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May 26[edit]

The problem with gravity.[edit]

Not sure if this is taken care of by GR, but I'll mention it anyway.

In the Newtonian universe, we have a G force acting perpendicularly to a planet orbiting a sun. But if this is so, then there is no force to decelerate the planet so that it can change direction whilst maintaining speed. So there is no way this force can be perfectly perpendicular - it would need to point in a direction a little backward from the motion of the planet. Byron Forbes (talk) 01:14, 26 May 2023 (UTC)[reply]

None of this requires GR. You're forgetting velocity is nonzero and perpendicular to the radius (or else conflating velicity and acceleration). One way to think about it is that the acceleration changes the velocity, so the planet's future position is angled compared to previous. It doesn't just affect the position directly (pulling it inward). Or another way to think about it is that there is forward motion and sideward pull, so the net effect is motion "forward and a little sideward". Because "sideward" is towards the sun, the long-term effect that it keeps curving around. Or yet another way, one way it's explained to kids, is that the planet is falling towards the center but moving forward, so it keeps "missing" and instead goes around. DMacks (talk) 01:25, 26 May 2023 (UTC)[reply]

Lets look at this at a point in time where the velocity of the planet is due north at the 3 o'clock position. Now look at time t' just a moment later. In order for both direction to change and speed be maintained, the velocity component in the Nth direction must decrease. But if the G force was/is always perpendicular, then how can this happen? Byron Forbes (talk) 02:58, 26 May 2023 (UTC)[reply]

The velocity component in the West direction increases because the force of gravity is pulling due-west (from 3 o'clock towards the center). So the direction changes (the angle of the velocity but not its magnitude, so speed is constant). DMacks (talk) 03:10, 26 May 2023 (UTC)[reply]

The speed remains constant because momentum is conserved. The westward acceleration due to the sun's gravity increases the velocity in the westward direction, so the velocity in the northward direction must decrease to keep the total speed unchanged. As a simple analogy, imagine a stick lying on the ground pointing northward. Its projection in the north-south direction is equal to its length, and its projection in the east-west direction is zero. If you move the stick so that it's pointing slightly northwest, the length of the stick remains unchanged. But its projection in the north-south direction is smaller that it was originally, and its projection in the east-west direction is larger. CodeTalker (talk) 04:36, 26 May 2023 (UTC)[reply]

Try putting a weight at the end of a string and spinning it round. You can stop moving your hand and the weight will continue to go round in a circle until frictin stops it. The string can't exert a force except inwards towards your hand. NadVolum (talk) 08:40, 26 May 2023 (UTC)[reply]

Have a look at orbit. Kepler (1571–1630) first realised that the orbits are not circular, they are ellipses. His laws included the "equal areas" observation (his second law) though his assumption that the sun was at exactly one focus of the ellipse was subsequently modified by Newton to use the combined centre of mass of the whole system. To return to the OP's question, gravity will always act towards the focus, and so for half the orbit it is ahead of the perpendicular to the planet's path, thereby accelerating it. For the other half it is behind the perpendicular as so retards the orbital speed. Martin of Sheffield (talk) 09:16, 26 May 2023 (UTC)[reply]

True. But I think the OP's question is also germane to near circular orbits such as those of our geostationary satellites which orbit with near constant speed. In this case, the change in orbital speed is negligible and one can note that the OP's Nth component deceleration is balanced by a perpendicular Wst component of acceleration (as pointed out above by DMacks), assuming a uniform inward force exists... Modocc (talk) 15:41, 26 May 2023 (UTC)[reply]

It was the OP's use of the phrase "there is no way this force can be perfectly perpendicular" that rang an alarm bell. It's correct, but not for the reason argued. Is it possible that the OP is thinking of simple vector diagrams which show neat little triangles and hasn't let the calculus go to the infinitesimal calculus? I'm neither a mathematician nor a teacher so having sown the thought I'll beat a hasty retreat! Martin of Sheffield (talk) 15:54, 26 May 2023 (UTC)[reply]

