h2g2 The Hitchhiker's Guide to the Galaxy: Earth Edition

Hallo! Is anyone there? It has gone very quiet these last few days.

I was hoping, once the principle had been clarified, we might take a look at some of the implications, which might suggest simple answers to some of the anomalies created by the theory of a pull.

Would that be going further than you have time to take the matter?

Hello antimather, yes there is at least one other person on this thread. It's just that work has taken off suddenly. I return here every day, but it's not easy to do so for more than a few minutes at the moment. I will say something constructive on the weekend, I hope.

Something that could really help the argument for the general reader, as you said earlier, would be a diagram of the two bodies illustrating the shielding effect. I have not been able to find one anywhere to link to.

I remember seeing just the sort of illustration of the mutual shielding effect we need, in an entry on the Web. What I can't remember is where, and my search for it has been unavailing.

By way of practice I have managed to assemble a diagram illustrating how the inverse square ratio fits into the picture, so I am now trying to see if I can do the same for the shielding effect. The principle is the same but the practical problems are of a quite different order.

The biggest problem with creating diagrams is how to get them into formats that can be transmitted under restricted protocols.

Even assembling in Windows Visio for incorporation into Windows Word text has been troublesome, but I think I have now got simple diagrams to illustrate the geometry of the inverse square, and of mutual shielding by celestial bodies from an external force, too.

They are more practical explanations than dramatic eye-catching pictures and to that extent may help get the message across to anyone sufficiently interested. Whether anyone really wants to know is another matter.

Perhaps Michael Flanders was right in saying that it is the function of satire to strip away the veneer of comfotable illusion and cosy half-truth! Unfortunately that is not my scene.

<>

Ah, but he added in almost the same breath that it was his job (and Donald Swann's) to put them right back again!

Anyway, I am looking for a diagram: I think it is key, as an image is worth much more than a thousand words in this case.

Look no further!

In default of a ready-made drawing of the mutual shielding effect, I have contrived one, after much labour on my computer, which I believe gives a clear picture of it. In conjunction with my earlier diagram of the geometry of the inverse squares, they show how the one involves the other, leading to a general statement of the law and a numerical evaluation of it for Earth's gravity. That seems to be enough for Chapter 1.

Chapter 2 is the more daunting one of examining all the implications arising out of it, which you may feel is taking matters further than either of us could manage and too far into the field of academic research. What do you think?

Congratulations! Can you give me instructions to construct the same diagram?

Thanks! but the process involved so much trial and error using Microsoft Visio to turn a flat schematic diagram into a passible semi-transparent perspective drawing that I cannot begin to describe it.

All I can say is that it involves a constant stream of expletives and curses on a system that seemed incapable of following simple commands, which finally yielded the desired result. To quote the same source again: 'You can try it if you like, but you'd far better notta'!

Hmmm. Ok, what does the result look like? Can you begin to describe it? I know it's not easy at the drop of a hat...

I am in something of a quandary trying to describe in words a drawing that is meant to illustrate a concept that is itself hard to convey without resort to a drawing, without referring back to the original concept that needs illustrating. I feel like a dog chasing its own tail!

However if one imagines a body with a satellite, like the Earth and the Moon, and the radii of each extended as tangents to the periphery of the other, enclosing identical areas at their own surfaces as in describing the inverse square ratio to distance apart, and then extends the radii as diameters in each direction to 'infinity' from which a universal force is received, it may make it possible to envisage an imbalance in it along the line joining their centres, varying in the correct proportion.

A single drawing to illustrate this gets rather cluttered, the more so when pursuing the argument that the resultant force is most simply calculated in terms of a notional sphere on the periphery of which the second body lies. This, though, makes the point quite clearly that the same force of gravity applies, whether at the surface or at a distance, at the same time illustrating the connection with Kepler's law.

I hope you get the picture but there are just too many interrelated factors to be accommodated in one two-dimensional drawing, calling for different projections, sectional, isometric and perspective, to convey the whole message. It is not very much easier in words!

