(x,y,z) = -\frac {G M_s}{r} - \phi_S(x,y,z)[/math] The gravitational forces on a test point mass of mass m located at (x,y,z), with [math]x^2+y^2+z^2<r^2[/math] are thus [math]\aligned \boldsymbol a_S &= -m\,\nabla \phi_{S}(x,y,z) \\ \boldsymbol a_{S_C} &= -m\,\nabla \phi_{S_C}(x,y,z) \\ &= -m \nabla\left(-\frac {G M_s}{r} - \phi_S(x,y,z)\right) \\ &= m \nabla \phi_{S}(x,y,z) \\ &= - \boldsymbol a_S \endaligned[/math]

The Senate didn't vote down a bill. They rejected an amendment that would have poisoned a bill they really want to get passed.

Note: I was talking about points interior to the spherical shell. It is a direct consequence of Gauss' law of gravitation and the additivity of Lebesgue integrals for measurable subsets of a space.

Prove this claim, with math. Hint: You can't, because its wrong. The forces exerted by each of the two shells are equal in magnitude but have opposite signs. In fact, if you split the spherical shell into two measurable parts, the forces exerted by the two parts will be equal in magnitude and opposite in sign. Your entire argument against the shell theorem is based on this faulty assumption.

Gotta love that site.

The problem here, Uri, is that you are stuck on the idea of weight being mass times gravitational acceleration. That is not the only definition. See post #9. From http://www.thefreedictionary.com/weight (emphasis mine): Note well: Gravitational weight cannot be measured. These definitions are alluding about some other aspect of the word "weight". They are in fact alluding to what I called in post #9 "scale weight". Gravitational weight obviously is not the meaning of weight used in the article you cited in the original post.

The answer depends on what you mean by "weight". Weight has multiple meanings. Legal weight. Legally (and colloquially), weight is a synonym for mass. A one pound can of beans weighs 10 pounds on the surface of the Earth, on the Moon, and on the space station. Gravitational weight: [math]W_g=m*g[/math], where m is the object's mass and g is the gravitational acceleration of the object. This is the definition of weight used in most introductory physics texts. One problem with this definition: It is unmeasurable. Scale weight: [math]W_s = \sum F_{\text{ext}}-W_g[/math], where Fext are the external forces (including gravity) acting on the object. This is the quantity measured by an ideal scale. This is the definition of weight used in some undergraduate physics texts very and is closely aligned with the definition of weight used in general relativity. So, if you use gravitational weight as your definition, two objects with the same mass have the same weight. If you use scale weight as your definition, two objects with the same mass can have vastly different weights. One of the external forces acting on an object is buoyancy, and buoyancy is a function of the object's volume.

I beg to differ. Wikipedia, as Swansont noted, is a great starting point. However, as a reference, it stinks. As a quality reference, it really stinks. Just a few of the problems: Wikipedia is constantly changing. Suppose you write a paper and use wikipedia as a reference. Five years later, I read your paper. Is your Wikipedia page still there? Maybe, maybe not. Is it the same reference you used? Probably not. Compare this to an encyclopedia reference. A proper reference includes the publication date. Five years later, I can go to the library and find the exact same reference you used. The writing quality varies a lot, all the way from terrible to excellent. The graphics quality varies a bit, too. Wikipedia graphics do not vary as much in quality as does the writing -- but that is because most of the graphics in Wikipedia are lousy. Wikipedia is free. The writers may or may not be experts in the field, Wikipedia doesn't hand out graphics to people who are good at creating graphics (most technical writers are not), and the peer review is haphazard at best. How does a naive reader know if they have just read one of the Wikipedia pages that would never make its way into an encyclopedia because it was written by and for crackpots? They answer is, they don't. The process by which encyclopedias are written filter out the garbage. Wikipedia is chock full of garbage. The comparison against encyclopedias is a bit of a red herring. By the time one gets to high school, and certainly college, encyclopedias are not good references. Prohibiting college-age students from using Wikipedia makes sense. College-aged students shouldn't be referencing encyclopedias, period. Prohibiting fourth graders? That might be a bit extreme. Who is going to read a fourth grader's report five years from now?

This is the subject of several physics FAQ entries. Does Gravity Travel at the Speed of Light? How does the gravity get out of the black hole? Gravitational Radiation Some Frequently Asked Questions About Virtual Particles In particular, see questions 4 and 5.

This is gobbledygook. For starters, there is only one sphere involved here. Next, before you advance to looking at the sphere as a whole you really need to understand the gravitational field induced by a ring of mass. You do not understand that result. Who made that claim? If anyone here made it, its wrong. Split a sphere in half. In fact, split a sphere any way you want into two separate parts. Now compute the gravitational force induced by the two parts of the sphere at some test point inside the sphere. No matter where you place the test point and no matter how you split the sphere into two parts, the forces from these two parts on the test point will be equal in magnitude and opposite in sign. Think of it this way. If a sin(theta) term does creep in, that means half the ring has negative mass. There is no sin(theta) term. I have asked you many times now to compute the gravitational force exerted by a thin hoop of mass on a test point mass. You have yet to do this.

