## 1547: "Solar System Questions"

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Eoink
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### Re: 1547: "Solar System Questions"

rmsgrey wrote:
Neil_Boekend wrote:Also: the radiation pressure depends mostly on your surface area while the gravity depends on your mass. Whether they balance out depends on your surface/mass ratio. A solar sail will have winning radiation pressure (on all sane distances) while a uranium sphere will have winning gravity.

A uranium needle will have winninger gravity.

But as a physics graduate, I will immediately assume your needle is a sphere in a vacuum.

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### Re: 1547: "Solar System Questions"

rmsgrey wrote:
Neil_Boekend wrote:Also: the radiation pressure depends mostly on your surface area while the gravity depends on your mass. Whether they balance out depends on your surface/mass ratio. A solar sail will have winning radiation pressure (on all sane distances) while a uranium sphere will have winning gravity.

A uranium needle will have winninger gravity.

Some maverick fashion designer is right now specifying that their new collection be stitched using D.U. needles.
xtifr wrote:... and orthogon merely sounds undecided.

keithl
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### Re: 1547: "Solar System Questions"

Neil_Boekend wrote:
keithl wrote:
DHT wrote:Yes, the Oort cloud is real. It is the region where the gravitational pressure of the sun (inward) and the radiative pressure of the sun (outward) roughly equalize.
Both gravity and radiation pressure from the Sun are inverse-square - they balance at all >1 AU distances. ...
Also: the radiation pressure depends mostly on your surface area while the gravity depends on your mass. Whether they balance out depends on your surface/mass ratio. A solar sail will have winning radiation pressure (on all sane distances) while a uranium sphere will have winning gravity.
That is true, but irrelevant to the Oort cloud. Of course, the balance depends on the sun-facing area to mass ratio - at ALL distances, not just the Oort cloud. It also depends on velocity; centrifugal acceleration compensates for gravity[citation needed]. What is "out there" is Really Cold hydrogen and water molecules, and tiny amounts of heavier elements, mostly ejected by light pressure unless the particles are big enough for the mass-to-surface area to approximately balance. Albedo and shape is important - reflected light increases light pressure somewhat (for a perfect sphere, perhaps 45%).

Light pressure for a black particle is power flux divided by c, and the power flux is inverse square. At earth radius, the power flux is 1367 W/m², the light pressure is 4.56e-6 N/m², and the solar gravitation is 5.93 mm/s². That supports a black particle with a mass to area ratio of 0.77 g/m², or a reflective spherical particle with a mass to area ratio of 1.12 g/m². And that is true all the way out to the borders of other star systems, it is not sensitive to distance, the first point I hoped to make in my posting.

The numbers I gave apply today, 4.6 billion years after the sun formed, but solar luminosity ramped up from zero over time, so the "balance" occured for smaller particles in the past, and larger stuff fell inwards. A competent stellar astronomer can tell you what that means for the evolution of the Kuiper belt and (hypothetical) Oort cloud.

The (hypothetical) Oort cloud contains few solar sails, uranium spheres, uranium needles, unicorns, or Russell's Teapots. And it remains hypothetical because if it does exist, it is so cold and diffuse and distant that it is virtually impossible to observe. The much easier-to-observe Kuiper belt is still mostly unexplored - there may be dwarf planets larger than Pluto and Eris out there. But then, the vast majority of the matter in the universe is unobservable with current technology.

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### Re: 1547: "Solar System Questions"

rmsgrey wrote:
Neil_Boekend wrote:Also: the radiation pressure depends mostly on your surface area while the gravity depends on your mass. Whether they balance out depends on your surface/mass ratio. A solar sail will have winning radiation pressure (on all sane distances) while a uranium sphere will have winning gravity.

A uranium needle will have winninger gravity.

I was originally going to go with a black hole, but discussing black holes and the influence the gravity of another object (in this case the sun) upon it gets confusing fast. So I changed it to a uranium sphere.
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danegraphics
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### Re: 1547: "Solar System Questions"

Why are all the blotches on the near side?

I supposed gravitational pull has a great deal to do with the way that the volcanos on the moon were oriented (back when it was developing). Considering the fact that only one side of the moon ever faces earth, that is the side that, if there were water, would have a constant "high-tide" due to the gravitational pull of the earth.

My theory (which I supposed to be the only logical explanation so far) is that earth's gravity simply created enough pull that all the volcanos on our side erupted a heck of a lot more than the volcanos on the other side. Perhaps the volcanos were created by the gravitational pull of earth. Who knows?

But all in all, I blame gravity.

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### Re: 1547: "Solar System Questions"

silvermorph wrote:Q: Why are all the blotches [on the moon] on the near side.

Speculative answer: They aren't. The far side of the moon collects small impacts much more frequently than the near side, because the near side is generally protected from small impacts by the earth. These small impacts break up the relatively smooth lava beds, and over time obfuscate the splotches. You can still see the vague outline of splotches on the far side in google earth's moon view, they're just dusted over with lighter impact dust.

