J.C.MacSwell

Neither sounds like you will be throwing away your future. What would you most like to do for the next few years?

The bold is false. You are in other reference frames relative to which you are close to the speed of light, yet you are in no danger of collapsing into a black hole in those frames any more than you are in your rest frame.

That is correct. So you see that, yet you are continuously treating the sphere as if the mass was all concentrated along the same axis when you do your calculations. This causes your error.

LOL. If you are having us on that is brilliant. Answer: If he arrives at speed with is body aligned in the direction of travel (effectively a fly by)...yes. If he arrives by coming to rest...no.

Geiestke. Imagine a sphere lying with it's COM directly along the x axis from a test mass Now replace the sphere with a magic stick. 1. The magic stick has the same COM as the sphere. It's COM is positioned in the same place as the COM of the sphere was. 2. The axis of the stick lies along the x axis 3. The magic stick has the same length as the sphere's diameter 4. The magic stick has the same mass as the sphere 5. The magic stick has the same mass distribution as the sphere in the x direction but it is magically concentrated along the x axis Now the question is: Is the force of gravity between the test mass and the magic sphere greater, lesser, or the same as it was between the test mass and the sphere? Do you see the difference?

I think that is his answer right there if he thinks about it carefully, even without doing the math he should see where he is erring. In the above the closer "twin" contributes more to the resultant force though, correct? The net force along the axis is greater for the closer twin. The only reason I ask is that I'm not sure Geistke feels we see this.

From the laws of physics proven by experiment.

27 for me. What do I win?

What makes you arbitrarily do this? It is wrong. Segmenting further into quarters with m1 at 8.5, m2 at 9.5 m3 at 10.5 and m4 at 11.5. is more wrong The more segments the more wrong it gets even if it converges to limit you error.

LOL. You can say that again! (just not in this thread )

Not sure if If I follow what you are saying exactly, but by continuously using the COMs along the x axis in your calculations you end up with higher forces, because you are generally picking a point closer to the test mass than the average distance from the test mass of all the points it represents. If, say, your segments were an infinitesimally thin set of discs perpendicular to the x axis, and you used the COM of each disc, each point chosen to represent the disc would be the closest point on each disc to the test mass and lead to an overestimation of the force, since you continuously underestimate the effective distance. For some shapes this effective distance is further than the COM, for others it is closer, but for the sphere it is exactly the right distance.

I have some chubby checkers. They stay on the board better than my thin checkers, especially outside on a windy day.

Geistke. The half spheres cannot be modeled as mass points at their respective centers of mass for this purpose (the inverse square law). You agree with that, correct? Yet you are using that assumption to prove that it cannot be true for the sphere itself. The half spheres, not being spheres or spherical shells, cannot be modeled this way. Spheres and spherical shells can be modeled this way. Look in particular at the closer half sphere, the half that is responsible for the majority share of the force. If it had to be replaced by a point mass (of equal mass) that would bring about an equivalent share of the force, this point mass would have to be positioned further from the test mass than the center of mass of the half sphere is positioned. I don't think you realize this point. It is not so much that your logic is wrong, as it is the assumptions you are making.

How about a bucket of dirt. Really dirty dirt. I don't want to talk about it. And I have no pictures of it.

So my mum's threat some 40 years ago about growing potatoes in my ears - nothing to worry about?

1. abuse (hacking etc,) 2. use (wear and tear) 3. neglect (lack of maintenance over time) I think that is a common term you might be looking for. Possibly decay or weathering as well.

I'm pretty sure GDG had it right for a hot air balloon. Your right that it doesn't or shouldn't leak out, but it certainly is free to get out, as it is open at the bottom. http://www.inhabitat.com/wp-content/uploads/hotairbal-ed01.jpg

101%! I think we are assuming the majority are liquid hippos.

So...how does that differentiate time dilation due to expansion displacement vs movement displacement? Not disagreeing or agreeing there's a difference, but I don't see it in those paragraphs.

Ouch.

Moist air is heavier though, when it has liquid water suspended in the air. I think a lot of people consider that moist air.

Nice Link, but can you direct us to where it makes the time dilation distinction between expansion recession and recession due to movement otherwise? If we are both on the CMB rest frame but very far apart and I send a probe that accelerates toward you, closes the gap, decelerates less than it accelerated originally to come to a halt in your lap, is it then in the same time frame it started? I don't know, and saw nothing in that link that would indicate either way when I scanned it. Edit: I hadn't read NTWK's latest post. I will re-check and try and see if it makes sense to me. Re-edit: Checked it rather quickly but not sure that's the key. Isn't the expansion of the Lorentz interval not equal in all directions if you are not at rest wrt the CMB? How can it be invariant for GR if that is the case? (I can see how it is invariant for SR)

How about a mole of moles?The furry kind that can't see very well. That must be closer to the moon's size.

Fair enough, you have used pretty much the same definition of weight in each case. However, if you used a scale, 10 kg of lead would weigh more than 10 kg of feathers due to the greater buoyant force of the atmosphere on the feathers. That is more or less the definition some others have used in this thread. They have allowed buoyancy to be factored into the terms heavier or lighter. Still valid, but they are weighing things differently. OTOH, mass is always defined the same way. So on the moon, with no buoyant forces from the atmosphere 10 kg of lead or feathers weigh the same, even on a scale. The difference might be subtle, but become quite significant with respect to something as "light as air".

Not if while and when are defined in the same frame.