Do you understand the notion of an orbital plane? The Earth orbits the Sun in a plane (more or less). Planes in three dimensional space can be characterized in terms of a point on the plane and a normal to the plane. In the case of the Sun-Earth orbital plane, the Sun and Earth are points on the plane. Now think of what this orbit looks like from the perspective of the Sun. Hold your right hand with your thumb extended and fingers curled. Orient your hand so that the your curled fingers represent the Earth's motion about the Sun. Your thumb is pointing normal to the orbital plane. Call this direction in which your thumb is pointing "+z". Now imagine you have a magical spacesuit that lets you stay at the L2 point indefinitely. Imagine orienting yourself such that the line from your feet to your head is pointing in the +z direction and the Earth and Sun are behind you. Now imagine you are rotating (with respect to the fixed stars) so that the Earth and Sun are always behind you. You are the origin of a reference frame, with +x forward, +y to your left, and +z up. This is the rotating reference frame in which the diagram in the article is represented. If you want to learn more about reference frames, google the term "reference frame". You really need to understand the concept if you want to have an explanation beyond "that's how it works".

That you think there is any distinction between the two terms is part of your problem. It is not a good idea to invent your own jargon that is contradictory to conventional notation. Doing so puts you dangerously close to the realm of crackpots. Total force and net force are synonyms. Good, but what happens to the components of the force normal to the ring axis? (Hint: It vanishes upon integration. However, since you have a hard time comprehending the vectorial nature of forces, it would be good for you to show this is the case.) Nope. First off, the mass of the ring is given. You should have computed the density in terms of the mass of the ring. More importantly, your equation is incorrect. In terms of the given mass m of the ring, the mass of the infinitesimally small portion of the ring within some infinitesimally small angle subtended from the center of the ring is [math]\frac m{2\pi}d\theta[/math]. If instead you want to treat the ring as a torus of density ρ with radii R and r, where R is the distance from the center of the torus to the center of the tube and r is the radius of the tube, then the mass of the torus is [math]m=\rho(\pi r^2)(2\pi R)[/math]. The mass of an infinitesimally small portion of the torus within some infinitesimally small angle subtended from the center of the torus is [math]dm=\rho(\pi r^2)Rd\theta[/math] Or, in terms of the total mass, [math]dm=\frac m{2\pi}d\theta[/math]. There is no sin(theta) term. Your mistake in the calculation of the differential force made you calculation of the net force incorrect. When you correct your errors, finish off the math. In other words, I want you to evaluate the integral. The rest of your post is rambling gobbledygook.

The math gets pretty hairy. This paper, http://www.aoe.vt.edu/~cdhall/papers/KIHAPaperAAS.pdf goes into the math in all its hairy detail. The easiest way to get a picture of what is going on is to look at things from the perspective of a rotating reference frame with origin at the L2 point. The reference frame rotates at the Earth's orbital rate. Ignoring the eccentricity of the Earth's orbit, the Earth and Sun will have fixed positions in this frame. The Sun, Earth, and L2 point are collinear. Call this line the rotating frame's x axis, with positive x pointing away from the Sun and Earth. The rotating frame's z axis is normal to the Earth's orbital plane, with positive z pointing along the the Earth's angular momentum. Finally, the y axis is normal to x and z, choosing the direction to form a right-handed coordinate system. Now consider what happens to an object located between the L2 point and the Earth (x coordinate is negative, y and z are zero) that is moving in the +y direction in the rotating frame. If the object's instantaneous velocity (in this reference frame) was zero, the object would experience an apparent force directed away from the L2 point (i.e., the -x direction). However, if the object's velocity is sufficiently large velocity, the coriolis force will counter this null velocity force, making the total apparent force be in the +x direction. The object will curve inward. A tiny bit of time later, the object will have moved a tiny bit in the +y direction. It's velocity vector will have a small +x component. The gravitational force will have a small -y component. The x component of the acceleration will have changed a little, but is still pointed in the +x direction. The object again curves inward. If you make that initial velocity just right, the object will follow a curved path around the L2 point and will come right back to where it started. In other words, it follows an orbit. This is called a Lyapunov orbit. The next step in complexity is to make the initial position below L2 and a bit out of plane. An initial velocity can once again be found that makes the object follow a closed path, this time called a halo orbit. The paths followed by Herschel and Planck are even more complex: They are called Lissajous orbits. These are not true orbits in the sense that they aren't a closed path. They are orbits in the sense that the state remains within some bounded region. An object located exactly at L2 and moving exactly with L2 will stay there forever -- but that placement has to be exact and ignores perturbations induced by the Moon and the other planets, by non-circular nature of the Earth's orbit. The object could stay close to L2, but this would require constant application of thrust. Just as L2 is unstable, so are Lyapunov orbits, halo orbits, and Lissajous orbits about it. However, the energy needed to maintain those orbits is less, a lot less, than the energy needed to stay exactly at L2. All vehicles that operate in the vicinity of one of the collinear libration points are placed in some kind of orbit about the point.

