Elenin Orbital physics

Re: Elenin Orbital physics

Wed, 20 Apr 2011 17:31:59 +0000

That is what I like! Peer review!
OK I am checking..

Re: Elenin Orbital physics

Tue, 19 Apr 2011 22:24:33 +0000

I don't think your Earth and Elenin coordinates can be right. They put Elenin at 1 AU from the Sun and Earth at 0.5 AU.

Re: Elenin Orbital physics

Tue, 19 Apr 2011 19:40:45 +0000

Fun stuff for those that want to calculate the gravitational forces on Elenin.
Object's move in ellipsis or hyperboles or parabolas, but in reality they have no knowledge of these curves.
These curves are simplified models of the orbit and predicts it pretty accurate, but in the case of close encounters of other gravitational forces it gets more complicated.

So for those that want to have fun I give you the starting speeds and vectors of the Sun, Elenin and Earth at a certain date. With the gravitational formula you can calculate its trajectory in paper, including Earth's pulled by the Sun.

Position in AU, speed in km/s for 9 sept 2011 at 15:54

SunPosition = 0,0,0 Speed 0,0,0
Elenin position = 0.2545x, -0.4092y, -0.005536z Speed 49.83x, 34.43y, 1.843z
Earth position = 0.9753x, -0.2378y, 9.29e-6z Speed 6.523x, 28.82y, -0.001567z
Venus position = -0.7222x, -0.02864y, 0.04109z Speed 1.155x, -35.16y, -0.548z

Of course I cheated and be lazy, I used a simulator to get these numbers.
But it is really fun if you just create a hand drawn graph by calculating this because these are the real coordinates and speeds.

Edit: Small mixup with x and Y coordinate of Elenin and Earth. Now corrected.

Re: Elenin Orbital physics

Thu, 14 Apr 2011 21:09:25 +0000

I was wondering if a comet can be easily captured by Jupiter.
The bigger mass does have an effect but also the fact that Jupiter moves slower and also does the comet.

Elenin appears to move at 18.6 km/s at Jupiter's distance. Jupiter itself moves at 13.5 km/s.
The angle they cross is at about 70 degree seen from above. It is moving above the plane of Jupiter.

Jupiter's escape velocity is 59.5 km/s at the surface.
At 10 times Jupiter's radius (714,920 km = 1.8 times Earth-Moon) it drops to 18,828 km/s which will capture the comet in Jupiter orbit it it moves that close.

Below 5 Jupiter radius it will not stay in a circular orbit and move closer to Jupiter. I need to fine tune this math part to check at what speed it would eventually hit Jupiter.

Re: Elenin Orbital physics

Wed, 13 Apr 2011 18:22:54 +0000

It is a shame, I have been testing is Elenin gets the tail pointed towards Earth's orbit and we get through it but it seems that we are missing it. So no additional beautiful falling stars I am afraid.

Re: Elenin Orbital physics

Mon, 04 Apr 2011 18:38:13 +0000

On the car on the way back I realised the comet get 2 lucky shots. And this chance calculation is only for comets coming closer than 1 AU.
But something else that I did not expect during simulations is that it seems that Earth is unable to gravitational catch a comet at all. They all move too fast relative to Earth.

Jupiter has a better chance to capturer a comet. Not only because of its mass but because comets move more slowly.
Note: I still need to confirm this,

Re: Elenin Orbital physics

Sun, 03 Apr 2011 00:18:37 +0000

OK the second part of my Elenin debunking.
In my previous example I showed that Elenin can only get attracted by Earth and spiral towards Earth when its speed relative to Earth is below 7.8 km/s at a height of 100 km above Earth's surface. At Moon distance it should even be lower than 1.018 km/s to spiral towards Earth and below 1.44 km/s to orbit Earth.

But is there another way to fall towards Earth?
Well yes there is, when Earth happens to be at the wrong palace at the wrong time.
In this case gravitational attraction would not be needed to hit Earth.
A bit like a target you shoot a bullet at and get lucky.

Now how big is this target? Well the area of a circle with the radius the size as Earth's radius.

(1)

\begin{align} A=\pi \times r^2 = \pi \times \text{6,367,470}^2 \text{ = }1.27375*10^{14} \text{ m}^2 \end{align}

Sounds big but comets come from all directions so the surface it can hit will be the surface of a sphere with a radius of 1 AU!

