Supermoon Physics

Summary:

Some maths involved

Supermoon Physics

obaeyens 19 Mar 2011 18:26

Ok lets debunk the Super moon with real numbers.
I already said that 90% of the 2012 hoaxes are easily debunked with only with $+,-,\left/,*,\sqrt{x},x^2\right.$

This is the formula for gravitational attraction:

(1)

\begin{align} \text{Force}=\frac{G\times M_1\times M_2}{(r)^2} \end{align}

G =$6.67428*10^{-11}$ = Gravitation constant.
M1 = $7.3477*10^{22}$ kg = Moon mass
M2 = 1 kg of mass on Earth the Moon will act on.
r = Moon Earth distance.

Filling in the numbers and we get

(2)

\begin{align} \text{Force}=\frac{6.67428*10^{-11}\times 7.3477*10^{22}\times 1}{(r)^2} \end{align}

Simplified we get this:

(3)

\begin{align} \text{Force}=\frac{4.90406*10^{12}}{r^2} \end{align}

Now we choose r = Furthers Moon distance = 405,696,000 meters minus Earth radius to know force at Earths surface level.

(4)

\begin{align} \text{Force}=\frac{4.90406*10^{12}}{(405,696,000-6,371,000)^2} = 0.0000307541 \text{ Newton} \end{align}

Now we choose r = Closest Moon distance = 356,000,000 meters minus Earth radius to know force at Earths surface level.

(5)

\begin{align} \text{Force}=\frac{4.90406*10^{12}}{(356,000,000-6,371,000)^2} = 0.0000401182 \text{ Newton} \end{align}

The attraction force from earth on that same 1 kg of mass

G =$6.67428*10^{-11}$ = Gravitation constant.
M1 = $5.9742*10^{24}$ kg = Earth mass
M2 = 1 kg of mass on Earth the rest of Earth will act on.
r = 6,371,000 m Earth radius

Filling in the numbers and we get

(6)

\begin{align} \text{Force}=\frac{6.67428*10^{-11}\times 5.9742*10^{24}\times 1}{(6,371,000)^2} \text { = 9.82356 Newton} \end{align}

Yes anyone noticed the 9.82 Newton? Earth's acceleration constant? ;-)

This means that the attraction force of Earth will pull any mass on the surface with 9.82 newton force down, while the moon will pull with 0.0000307541 Newton force up at its furthest distance.
This is $\frac{0.0000307541}{9.82}\times100$ = .000313178 % difference = 0.3 milli pro-cent!
The moon has 0.3 milli pro-cent influence on Earth's mass.

The closest distance would be
This is $\frac{0.0000401182}{9.82}\times100$ = 0.000408536 % difference = 0.4 milli pro-cent!
The moon has 0.4 milli pro-cent influence on Earth's mass.

I don't know the plate tectonics force but these will be much higher than 0.4 milli procent.

[Edit: small addition and fix in the calculation]

Last edited on 19 Mar 2011 18:36 by obaeyens Show more

Re: Supermoon Physics

DieselHorseLAW 19 Mar 2011 18:29

Good job on the math again Obae. I for one will be bringing binoculars with me tonight! :-P lol

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“In all matters of opinion, our adversaries are insane.” - Oscar Wilde

Re: Supermoon Physics

obaeyens 19 Mar 2011 18:41

I hope it motivates people to realize that what we claim here are not false claims that the Supermoon is a hoax, but also understand that this can easily be debunked by a 15 year old and a calculator.

Re: Supermoon Physics

DieselHorseLAW 19 Mar 2011 21:30

Figured I would add this from the USGS page about celestial objects and seismic activity.

"Q: Can the position of the moon or the planets affect seismicity?

A: The moon, sun, and other planets have an influence on the earth in the form of perturbations (small changes) to the gravitational field. The relative amount of influence is proportional to the objects mass, and inversely proportional to the third power of its distance from the earth.

The stresses induced in the earth by an extraterrestrial mass are proportional to the gravitational field gradient dg( r ) / dr
and NOT to the gravitational field g( r ).

g( r ) = GMm / r^2

thus:

dg( r ) / dr = -2 * g( r ) / r = -2GMm / r^3

Many studies in the past have shown no significant correlations between the rate of earthquake occurrence and the semi-diurnal tides when using large earthquake catalogs.

Several recent studies, however, have found a correlation between earth tides (caused by the position of the moon relative to the earth) and some types of earthquakes.

