Please provide a Geocentric diagram
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05-02-2013, 06:28 AM (This post was last modified: 05-02-2013 06:32 AM by AtheismExposed.)
RE: Please provide a Geocentric diagram
Ok then answer me this:

Using Newton's equation to work out the gravitational attraction between two bodies (and putting in the heliocentrist's own numbers for the mass and distance of the sun), the gravitational force between the earth and sun is two orders of magnitude greater than the force between the earth and the moon.

Why, then, does the moon cause much larger tides on the earth's oceans than the sun? Given that the sun is tugging at us 100 times harder?

Try it yourself,

F = G(m1*m2 / r^2)

where m1 and m2 are the masses of the bodies and r is the distance between them.

For the earth and moon, the magic number is 1.982E20, for the earth and sun it's 3.529E22. Surely the sun should be causing tidal surges almost 200 times larger than what we get from the moon?

Or can we safely conclude that the Atheist's numbers for the mass and distance of the sun are RUBBISH?

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05-02-2013, 07:19 AM
RE: Please provide a Geocentric diagram
(05-02-2013 06:28 AM)AtheismExposed Wrote:  Ok then answer me this:

Using Newton's equation to work out the gravitational attraction between two bodies (and putting in the heliocentrist's own numbers for the mass and distance of the sun), the gravitational force between the earth and sun is two orders of magnitude greater than the force between the earth and the moon.

Why, then, does the moon cause much larger tides on the earth's oceans than the sun? Given that the sun is tugging at us 100 times harder?

Try it yourself,

F = G(m1*m2 / r^2)

where m1 and m2 are the masses of the bodies and r is the distance between them.

For the earth and moon, the magic number is 1.982E20, for the earth and sun it's 3.529E22. Surely the sun should be causing tidal surges almost 200 times larger than what we get from the moon?

Or can we safely conclude that the Atheist's numbers for the mass and distance of the sun are RUBBISH?

I don't have to prove shit, Tidal Forces have already been established as the most likely and mathematically sound basis for the tides. It's up to you to prove not only how they are wrong, but to come up with something that has better precision and explanatory powers than our current models. Then get that verified in peer review.

You're mathematical and scientific ignorance does not prove that the Sun revolves around the Earth...

http://en.wikipedia.org/wiki/Tidal_force

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05-02-2013, 07:36 AM
RE: Please provide a Geocentric diagram
I guess he didn't even bother trying to search for that answer.

One quick search input and I immediately found 6 different sites with the same equation that explains the reason why the moon has a greater effect than the sun on tides.... Undecided

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05-02-2013, 01:35 PM (This post was last modified: 06-02-2013 12:39 AM by Phaedrus.)
RE: Please provide a Geocentric diagram
Tides are based on the change in gravitational field as you move outward from the moon or sun, and diameter of the orbiting body. Not on their net gravitational pull.


The Earth is 12,600km across. Across that huge distance, the gravity felt by the moon decreases a substantial amount, because the earth is close to the moon. Run the numbers yourself. This difference in gravity across the width of the earth means that objects closer to the moon want to orbit it in a slightly closer orbit, while objects further away from the moon want to orbit it in a slightly higher orbit. But the Earth's gravity holds everything together so it doesn't go flying off. This difference in gravity is too small for us to easily feel, but the oceans are free to flow where the change pulls them; so oceans facing the moon "orbit closer" and rise in a tidal bulge. Oceans facing away from the moon "orbit higher" and likewise rise, in the opposite direction.

The sun is farther away. This means that that 12,600km diameter is going to result in less change in its gravity over distance. There's less gradient, so there's less tides.

To put it another way, you've seen the "gravity well" pictures showing the Earth and/or Sun and/or moon with spacetime distorted around them like they were marbles on a rubber sheet? In this case, the moon's well is much smaller and weaker than the sun's, but it is very slightly steeper where the earth is, meaning more change in gravitational potential, meaning more tides.

This is Tides 101, by the way. "Gravity pulls harder!" is the kindergarten version.




What's your explanation of tides in a geocentric universe? I'm curious.

E 2 = (mc 2)2 + (pc )2
614C → 714N + e + ̅νe
2 K(s) + 2 H2O(l) → 2 KOH(aq) + H2 (g) + 196 kJ/mol
It works, bitches.
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06-02-2013, 03:33 AM
RE: Please provide a Geocentric diagram
Oh, by the way, before you say something stupid like, "herp but you said the moon orbits the earth not the other way around it cant orbit higher or lower herp"

The Earth and Moon actually orbit a common point called the Barycenter. This is an imaginary point on average 1700km below the earth's surface (out of a 12,600km diameter, so it's nowhere near the center) which represents the Earth-Moon system's center of mass. This is the point around which the Earth and Moon orbit each other.

I describe tidal forces in terms of the earth orbiting the moon for clarity. The best example for understanding tidal forces I can think of is Larry Niven's story Neutron Star, which describes a long, narrow, indestructible spacecraft moving in a parabolic orbit around an incredibly massive and dense neutron star. This allows clear, precise explanation of all tidal effects, including the longitudinal pulls away from the center of the body (what causes the oceans to bulge in Earth-Moon tides), and the resistance to rotation (the effect that slowed the moon's rotation until it always faced earth). He also covered a few neat effects relating to relativity. You should give it a read, it's a good story and you might learn something.

E 2 = (mc 2)2 + (pc )2
614C → 714N + e + ̅νe
2 K(s) + 2 H2O(l) → 2 KOH(aq) + H2 (g) + 196 kJ/mol
It works, bitches.
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10-02-2013, 09:17 PM
RE: Please provide a Geocentric diagram
Creationists always get scarce after a hard dose of reality... But most usually work up the courage to manage a salvo at least once a week. This is pathetic.

E 2 = (mc 2)2 + (pc )2
614C → 714N + e + ̅νe
2 K(s) + 2 H2O(l) → 2 KOH(aq) + H2 (g) + 196 kJ/mol
It works, bitches.
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13-02-2013, 01:35 AM
RE: Please provide a Geocentric diagram
Here is a picture to describe AtheismExposed's level of scientific understanding...

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13-02-2013, 02:54 PM
RE: Please provide a Geocentric diagram
No... That implies he knows what a chalkboard is.

E 2 = (mc 2)2 + (pc )2
614C → 714N + e + ̅νe
2 K(s) + 2 H2O(l) → 2 KOH(aq) + H2 (g) + 196 kJ/mol
It works, bitches.
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13-02-2013, 03:06 PM
RE: Please provide a Geocentric diagram
(13-02-2013 02:54 PM)Phaedrus Wrote:  No... That implies he knows what a chalkboard is.


Give it another decade and nobody new will.


I've not seen a chalkboard (we called 'em blackboards) for probably close to 10 years...

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13-02-2013, 05:57 PM
RE: Please provide a Geocentric diagram
*sleeps*

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