(2)

\begin{align} \text{Surface}=4\pi \times r^2 \text{ = } 4\pi \times 149,498,000,000^2 \text{ = }2.8123*10^{23} \text{ m}^2 \end{align}

This means that the number of Earth's that fit in the sphere surface:

(3)

\begin{align} \frac{2.8123*10^{23} \text{ m}^2}{1.27375*10^{14} \text{ m}^2} = 2.20789\times 10^9 \text{ times} \end{align}

This basically means that on average for every 2,207,890,000 comets that crosses Earth's orbit, one could get lucky.
But currently we have no comet that is that lucky on our radars yet.

This is basic physics folks! Something a 2012 hoaxer can never produce because it would expose their lies.

Re: Elenin Orbital physics

Sat, 02 Apr 2011 21:23:58 +0000

According to this ephemeris it will be at naked eye visibility during September and October. You might not be able to pick it up in the evening sky in September, but you should be able to see it in the morning sky in October. Of course, predicting the brightness of comets is a tricky business. But it isn't going to brighten and fade in the space of a few nights when the Moon happens to be close, so I wouldn't worry too much about that.

Re: Elenin Orbital physics

Sat, 02 Apr 2011 20:18:24 +0000

A productive day today.
After someone claiming that Elenin caused the Japanese Earth quake, I created the same simulation as above but this time with 3 different viewpoints and put the number how big the influence of Elenin had compared to the moon on that given day.

http://www.youtube.com/watch?v=4maEe8Ikxbg

Re: Elenin Orbital physics

Sat, 02 Apr 2011 17:23:26 +0000

There's a forest near my city, where they have a planetarium, think i could have any luck seeing it there? :)

Re: Elenin Orbital physics

Sat, 02 Apr 2011 17:03:37 +0000

A pair of binoculars will help :-)
Or join an astronomy club. I bet they ill be out with big telescopes.

Re: Elenin Orbital physics

Sat, 02 Apr 2011 17:00:51 +0000

Oh damn, i really wanna see it! :)

Re: Elenin Orbital physics

Sat, 02 Apr 2011 16:58:24 +0000

For Denmark
Bad luck the moon is in the sky when it is nearest and when it is most visible, just barely with your eyes.
http://img268.imageshack.us/img268/4392/eleninviewdenmark.jpg

You might have a bit more luck a few days later, but the comet gets less and less brighter.

Re: Elenin Orbital physics

Sat, 02 Apr 2011 16:47:27 +0000

I live in Denmark :)

Re: Elenin Orbital physics

Sat, 02 Apr 2011 16:32:03 +0000

I took a snapshot of the closest encounter of Elenin to give some sense of scale:
http://img19.imageshack.us/img19/3882/elenin20111017earthview.jpg

Re: Elenin Orbital physics

Sat, 02 Apr 2011 15:57:44 +0000

It depends on your location.
I have calculated for me that would be around the 29th October that for a couple of days it will become barely visible, magnitude 5.5 from then on it fades pretty rapidly.

I could check the position for you when I know your general location. The more South the more chance you see it.

Re: Elenin Orbital physics

Sat, 02 Apr 2011 15:04:25 +0000

This comet will be close in October, right? (That's so awsome, 'cous its around my birthday!)
Does anyone know yet where in the world it will be visible?

Re: Elenin Orbital physics

Sat, 02 Apr 2011 14:24:19 +0000

I just posted the simulation how the orbit of Elenin would fly through the Solar system on my YouTube account.
Now you have a good reference where it is at what time and how far it is from Earth.

http://www.youtube.com/watch?v=9KBUlr0HTXI

Re: Elenin Orbital physics

Sat, 02 Apr 2011 13:24:40 +0000

I just did some simulation, Elenin flies by Earth at a speed of 23,000 m/s.
And since Elenin would only to be able to be captured by Earth and crash into it at a mere speed of 108.294 m/s it is flying 212 times too fast.

Re: Elenin Orbital physics

Sat, 02 Apr 2011 02:41:56 +0000

Now I am wondering of anyone of the 2012 hoaxers is brave enough to point out my mistake into the maths part. LOL
Of course I will answer with more maths that they can check themselves LOL

Re: Elenin Orbital physics

Sat, 02 Apr 2011 02:26:13 +0000

Now any comet might get a lucky shot, but that lucky shot would be the surface of Earth's circle cut in halve divided by the complete surface area of a sphere with a radius of 150,000,000 km!!!