One study, for example, concludes that during times of higher earth and ocean tides, such as during times of full or new moon, earthquakes are more likely on shallow thrust faults near the edges of continents and in (underwater) subduction zones. Lunar or solar eclipses represent, of course, special cases of full and new moon, but do not cause any special or different tidal effects from full and new moon. Earth tides (Earth's surface going up and down by a couple of centimeters) and especially ocean tides (surface of the ocean going up and down by a meter or more) raise and lower the confining pressure on shallow, dipping faults near continental edges and in subduction zones. When the confining pressure is lessened, the faults are unclamped and more likely to slip. The increased probability is a factor of ~3 during high tides. But you must stop are realize that the background probability is, in general, very low in a given place and year (fractions of a percent), so that raising this tiny probability by a factor of 3 during high tides still results in a very tiny probability.

There have also been some small but significant correlations reported between the semi-diurnal tides and the rate of occurrence of aftershocks in some volcanic regions, such as Mammoth Lakes. (UC Berkeley) "

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“In all matters of opinion, our adversaries are insane.” - Oscar Wilde

Re: Supermoon Physics

obaeyens 20 Mar 2011 01:52

The stresses induced in the earth by an extraterrestrial mass are proportional to the gravitational field gradient dg( r ) / dr
and NOT to the gravitational field g( r ).

The calculations I showed was to show how much force difference there would be between the Moon at its furthest distance and it closest distance on a surface mass. That force is tiny compared to the local force of Earth itself.

This means the complete surface is pulled with this force equally. No Earth quake can happen if everything is pulled equally.

The gravitational field gradient dg, which basically means the difference in force between 2 points and really could cause something , for example would be even more less.

Lets take back the gravitational attraction formula:

(1)

\begin{align} \text{Force}=\frac{G\times M_1\times M_2}{(r)^2} \end{align}

G =$6.67428*10^{-11}$ = Gravitation constant.
M1 = $7.3477*10^{22}$ kg = Moon mass
M2 = 1 kg of mass on Earth the Moon will act on.
r = Moon Earth distance.

Filling in the numbers and we get

(2)

\begin{align} \text{Force}=\frac{6.67428*10^{-11}\times 7.3477*10^{22}\times 1}{(r)^2} \end{align}

(3)

\begin{align} \text{Force}=\frac{4.90406\times 10^{12}}{\text{r}^2} \end{align}

Now we have r= which is the distance Moon and your target.
The upper crust would be the distance Earth-Moon minus the radius of the Earth.
We use the minimal distance in our example.
r = 356,000,000 - 6,371,000 = 349,629,000 m

(4)

\begin{align} \text{Force}=\frac{4.90406\times 10^{12}}{\text{349,629,000}^2} \text{ = 0.0000401182 Newton} \end{align}

Now lets calculate for 50 km deeper below the crust.
r = 356,000,000 - 6,371,000 + 50,000 = 349,579,000 m

(5)

\begin{align} \text{Force}=\frac{4.90406\times 10^{12}}{\text{349,679,000}^2} \text{ = 0.0000401067 Newton} \end{align}

The difference = 0.0000401182- 0.0000401067 = 0.0000000115 Newton.

Now compare this with the 9.82 gravitational force of Earth
This is $\frac{0.0000000115}{9.82}\times100$ = 0.000000117108 % difference = 0.117108 micro pro-cent!

Basically, local forces near the Earth quake are many many many times stronger than the influence of the Moon.
One could claim that slamming your car door would cause the big Earth quake if you claim the moon caused the Japan Earth quake.
It is that foolish.

Now how come you see the tides go up and down so much?
Simply because the water can freely move whereas the rest of Earth is pretty stuck.

_{A small note: The gravitational field gradient $-\frac{2 \text{g1} M_1 M_2}{r^3}$ is basically the slope of a curve when you would draw the gravitational forces in a graph But is equivalent what I did in the calculation in order to make it simpler to understand.}

Last edited on 20 Mar 2011 01:52 by obaeyens Show more

Re: Supermoon Physics

Lindsay88 20 Mar 2011 04:06

Well, it's officially the 20th here..haha

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*live*laugh*love*dance*

Re: Supermoon Physics

Lindsay88 20 Mar 2011 04:07

…and I could see not much of a difference in the moon..haha

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*live*laugh*love*dance*

Re: Supermoon Physics

Astrogeek 20 Mar 2011 04:10

The difference in size would be tough to spot without a fixed frame of reference. Side-by-side photographs, however….

http://www.starrynightphotos.com/moon/perigee_apogee.htm

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"Do you ever think about things you do think about?" - Henry Drummond to Matthew Harrison Brady in Inherit the Wind

Re: Supermoon Physics

Lindsay88 20 Mar 2011 04:19

Cool, thanks astro!

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Re: Supermoon Physics

obaeyens 23 Mar 2011 21:41

The last deadline was the 26th.
They are also running out of that dead line. Two more days left.
I predict also nothing related to the supermoon.

Re: Supermoon Physics

obaeyens 29 Mar 2011 19:20

The did postpone the 26th to the 29th but still nothing.
Another failed end of the world claim.