Going off of that, I think it might be a combination of that along with the gravitational pull of earth affecting the eruption frequency of volcanos on the moon. Considering that only that side face earth, that side would be in a constant "high-tide" and the opposite side would be in "low-tide", and the volcanos on the opposite side would be less active than those on the side facing earth.

orthogon
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### Re: 1547: "Solar System Questions"

danegraphics wrote:
silvermorph wrote:Q: Why are all the blotches [on the moon] on the near side.

Speculative answer: They aren't. The far side of the moon collects small impacts much more frequently than the near side, because the near side is generally protected from small impacts by the earth. These small impacts break up the relatively smooth lava beds, and over time obfuscate the splotches. You can still see the vague outline of splotches on the far side in google earth's moon view, they're just dusted over with lighter impact dust.

Going off of that, I think it might be a combination of that along with the gravitational pull of earth affecting the eruption frequency of volcanos on the moon. Considering that only that side face earth, that side would be in a constant "high-tide" and the opposite side would be in "low-tide", and the volcanos on the opposite side would be less active than those on the side facing earth.

There are tides on both sides! *

What I mean is, there is a high tide on the side facing the other body, and on the opposite side, and low tides halfway between these, hence the two high tides per day on Earth. But perhaps you didn't mean tides literally.

* continue in the style of Spike Milligan / Dr Seuss
xtifr wrote:... and orthogon merely sounds undecided.

danegraphics
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### Re: 1547: "Solar System Questions"

orthogon wrote:
danegraphics wrote:
silvermorph wrote:Q: Why are all the blotches [on the moon] on the near side.

Speculative answer: They aren't. The far side of the moon collects small impacts much more frequently than the near side, because the near side is generally protected from small impacts by the earth. These small impacts break up the relatively smooth lava beds, and over time obfuscate the splotches. You can still see the vague outline of splotches on the far side in google earth's moon view, they're just dusted over with lighter impact dust.

Going off of that, I think it might be a combination of that along with the gravitational pull of earth affecting the eruption frequency of volcanos on the moon. Considering that only that side face earth, that side would be in a constant "high-tide" and the opposite side would be in "low-tide", and the volcanos on the opposite side would be less active than those on the side facing earth.

There are tides on both sides! *

What I mean is, there is a high tide on the side facing the other body, and on the opposite side, and low tides halfway between these, hence the two high tides per day on Earth. But perhaps you didn't mean tides literally.

* continue in the style of Spike Milligan / Dr Seuss

Exactly. Hence the quotes.

I was specifically referring to the gravitational "tide" on the side facing earth, which, no doubt, is much greater than that on the other side, leading to more active volcanos.

speising
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### Re: 1547: "Solar System Questions"

No, it isn't greater. The tidal forces are the same on both sides.

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### Re: 1547: "Solar System Questions"

speising wrote:No, it isn't greater. The tidal forces are the same on both sides.

I was just coming to a similar conclusion. Also, because the Earth is always on the same side, the Earth's gravity effectively is part of the Moon's "geoid" (lunoid?); has the Moon had enough time to reach equilibrium? I.e. is it stretched slightly in the direction of the Earth such that there aren't any tidal forces?
xtifr wrote:... and orthogon merely sounds undecided.

danegraphics
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### Re: 1547: "Solar System Questions"

speising wrote:No, it isn't greater. The tidal forces are the same on both sides.

Actually no, they're not, and mathematically could not be. Through an odd combination of the moon's gravity, the sun's gravity, and earth's spinning forces, there are 4 tides on earth. A greater (near) and lesser (far) sun tide (not very big due to distance), and a greater and lesser moon tide, and these overlap differently depending on the position/phase of the moon. The spinning earth passes through these tides and the spinning itself contributes to the shape and size of the tides. Needless to say, earth's tides have many changing variables that create them.

The moon though, is only spinning around the earth, always facing it. The only way the tidal forces would be equal on both sides is if the moon were a point. It is not. It is a sphere, with one side being more affected by earth's gravity by proximity than the other. The one side always facing earth feels earth's gravitational pull to a greater degree than the other side, and the moon's lack of spin relative to the gravitational pull of earth and the distance from the sun prevent the other side of the moon from feeling any significant (relative to the side facing earth at least) "lifting"/"high-tide" effects gravitationally.

speising
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### Re: 1547: "Solar System Questions"

What happens when two opposite, but not equal forces act on a body?

rmsgrey
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### Re: 1547: "Solar System Questions"

danegraphics wrote:
speising wrote:No, it isn't greater. The tidal forces are the same on both sides.

Actually no, they're not, and mathematically could not be. Through an odd combination of the moon's gravity, the sun's gravity, and earth's spinning forces, there are 4 tides on earth. A greater (near) and lesser (far) sun tide (not very big due to distance), and a greater and lesser moon tide, and these overlap differently depending on the position/phase of the moon. The spinning earth passes through these tides and the spinning itself contributes to the shape and size of the tides. Needless to say, earth's tides have many changing variables that create them.

