Round Earth "Gravity"

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6strings

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Round Earth "Gravity"
« on: January 14, 2006, 07:51:42 PM »
For my next magic trick, and to prove I'm smarter than all you other "real" flat-earthers, I'll point out a chink in the round earth gravity theory, which none of you have been able to do as of yet, despite your attempts to.

Ok, so I've got a question for all you round earthers who fanatically cling to your gravity theory: why isn't the moon attracted to the sun rather than the earth?  Let's do the math (ask Mundi to check it for you).

The equation to calculate the force gravity is exerting on an object is this:
Force= (Mass 1 * Mass 2)/ distance between masses^2

Now, let's calculate the force between earth and the moon:
F=(7.347 673×10^22 kg (moon's mass) x 5.9736×10^24 kg (earth's mass))/384,403 km ^2
F=43.8921x10^46 kg^2 / 1.477566641x10^11 km^2
F=29.7057x10^35 kg^2/km^2

Now between the moon and the sun:
F=(1.9891×10^30 kg (sun's mass) x 7.347 673×10^22 kg)/ (149.6×10^6 km (sun's distance from earth)+384,403 km(moon's distance from earth))^2
F=14.6152x10^52 kg^2 /2.2495321143x10^16 km^2
F=6.4970x10^36 kg^2/km^2

Now, I think it's safe to state that 6.4970x10^36>29.7057x10^35
So...what explains this?

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Erasmus

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Round Earth "Gravity"
« Reply #1 on: January 14, 2006, 11:00:47 PM »
You're math is correct but you've left a few things out.

First, you set up the problem so that you're considering the moon at opposition (i.e. to get the moon-sun distance you're adding the moon-Earth distance to the Earth-sun distance).  So what if the force attracting the moon to the sun is stronger?  The Earth is "in the way", so when the moon falls towards the sun, it's falling towards the Earth as well.

More interesting to consider is what happens at conjunction, when the moon is directly between the Earth and the sun.  At that moment the Earth and moon are travelling in parallel directions.  Which way the moon goes from there would seem to depend on who wins out -- Earth or sun -- in the gravitational tug-o-war.

Next: probably you should not be talking about gravitational force, but acceleration.

So, acceleration of moon towards sun at conjunction:
  Ams = 0.8933634046 x10^14.
Acceleration of the moon towards the Earth:
  Ame = 0.4042680134 x10^14.

(NB: I don't include units because I never multiplied by the gravitational constant... you can just pretend it's there, and treat these values as constant coefficients.)

Yes, Ams > Ame.  In fact, Ams ~ 2Ame.  So why doesn't the moon fall towards the sun?

Well, it is!  But what you forgot to calculate is the force with which the Earth is pulled towards the sun, and hence, the acceleration of the Earth towards the sun.  As it turns out, Aes = 0.8887782751 x10^14, almost enough to completely cancel out the rate at which the moon falls.  The discrepancy is more than made up for by the 0.404 x10^14 moon-Earth acceleration.  What's happening is the Earth and moon are both being pulled towards the sun at almost identical rates, and the Earth gives a comparable (though smaller) tug on the moon in the opposite direction, which keeps the moon in orbit around the Earth.

To consider an analogy, imagine dropping a magnet and a small piece of iron off the roof of a tall building, at the same time, one an inch above the other.  The magnet will still attract the iron on the way down, even if the magnetic pull on the iron is much less than the gravitional pull of the Earth on the iron.

Basically, the Earth isn't trying to keep the moon away from the sun.  They're falling down together, and the Earth is doing a little bit on its own on the way down.

-Erasmus
Why did the chicken cross the Möbius strip?

Round Earth "Gravity"
« Reply #2 on: January 14, 2006, 11:36:42 PM »
I concur with Erasmus

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6strings

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Round Earth "Gravity"
« Reply #3 on: January 15, 2006, 06:42:37 AM »
Heh, way to contribute Crass.
And as for Erasmus: "I would have gotten away with it too, if it weren't for you meddling round-earthers".  
*Sigh* just another of my flat-earth schemes defeated by logic and reasoning...it's almost like logic and reasoning are easier to use if they have actual facts to back them up.  Oh well...

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6strings

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Round Earth "Gravity"
« Reply #4 on: January 15, 2006, 05:18:13 PM »
Waiiiit a minute...I just realized something:
Acceleration due to gravity=((Mass 1*Mass 2)/distance^2)*gravitational constant
Wherein gravitational constant=6.67300 × 10-11 m^3 kg^-1 s^-2
Right?