Remember I just showed that Elenin could not have any gravitational influence because it moves too fast.

Re: Elenin Orbital physics

Sat, 02 Apr 2011 02:11:28 +0000

To continue with the previous maths.

Elenin does not come within 100 km from Earth's Surface, it will be 34,000,000 km from Earth's surface at its closest point.
So r = 34,000,000,000 m

Fill in the numbers:

(1)

\begin{align} V_{\text{Orbital}}=1.99683\times 10^7 \sqrt{\frac{1}{34,000,000,000}}\text{ m/s = 108.294 m/s} \end{align}

(2)

\begin{align} V_{\text{escape}}=2.82395\times 10^7 \sqrt{\frac{1}{34,000,000,000}}\text{ m/s = 153.15 m/s} \end{align}

So in order to get caught by Earth's gravity and fall to Earth it cannot move faster than $V_{\text{Orbital}}\text{ = 108.294 m/s}$ !
And moving faster than $V_{\text{escape}}\text{ = 153.15 m/s}$ means it would never be able to orbit Earth!

Now look at the facts. Elenin has about 42,000 m/s speed when it crosses Earth's orbit. It is basically waaaaaay too fast to even get influenced by Earth at all!

Elenin Orbital physics

Sat, 02 Apr 2011 02:03:13 +0000

Some people claim that comet Elenin might get attracted by Earth and crash onto Earth.
I tell you that it is baseless claim and I am going to show it mathematically.
Again it is just simple basic mathematics a 15 year old learns in school.

The orbital velocity an object has to be tangential to the Earths horizon is calculated with this formula:

(1)

\begin{align} V_{\text{Orbital}}=\sqrt{\frac{G M}{r}} \end{align}

The escape velocity would be this formula:

(2)

\begin{align} V_{\text{escape}}=\sqrt{2}V_{\text{Orbital}} \end{align}

We know G= $6.67428*10^{-11}$
And we know that M = Earths mass = $5.9742*10^{24}$ kg
r = the distance from Earths center

The maths get simplified like this:

(3)

\begin{align} V_{\text{Orbital}}=1.99683\times 10^7 \sqrt{\frac{1}{r}}\text{ m/s} \end{align}

(4)

\begin{align} V_{\text{escape}}=2.82395\times 10^7 \sqrt{\frac{1}{r}}\text{ m/s} \end{align}

Now what would the orbital speed need to be when you orbit earth just 100 km above the surface?
Well r=6,367,470m + 100,000 m = 6,467,470 m

(5)

\begin{align} V_{\text{Orbital}}=1.99683\times 10^7 \sqrt{\frac{1}{6,467,470}}\text{ m/s = 7,851.9 m/s} \end{align}

(6)

\begin{align} V_{\text{escape}}=2.82395\times 10^7 \sqrt{\frac{1}{6,467,470}}\text{ m/s = 11,104.3 m/s} \end{align}

So what does this mean?
Well $V_{\text{Orbital}}=\text{ 7,8519 m/s}$ means that if an object travels 7,851,9 m/s tangential with Earth's surface it stays at 100 km height.
If it comes below 7,851.9 m/s then it falls to Earth.
If it moves above 7,851.9 m/s and below $V_{\text{escape}}=\text{11,104.3 m/s}$ then it stays in orbit but never falls towards Earth. And above 11,104.3 m/s it completely gets ejected out the Earth's orbit.

What does this mean for the comet Elenin.
Well if Elenin wants to be captured by Earth's gravity when it would pass Earth at a mere 100 km from the surface then it must be a speed below $V_{\text{escape}}=\text{11,104.3 m/s}$ . This does not mean it will crash onto Earth, just orbit!
It can only crash onto Earth when it gets below $V_{\text{Orbital}}=\text{ 7,851.9 m/s}$ relative to Earth!!!

It basically means that it must slow down that much otherwise Earth's gravity does not get a chance to pull it in, since it is too fast.

Note: 7,851.9 m/s = 28,266.8 km/h (space shuttle speed)
Note: 11,104.3 m/s = 39,974.4 km/h (Apollo to the Moon)