The moon though, is only spinning around the earth, always facing it. The only way the tidal forces would be equal on both sides is if the moon were a point. It is not. It is a sphere, with one side being more affected by earth's gravity by proximity than the other. The one side always facing earth feels earth's gravitational pull to a greater degree than the other side, and the moon's lack of spin relative to the gravitational pull of earth and the distance from the sun prevent the other side of the moon from feeling any significant (relative to the side facing earth at least) "lifting"/"high-tide" effects gravitationally.

The radius of the Moon's orbit is pretty close to 200 times the Moon's radius. The difference in tidal forces between the nearest and farthest points of the moon's surface is on the order of 1%. If the far side only felt 90% of the force the near side does, it would still be a significant high tide effect.

As for the solar tides on Earth, the difference between near and far tides is even smaller.

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### Re: 1547: "Solar System Questions"

The far side of an orbiting body doesn't feel "lifting" effects due to the gravity of the primary, but it does feel them due to inertia, and as far as the moon itself "knows" those two things feel the same.
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### Re: 1547: "Solar System Questions"

Yes. For analogy, imagine that you and two other people are parachuting. All three of you are falling, but due to different weights or parachutes or whatever, one of your friends is falling faster than you, and one is falling slower than you. From your perspective, both of them are drifting away from you in opposite directions, even though in reality (rather, from the Earth's perspective), all three of you are drifting in the same direction, but at different rates.

The friend drifting away above you is the far side of an object experiencing tidal forces, in this analogy. The friend drifting away below is the near side of it, and you are the object itself.
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danegraphics
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### Re: 1547: "Solar System Questions"

Water on the opposite side of Earth facing away from the Moon also bulges outward (high tide), but for a different and interesting reason: in reality, the Moon and the Earth revolve together around a common gravitational center between them, or center of mass. Here's a rough but helpful analogy: picture yourself swinging a heavy object attached to a rope around your body as you rotate. You have to lean back to compensate, which puts the center of mass between you and the object. With the Earth-Moon system, gravity is like a rope that pulls or keeps the two bodies together, and "centrifugal force" (inertia) is what keeps them apart. Because the "centrifugal force" (inertia) is greater than the Moon's gravitational pull, ocean water on the opposite side of the Earth bulges outward.

^this effect also works on the moon^

Gravity is not what causes the pull on the opposite side. And though the gravitational difference in distance is small, it is still a difference and is enough to be observable. If anything, the gravity would help to pull in the opposite side of the "tidal forces".

As I said, my theory is that this, combined with the fact that the earth blocks asteroids that would hit the side of the moon facing us, is what has left a lot more visible cooled lava fields on the near side compared to the far side.

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### Re: 1547: "Solar System Questions"

danegraphics wrote:If anything, the gravity would help to pull in the opposite side of the "tidal forces".
No, because the tidal force is caused in the first place by the fact that Earth's gravity pulls harder on the near side of the Moon than on the far side.
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xtifr
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### Re: 1547: "Solar System Questions"

gmalivuk wrote:
danegraphics wrote:If anything, the gravity would help to pull in the opposite side of the "tidal forces".
No, because the tidal force is caused in the first place by the fact that Earth's gravity pulls harder on the near side of the Moon than on the far side.

More precisely, it's caused by the difference in gravitational attraction on the different parts of the moon. The near side experiences a stronger gravitational pull, so it wants to fall faster than the moon as a whole, while the far side experiences a weaker pull, so it wants to fall slower. Both result in a net force directly away from the center. In other words, the tidal force is basically what's left when you subtract out the average gravitational pull experienced at the center of mass.

(Calculating the tidal force at any point other than on the direct line from nearest point to farthest point is a little trickier, since you have to subtract the directional vectors, but it still basically boils down to the same thing.)
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### Re: 1547: "Solar System Questions"

Gravity pulls on the near side more. That's why there's a tide there.
Gravity then also pulls the rest of the earth more than the far side, pulling it away from the far side, which gets left behind. That's why it looks like there's a tide there. It's not "really" a tide, but the leftover water sloshing around where the planet tried to abandon it for the moon.

A competing theory is that of gravitational shielding (GS). Under GS, the planet shields the far side from gravity, and it floats out into space, held in place only by the surface tension of the atmosphere. The stress causes electrical polarity, and is the primary cause of thunderstorms. This is why lightning is more visible at night than during the day.

This can be expanded with Binomial Smoothing (BS). The earth exhibits normal modes, and one of the simplest nondegenerate cases is oblate expansion. This harmonic mode is excited by gravitational waves radiating from the moon.
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Reflected really. The moon isn't massive enough to radiate much, but the sun emits large quantities of gravitation, and the moon acts as a reflector. Because of the unique properties of gravitation, the moon can also act as a gravitational lens, which is why solar tides are amplified when the moon is in front of the sun, or in opposition to it.
Tides are actually ahead of the moon, because of the earth's spinning. This causes a dipole gravitational moment which "pulls" the moon along in its orbit. If it weren't for this, the moon would eventually stop, and come crashing down to earth. Some theorize that this is what happened to the planet between Mars and Jupiter.

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