i am basically trying to quantify the influence of cosmic forces on a person on earth... so as to discredit or credit theories such the influece of the moon on mood or other astrological theories. it seems like the moon and sun are the only cosmic entities the really have any substantive effect on the earth (venus is the only planet that really does but it's effect is puny compared to the sun and moon so i don't think it deserves consideration /other planets have a very small effect but it is nearly zero/).

distance of sun to a person standing on earth - 93 million miles on average
distance of moon to a person standing on earth - 240,000 miles on average
distance to center of earth - 3963 miles

BTW - it's easier if you convert to metric units...

Mass of Earth - 5.98x10^24 kg
Mass of Sun - 1.99x10^30 kg
Mass of Moon - 7.36x10^22 kg
Radius of Earth - 6.37x10^6 m
Distance to Sun - 1.50x10^11m
Distance to Moon - 3.83x10^8 m

Hurkyl

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but the moon affects tides more than the sun by factor of 2... so why wouldn't the same dynamic be true for a person standing on the earth (which has far less mass and volume than the oceans)?

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It is. Though its true that the sun exerts a greater force on the earth and objects on it, the earth is in orbit around the sun - so those forces cancel out to zero since we are essentially in free-fall. The sun does NOT affect your weight. But since the moon orbits the earth (or rather they orbit a common cg just below the surface of the earth) the moon DOES affect your weight. I don't feel like running the calculation, but that effect is far lower than the effct of weighing yourself upstairs as apposed to downstairs.

Tidal forces are what is significant. And since tidal forces on the earth depend on differences in distance between one side of the earth and the other and the distance to the sun (or moon) being further away from the sun makes the difference in distance from one side of the earth to the other insignificant.

Now to debunk tidal effects on human emotions is simple. The tides on earth depend on the diameter of the earth. Tidal forces on YOU (or your brain) depnd on your height (or the diameter of your brain). Since the diameter of your brain is pretty small compared to the distance to the moon (and even smaller compared to the distance to the sun) the tidal forces on your brain are insignificant.

Now all this is nice, but there *ARE* well documented effects on human emotions and behavior from the heavens. The most pronounced is sunlight. Your sleep cycle depends on sunlight and the difference in sunlight between summer and winter has an effect on both mood and your body's regulation of its functions.

Tidal forces scale (to first order) as M*h/r^3 where M is the affecting body, r its radius, h the height of the object.

russ the only thing that i don't understand is if the pull of the sun is nullified by orbital forces (which makes sense) what causes solar tides?

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russ the only thing that i don't understand is if the pull of the sun is nullified by orbital forces (which makes sense) what causes solar tides?

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Take two objects that weigh 1 pound at a certain distance from the sun. Now move one object slightly closer and the other slightly further away. The object that is closer now weighs more: say 1.01 pounds. The object that is further away weighs less: say .99 pounds. The difference, .02 pounds, is the tidal force and if you connect these two objects with a wire, that wire is under .02 pounds of tension. And when you ADD the two together, .01+(-.01) = 0. The total weight of the system is still zero.

When you put this system in motion, the motion is IRRELEVANT to the tidal force as long as those distances remain the same. If the system is in orbit, the total weight that was 2 pounds is now 0 pounds. But the individual weights of the objects are 1.01-1 = .01 pounds and .99-1 = -.01 pounds. .01-(-.01)=.02 pounds.

Now here's the interesting thing about that: -.01 means the object that is on the far side is being "pulled" AWAY from the sun. Why? The orbit is defined at the center of gravity between the two objects and by moving them apart, you screw up the orbit. The object further away is moving too fast for its orbit so it feels like it should fly off into space and the object closer to the sun is moving too slow for its orbit and feels like it should sprial in towards the sun. The wire keeps them in orbit. And in fact if you cut the wire, one object will spiral away from the sun and the other will sprial towards it.

This force really does try to rip objects apart. Its part of the reason the larger plantets have rings and lots of moons instead of one big moon - the tidal forces rip apart the moons. And the tidal forces on the earth are trying to pull the oceans away from the earth. The ocean on one side of the earth is pulling toward the sun and the ocean on the other side is pulling away from the sun. But since -.01+.01 = 0, the earth is still weightless.

I hope this was clear, but I'm not at all sure if it was.

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Actually, the sun and earth revolve around a common point inside the sun, just as the earth and moon revolve around one inside the Earth

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so in terms of peak gravitational effects which is greater on a person standing on the earth, the moon or the sun gravity, considering that at peak the moon effects tides 2x more than the sun's peak effect.

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Originally posted by russ_watters:
And in fact if you cut the wire, one object will spiral away from the sun and the other will sprial towards it.

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Your explanation is fine up to this point. The objects won't "spiral" in or out, they'll just take up new orbits. Both objects' orbits will be slightly eliptical (assuming the original orbit was circular). The outer one's orbit would have its point of release as its perihelion, and the inner one would have its aphelion be the release point. Each object will return to these respective points once every orbit (though not at the same time, as the periods of the orbits will be different).