Moved to pseudoscience. Reality doesn't care what you believe. Reality is what it is -- and as far as we can tell, Einstein is right. Space and time are not nearly as simple as you would like them to be. There is overwhelming evidence in favor of relativity. If Einstein's theories are wrong (and they probably are at some level), whatever comes along to replace them must still display things like time dilation and length contraction because those phenomena are very real.

J.C., I never said that the single hoop can be modeled as a single point mass. The force can nonetheless be calculated. Calculating the force for a point along the hoop's axis is not that hard. Knowing how to do that is central to understanding the shell theorem. I am asking geistkei, and not you, to provide this calculation.

The reason I asked the question is because knowing the answer to that question is central to knowing how to calculate the total force exerted by a spherical shell. You have to integrate over a surface. In short, you have to do a double integral. By far the easiest way to do this double integral is to slice the spherical shell into infinitesimally small circular hoops.

You have it backward, twice around. The spaceman sees the Earth as being subject to length contraction. He sees himself as the same height he has always been. People on the Earth see the spaceman as being subject to length contraction. They see themselves as the same height they have always been.

I am one of this site's physics experts, which means I have limited moderation capability. By taking active part in this thread I can no longer apply these capabilities to this thread. I can however ask moderators who are not taking part to put this thread on a suicide watch pending your failure to answer the question posed in post #6. I want you to answer the question posed in post #6, That is not the answer. m is the mass of the ring, not the test particle. I asked for acceleration, not force. If you want to use force rather than acceleration (which is what I asked for), you need to introduce the mass of the test particle. Don't call that mass m because that is the given mass of the ring. That is a differential force, not the total force. You have the magnitude of the differential force correct, but in which direction does that differential force point, and how does this differential force relate to the total force? It's not right, even as a differential force. You don't have the magnitude right, and you don't even have the force represented as a vector. Hint: Do the math. The answer I am looking for will say two things: The direction of the total gravitational acceleration (or force) induced by the ring. The magnitude of this total acceleration (or force), in terms of the given quantities m, the mass of the ring; r, the radius of the ring; and l, the distance between the test point and the center of the ring.

Now we're getting there. Neither prediction nor explanation suffice. A conjecture might be strong in terms of explanatory capability, but if it those explanations don't conform with reality the conjecture is worthless. An ad hoc scientific law might be strong in terms of predictive capability, but the ad hoc nature of the law leaves something wanting: Specifically, why? A scientific theory explains why ad hoc scientific laws arise and when they are valid. Theory is the glue that binds prediction and explanation. I'll use blackbody radiation as an example. Wein's law (1896) does a good job of describing the high frequency spectrum of a blackbody but fails to describe the spectrum at low frequencies. Max Planck, in what was essentially an exercise in curve fitting, found an equation involving two rather arbitrary constants that yielded a much better fit. He presented this in October 1900, but quickly realized the solution was not unique. He rescued his law by inventing what was to him a purely mathematical construct, energy quantization. He presented the modern version of Planck's law in December 1900 (published 1901.) Planck himself saw this quantization as a mathematical fiction whose sole virtue was that it let him rescue his law and tie it to thermodynamics. Planck's law by itself is just that -- an ad hoc scientific law. It remained such for two decades. It took a decade for physicists to realize the true significance of Planck's quantization. It took another decade for physicists to come up with an explanation of why Planck's law is indeed valid. That explanation arises from statistical mechanics using the Bose-Einstein statistics for massless particles. Science is all about explaining. Those explanations must of course match reality, and aren't worth much if they can't predict outcomes. Scientific theory marries explanation, post-diction (matching reality), and prediction.

geistkie, in order to understand how Newton's shell theorem is just that -- a theorem, you first need to know how to compute the gravitational acceleration toward a thin hoop of mass. So, forget the shell for a bit. Suppose there exists a thin hoop with mass m of radius r. What is the gravitational acceleration, in math, not in words, at some point located on the hoop's axis at a distance l from the center of the hoop? Read the rules of this site. You need to answer this question if you have any hopes of keeping this thread alive.

Moved to pseudoscience pending moderation.

Not moving with respect to what? Keep in mind that there is no such thing as an absolute reference frame.