Well, let's see what affect this has on earth:
Acceleration due to gravity=(1.9891×10^30 kg (sun's mass) *5.9736×10^24 kg (earth's mass))/(149.6×10^6 km)^2)*6.67300 × 10-11 m^3 kg^-1 s^-2

Acceleration due to gravity=(11.8821*10^54 kg/149.6*10^15 m)*6.67300*10-11 m^3 kg^-1 s^-2

Acceleration due to gravity=0.0794*10^39 kg/m^2*6.67300 × 10-11 m^3 kg^-1 s^-2

Acceleration due to gravity=5.298*10^27 m/s^2

So due to the sun's gravitational effect on it, earth accelerates towards the sun at 5.298*10^27 m/s^2

The sun is 149.6×10^9 m from earth.  Now, we know that:
change in (delta) distance= velocity1*time+1/2 acceleration* change in (delta) time
or (delta)d=v1(delta)t+1/2*a*(delta)t^2

so let's solve for t, the point at which we all die because the earth hits the sun, assuming the gravitational force stays constant (it will actually increase, as distance decreases, but let's ignore that)

Also, let's assume that the earth has no velocity at this moment, shotening the equation to:
(delta)d=1/2*a*t^2

149.6×10^9 m= 1/2*5.298*10^27 m/s^2*t^2
149.6x10^9 m=2.649*10^27 m/s^2*t^2
56.48 x 10^-18 s=t^2
t=7.515 x 10^-9 s
or
t=0.000000007515 seconds.

So in the time you've been reading this, we died.  Several times.  I notice you're still reading this.  Anyone care to explain?

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Erasmus

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Round Earth "Gravity"
« Reply #5 on: January 15, 2006, 05:36:39 PM »
Instead of checking all your math, I'll point out that the first error is at the very top:

Quote
Acceleration due to gravity = ((Mass 1*Mass 2)/distance^2)*gravitational constant


In fact, the expression on the right hand side is not the acceleration due to gravity, but the force of gravitic attraction between the two masses, Fg.  To determine the acceleration, you use Newton's second law, F=ma, or in this case, a = Fg / Me.

I don't feel like doing the arithmetic (is it possible to use TeX in this forum?).

The other error is to assume that the Earth is not in orbit around the sun, but that Newton's Law still holds, and that the force of gravity is also as Newton defined it.  Even if you divided out the Earth's mass, you would still notice that the Earth should be acceelrating towards the sun.  Round-Earth cosmology claims it is.  If it were not accelerating towards the sun, it would fly off into space.  Round-Earth cosmology says that the Earth's velocity vector is always tangent to its elliptical orbit around the sun; Newton says that in the absence of a force, the Earth would continue moving the same direction (i.e., away from the sun).  It's the gravity between the Earth and the sun that keeps it in.

The done-to-death example of this is swinging a bucket of water or other heavy object over your head.  If the rope were to suddenly be cut, the object would go flying off.  If you suddenly stop spinning with the bucket almost directly overhead, it will come crashing down on you.  The rope creates a force (tension) which keeps the object accelerating towards you at all times, but the object's tangential inertia keeps it from coming crashing down.

[EDIT]Of course, you could try to claim that Newtonian mechanics is all wrong, but then you have to come up with some other explanation as to the bucket phenomenon.

-Erasmus
Why did the chicken cross the Möbius strip?

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6strings

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Round Earth "Gravity"
« Reply #6 on: January 15, 2006, 05:40:58 PM »
Yeah, but I'm really just trying to determine the point where we all die in a fiery inferno.  So, if anyone could tell me how fast we're moving towards the sun, and how they arrived at that answer, I'd be very grateful.  Assuming we don't die in the interim.

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Erasmus

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Round Earth "Gravity"
« Reply #7 on: January 15, 2006, 05:57:08 PM »
I think we'll have to worry about descending into Dante's inferno a lot sooner than the sun.

To answer your question, one would need to do an analysis of a somewhat complex system.  For one thing, since the round-Earth model has Earth's orbit as an ellipse, its tangential speed is not constant.  Also, this speed can be lessened by drag if the Earth moves through the sun's atmosphere (which it is currently doing, and is why radios sometimes go crazy and Apollo astronauts suffered early cataracts, or so NASA claims).  As the Earth falls towards the sun, the atmospher gets thicker and thus the drag force greater, but then, the sun is also pushing the Earth outwards with solar winds and simple radiative pressure.

Or, I could just point out that according to astrophysicists, the sun will turn into a red giant in five billion years, and engulf all the terrestrial planets in a fiery blaze of death and malice.

-Erasmus
Why did the chicken cross the Möbius strip?

Round Earth "Gravity"
« Reply #8 on: February 05, 2006, 12:01:02 AM »
God you suck balls at physics 6String.

The earth, moon, sun, everything else, aren't standing still, so their gravitational forces are for the most part orthogonal in at least 2 dimensions.

For the most part, the sun's gravity keeps us spinning in a circle around it. Think about if you connected a paint can to a rope and spun it in a circle around you head. You're exerting a massive force but the thing isn't coming closer to you, but you need that force to keep it going in a circle...

http://tutor4physics.com/motioncircular.htm