<>

I've constructed this diagram. The surface areas where the radii intersect the surfaces look plausibly equal - actually as it's a 3D diagram it's the lines that look equal in length - although I suppose there is a small error as they are not straight lines: is that right?

<>

This is the bit I cannot interpret. Extend radii in what direction (same as before?), and what is meant by "extend radii as diameters"?

My posting No.36 gives the trig showing the contained surface areas to be the same, being R x r/D in one case and r x R/D in the other, to within extremely fine limits for small angles that no drawing could disclose. The largest possible value where two spheres of equal size touch is 0.5, when symmetry confirms they must still be the same, though the inverse square ratio is no longer precise.

<> was a bit of careless phrasing for which please substitute <>; beyond the tangent point and through the centre and out to the further surface. The significance of diameters in place of radii is that they determine the relevant volume in which the force is absorbed to break the surrounding equilibrium.

Does that make the picture a bit clearer?

Well, I think I understand paragraph 2 now. The problem has been transferred to paragraph 1, because your posting no.36 mentions right angles, and *sigh* there aren't any on my diagram.

The right angle, you must remember, is between the tangent and a radius to the point of contact. Yes/no?

Well, yes, the thing is, when you actually draw the diagram, unless the two circles (2D planets) are small relative to the distance between them, there is a big margin of error here: the right angle does not drop automatically out of the diagram. (Or does it? Have I constructed it wrongly?)

I fear its it's a matter of 'back to the drawing board'.

The hypotenuse of the triangles is the line joining centres of the 2D planets. The other two sides being radii and tangents to the respective circles. The ends of the tangents that meet at the centres of the circles are then extended to the far rim as diameters of them to complete the picture.

I suspect this might be different from your earlier layout, if I understand you correctly. Am I right?

Aha! I think I'm lacking only one line on the construction.

I have a line extending from the centre of one circle to the periphery of the opposite one, lightly brushing it by. I also have a line extending from the centre of one circle to the opposite one. Then the missing line is from the interior of the opposite circle to the point where the first line lightly brushes it by, forming a tangent. And the right angle is where those two lines meet. Is that right?

This is just the first of four such constructions.

I think my problem was with the phrase "the radii of each extended as tangents to the periphery of the other". It didn't tell me to draw the final line.

Ok, is that the right diagram? Are there any lines missing before I try out the trig proof?

I think the geometry works. Take the triangle formed by i) the line from the centre of circle radius R to the edge [brushing it by] of circle radius r, ii) the radius r [at right angles to line i)], and iii) the line of length d connecting the centre of both circles. The sine of the smallest angle (which appears at the centre of circle radius R) is the opposite over the hypoteneuse, which = r/D. Then to find the length of the short side at the surface of circle radius R, one scales up by a factor of R. So the length is (r/D)xR.

And take the triangle formed by iv) the line from the centre of circle radius r to the edge of circle radius R, v) the radius R [at right angles to line a)], and vi) the line of length d connecting the centre of both circles. The sine of the smallest angle (which appears at the centre of circle radius r) is the opposite over the hypoteneuse, which = R/D. Then to find the length of the short side at the surface of circle radius r, one scales up by a factor of r. So the length is (R/D)xr.

Which means that they both = Rr/D, and are equal. QED. Of course, as you say, the curvature means that they are infinitesimally different.

So we've got that far at least! Well, I hope that's correct.

Drinks all round!

The geometry of the inverse square is exactly as described trigonometrically. Its significance is in relating relevant volumes in which the force is moderated, to the square of the distance between centres, which is a lot easier to imagine than to illustrate convincingly.

I see the way to an inverse square law, but how should one convert this line at the surface into an area - I mean, what shape should one use? The intuitive approach is to use a cone, but you said something about that earlier: the surfaces would not add up to the area of the sphere.

I can see that if you visualise lots of cones there is a what one might term a "packing problem", but is it actually a problem? Is there a need to visualise lots of them like that? Surely, if we are talking about two spheres, the relevant area must be a circle at the surface, so the relevant volume must be a cone (although the base has a slight curvature, which we are ignoring, I think).

Just a thought to throw in before the algebra...