Janus

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The objects won't "spiral" in or out, they'll just take up new orbits. Both objects' orbits will be slightly eliptical (assuming the original orbit was circular).

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Originally posted by russ_watters:

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The objects won't "spiral" in or out, they'll just take up new orbits. Both objects' orbits will be slightly eliptical (assuming the original orbit was circular).

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You sure about that?

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A circular orbit is just a special case of an eliptical orbit anyway. In an eliptical orbit, the speed varies with distance, being faster in closer and slower further out. In our case, the outer object is moving faster and the inner object is moving slower. So why isn't one below orbital velocity and the other above?

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The inner obejct does the same, but starts at its aphelion, swings in closer to the sun for perhelion and then swings back out to aphelion.

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The moon is spiraling away from the earth. Is that not due to its speed being too fast? Satellites in "decaying" orbit are decaying because of drag that slows their orbital velocity. Aren't these what we have here? I'm not positive on this one, but I'm pretty sure. Also, at the instant the string is cut, the new component of acceleration is DIRECTLY away from or toward the earth. I don't see how that can lead to a stable orbit.

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For an explanation of what is happening to the moon, see my answer at :

http://www.physicsforums.com/topic.asp?ARCHIVE=&TOPIC_ID=9764

Satellites can decay because the drag is constant.

The instant you cut the line, There is no component of acceleration anymore! ( not counting the inward centripetal accleration applied by the sun's gravity). The only true force (with the aforementioned caveat) acting on the two objects while they are tied together is that by exerted by the line in keeping the two objects from following the paths they should naturally take. Once it is cut, this force is gone and the objects are free to follow their natural orbits.

It is like whirling a stone tied to a string around your head. If the string breaks, the stone will fly off at a tangent. But not because some force or acceleration is acting on it, but because the force imparted to it by the string which kept it moving in a circle has been removed and it is now free to move in a straight line.

Janus

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The instant you cut the line, There is no component of acceleration anymore!

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It is like whirling a stone tied to a string around your head. If the string breaks, the stone will fly off at a tangent. But not because some force or acceleration is acting on it, but because the force imparted to it by the string which kept it moving in a circle has been removed and it is now free to move in a straight line.

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Lets lengthen the string. The radius of the earth is about 4000 miles, so lets say our orbital cg is at 4,200 miles. If we use a heavier anchor, we can keep move the outer object to 8,400 miles without the anchor dipping into the atmosphere. Since the orbital speed of the anchor is about 17,500 mph, and we doubled the radius, the outer object must have an orbital speed of 35,000 mph. Now since escape velocity is about 25,000 mph (where exactly I'm not sure) and our object is at 35,000 mph, if we cut the string the object will go flying off into space. Is there a critical string length below which the orbit is simply reshaped and above which the orbit is broken when the string is cut?

Please note, my position makes sense to me, but save for a high school astronomy course, I have no training in orbital mechanics. I'm a mechanical engineer. So I could quite easily be wrong. But I'm not convinced I am.

I finally went to the web and found some info on this. I couldn't find a discription of what exacly the difference is between the two, but, the equation for escape velocity is:

v=sqrt(gR)

g is the acceleration due to gravity at the surface of the earth and R is the radius of the earth. G will change by the law of gravity (1/d^2) so here's the full calculation for escape velocity at 150 miles (low earth orbit).

v = sqrt (32.2ft/sec^2 / (4150mi/4000mi)^2 * 4150mi * 5280 ft/mi)

v = 25602 ft/sec or 17,456 mph

So orbital velocity *IS* equal to escape velocity. And by definition that means that adding ANY velocity to an object above escape velocity makes it escape.

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Ok, all thats nice and I'll leave it up because I'm intellectually honest, but I just figured out an error in my reasoning: direction of motion. Escape velocity is perpendicular to the surface of the earth while orbital velocity is parallel (tangent). Doesn't that then mean there must then be a critical tangential speed at which an orbit can be broken and below which the orbit is simply reshaped? Otherwise ANY object moving at any speed on a course tangent to the earth (which is any object anywhere within earth's gravitational influence) would be captured. So what is that critical speed?

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Originally posted by russ_watters:
Escape velocity vs orbital velocity:

I finally went to the web and found some info on this. I couldn't find a discription of what exacly the difference is between the two, but, the equation for escape velocity is:

v=sqrt(gR)

g is the acceleration due to gravity at the surface of the earth and R is the radius of the earth. G will change by the law of gravity (1/d^2) so here's the full calculation for escape velocity at 150 miles (low earth orbit).

v = sqrt (32.2ft/sec^2 / (4150mi/4000mi)^2 * 4150mi * 5280 ft/mi)

v = 25602 ft/sec or 17,456 mph

So orbital velocity *IS* equal to escape velocity. And by definition that means that adding ANY velocity to an object above escape velocity makes it escape.