When I was at NCSU I improved a circle drawing algorithm and noted that every linear Cartesian coordinate step in one direction corresponded precisely one-to-one with steps in the perpendicular direction. In theory, one can model an infinite number of such steps, but pragmatically it's best to take just a few steps (for a circle) or even just one or two on account of the fact that smaller component steps sum to larger ones even if these are more abstract than real because the circles' arc primarily passes near finite rectangular coordinates. With a spherical coordinate system centered on the central body, there is less ambiguity regarding a particle's trajectory for which we do actually observe perpendicular inward forces that constrain masses to circular motion, such as the weight and string mentioned by NadVolum. Net forces can be perfectly perpendicular, including gravitational forces, but orbits often deviate from that. Modocc (talk) 17:01, 26 May 2023 (UTC)[reply]

Now that makes a lot of sense! Cheers! Byron Forbes (talk) 16:49, 27 May 2023 (UTC)[reply]

No reason a string cannot have shear forces, and I would suggest it must have. Byron Forbes (talk) 16:48, 27 May 2023 (UTC)[reply]

The string is only under tension and the linear force components are sinusoidal. At 3 o clock the magnitude of the Nth/Sth component of the tension crosses zero. Modocc (talk) 19:11, 27 May 2023 (UTC)[reply]

You could try putting the weight with a string attached onto a table and then flick the end of the string sideways with your finger. It will be interesting to see how much shear force you can impart to the weight. You could try with a thread as well. NadVolum (talk) 20:07, 27 May 2023 (UTC)[reply]

Mathematically orthogonality of the acceleration to the velocity if the absolute value of the latter is kept constant can be easily proved. Suppose that

\mathbf {v}

is the velocity, then the acceleration is

d\mathbf {v} /dt

. Now, if

|\mathbf {v} |^{2}=constant

, we have

d|\mathbf {v} |^{2}/dt=d(\mathbf {v} \cdot \mathbf {v} )/dt=2\mathbf {v} (d\mathbf {v} /dt)=0

, which proves the orthogonality. Ruslik_Zero 20:46, 26 May 2023 (UTC)[reply]

dv/dt = 0? The speed isn't changing, but the Velocity (direction) is always changing. That is a vector quantity you have there. Byron Forbes (talk) 16:56, 27 May 2023 (UTC)[reply]

The scalar product of the velocity and its derivative is zero, which means orthogonality. Ruslik_Zero 18:15, 27 May 2023 (UTC)[reply]

I do not understand the argument why the force cannot be perpendicular. Applying Newton's laws of gravitation and of motion to the two-body problem gives us a differential equation whose solutions have trajectories that are conic sections. Given the right initial conditions the bodies are in a circular motion around the centre of gravity, in which case the gravitational force is always perpendicular to the direction of motion. If the trajectory is a non-circular ellipse, the force is perpendicular only at the apsides and otherwise alternatingly either slightly forwards or backwards – forwards on the planet's way to the perihelion, and backwards on its way to the aphelion. --Lambiam 21:48, 26 May 2023 (UTC)[reply]

Can Vitamin K in say Spinach or Broccoli be destroyed[edit]

Does drying / cooking / freezing vegetables containing Vitamin K (such as Spinach or Broccoli) destroys the vitamin?

Thanks. 2A10:8012:17:CDC6:49DF:B235:F14:5969 (talk) 17:52, 26 May 2023 (UTC)[reply]

This page at the University of Rochester notes that it is not particularly temperature sensitive, but that light exposure can damage vitamin K, and recommends avoiding long exposures to strong light. --Jayron 32 18:20, 26 May 2023 (UTC)[reply]

Journal auto-skim[edit]

I'm looking to set up an auto search and email notify for recent scholarly articles on distributed cognitive architectures. How would I go about it? Google Scholar and Google alerts? Temerarius (talk) 23:18, 26 May 2023 (UTC)[reply]

Google scholar will be much more selective for scholarly articles, so recommended. Try here after you are logged on to Google:scholar_alerts Graeme Bartlett (talk) 00:24, 27 May 2023 (UTC)[reply]