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Ok, all thats nice and I'll leave it up because I'm intellectually honest, but I just figured out an error in my reasoning: direction of motion. Escape velocity is perpendicular to the surface of the earth while orbital velocity is parallel (tangent). Doesn't that then mean there must then be a critical tangential speed at which an orbit can be broken and below which the orbit is simply reshaped? Otherwise ANY object moving at any speed on a course tangent to the earth (which is any object anywhere within earth's gravitational influence) would be captured. So what is that critical speed?

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The equation you gave is for orbital velocity(for a circular orbit) not escape velocity. The more general form is

Vo = (GM/r)

with G being the gravitatonal constant and M the mass of the body.

This can be found by solving for GMm/r² = mv²/r

where GMm/r² is the force of gravity and mv²/r is the centripetal force needed to keep a mass m moving in a circle at velocity v.

Escape velocity is Vesc = (2GM/r) or always about 1.414 times greater than orbital velocity at any given r and M.
You can verify this by using your formula for the surface of the Earth, you will get a value .707 that of the accepted value of 11km/s(7mile/sec).

The escape velocity formula can be found by the following method.

The total energy of an orbiting body can be found by

Et = mv²/2 - GMm/r

Which is the sum of the objects kinetic energy and gravitational potential energy. (gravitational potential can be found by integrating GMm/r² with respect to r.

For an object moving at exactly escape velocity, Et=0 ( The total energy of an object in freefall is constant. Escape velocity is also equal to the speed an object starting at rest and falling from an inifinite distance would be traveling when it past a point a distance of r from the center of the mass.)

Thus for an object moving exactly a escape velocity:

0 = mv²/2 - GMm/r

solving for v:

GMm/r = mv²/2

GM/r = v²/2

2GM/r = v²

(2GM/r) = v

The escape velocity is the same no matter what its direction of its vector with respect to the gravitational field. (of course if it is straight down, it will hit the planet, but assuming it misses, then the direction doesn't matter.)

Janus

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The equation you gave is for orbital velocity(for a circular orbit) not escape velocity.

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It should be v=sqrt(2gR) which is just factored differently than yours.

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Originally posted by russ_watters:

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The equation you gave is for orbital velocity(for a circular orbit) not escape velocity.

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Oops, typo in my formula - i forgot the "2".
Link: http://www.physlink.com/Education/AskExperts/ae158.cfm

It should be v=sqrt(2gR) which is just factored differently than yours.

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In either case, escape velocity does not equal orbital velocity.

One point. 'g' is usually used to signify the value of Earth's surface gravity. Thus g = 9.8m/s[sup]2[sup] and has a fixed value. Which means the form you gave is only useful for Earth. The general form I gave can be used for any massive body, just by filling in the proper values for M and r

Janus

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Originally posted by russ_watters:
I'm still not convinced, though one thing I may be wrong about is the word "spiral." "Arc" might be a better word. Similar, but not quite the same. Anyway:

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The instant you cut the line, There is no component of acceleration anymore!

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The gravity of the planet is still causing an acceleration - but the acceleration isn't enough to keep the shape of the orbit. Isn't that the definition of escape velocity? Isn't orbital velocity = escape velocity? Ie, orbital velocity is about 17,500mph at about 150 miles, so at 17,501 wouldn't a satellite leave orbit? Or does it need to be going 17,505? Or 18,500?

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It is like whirling a stone tied to a string around your head. If the string breaks, the stone will fly off at a tangent. But not because some force or acceleration is acting on it, but because the force imparted to it by the string which kept it moving in a circle has been removed and it is now free to move in a straight line.

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Yes. This is exactly the effect we are describing. Except that in our example there are two things causing the acceleration: the string and the gravity. Either way, the total acceleration MUST be enough to hold the orbit. Also, when ANY object leaves orbit, it does so in an arc because of the gravity of the object it is leaving. And the lower the speed, the tighter the arc.

Lets lengthen the string. The radius of the earth is about 4000 miles, so lets say our orbital cg is at 4,200 miles. If we use a heavier anchor, we can keep move the outer object to 8,400 miles without the anchor dipping into the atmosphere. Since the orbital speed of the anchor is about 17,500 mph, and we doubled the radius, the outer object must have an orbital speed of 35,000 mph. Now since escape velocity is about 25,000 mph (where exactly I'm not sure) and our object is at 35,000 mph, if we cut the string the object will go flying off into space. Is there a critical string length below which the orbit is simply reshaped and above which the orbit is broken when the string is cut?

Please note, my position makes sense to me, but save for a high school astronomy course, I have no training in orbital mechanics. I'm a mechanical engineer. So I could quite easily be wrong. But I'm not convinced I am.

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If your string is long enough, sure, you could exceed orbital velocity enough at the outer end to reach escape velocity. For two reasons, 1. the velocity of the outer object increases, and 2. As the outer object moves outward, the escape velocityit needs to attain will decrease(due to the increased distance).

Janus