# Binary star?

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### #1 lemit

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Posted 02 September 2009 - 03:15 AM

I admit I haven't done a lot of research on this. I'm asking the rest of you to help me.

In another thread, reference was made to Walter Cruttenden and his book "Lost Star of Myth and Time." It seemed ridiculous, but when I searched it I couldn't find anyone who was dismissing it without saying it had some merit. (I did notice that most of the reviewers had their own books to promote, so maybe they had a collegial interest in giving the book at least some praise.)

Is there anybody on Hypography who can explain Cruttenden's theory to me and explain why it's not considered bunk--or if it is considered bunk? I don't want to have to buy the book. I'd prefer to waste my money on other things.

Thanks.

--lemit

### #2 modest

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Posted 02 September 2009 - 04:59 AM

Sirius is receding from us at 7,600 meters per second. How massive would Sirius need to be in order to be gravitationally bound to us? It's simply:

$m=\frac{v^2r}{2G}$

where:
• G = 6.67 x 1011
• r = 8.15 x 1016 m
• v = 7,600 m/s
The answer is 3.53 x 1034 kg or 17,748 times as massive as our sun. It's not possible for Sirius to be that massive.

It is also possible to use Kepler's 3rd law modified for binary orbits here: Earth Orbits to prove that the proper motion of Sirius is too great for a binary orbit. You should find that the proper velocity needed for Sirius relative to the sun is 80 m/s to maintain a circular binary orbit with us. The proper motion of Sirius is greater than that (it moves about a half a degree per 2,000 years).

~modest

### #3 modest

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Posted 02 September 2009 - 05:56 AM

[quote name='modest']It is also possible to use Kepler's 3rd law modified for binary orbits here: Earth Orbits to prove that the proper motion of Sirius is too great for a binary orbit. You should find that the proper velocity needed for Sirius relative to the sun is 80 m/s to maintain a circular binary orbit with us. The proper motion of Sirius is greater than that (it moves about a half a degree per 2,000 years).[/quote]

Here's a proof of that:

The proper motion of Sirius assuming it is in a circular binary orbit with us should be given by the following formula (solved algebraically from Earth Orbits):

$v=\sqrt{ \frac{G(m_1+m_2)}{r}}$
(this is the "If you are riding on one of the masses" formula from the link)

Given:
• G = 6.67 x 10-11
• m1 = 1.99 x 1030 kg
• m2 = 6.00 x 1030 kg
• r = 8.15 x 1016 m

Sirius needs a proper relative motion (i.e. tangent velocity) no greater or less than 80.8 m/s to be in a binary orbit with us.

The actual proper motion of Sirius (you can find on Sirius' wiki page) was found in 1718 by Halley (and many since) and is approximately one degree of arc per 4,000 years. [This, by the way, is a very fast moving star in the sky (perhaps the fastest, I don't know)] To find meters per second requires the trig function "tan". Sirius is 2.64 parsecs which is multiplied by tan(1 degree) = .04608. That is the distance in parsecs that Sirius moves in tangent motion per 4,000 years. 0.04608 pc / 4,000 years is:

11,264 meters / second (actual proper motion)

As 80.8 $\neq$ 11,264 by quite a lot we have proven that Sirius cannot be in a binary orbit with the sun. Given the mass of each system and the distance between us there is too great a tangent velocity between us to settle into an orbit. At this speed we will fly right by it (or it will fly right by us depending on how you look at it).

~modest

### #4 lemit

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Posted 03 September 2009 - 01:33 AM

Okay, now I'd like to have the same explanations done for dummies. In my sophomore year of high school I transferred from a rural school as a third child in possibly the only family to have had three college graduates in the school's history (I was the only one in my family not to get a terminal degree) to basically a college prep school, with many of my classmates getting scholarships I'd never heard of to Ivy League or British schools. Among the alumni of my new school were several members of congress and a Supreme Court Justice. (The first person I met there went on to become a senator.) The gap between what I was studying one week and what I was studying the next was enormous, probably the equivalent of a year of school, possibly more. My new classmates suffered me because I amused them. I knew how to write parodies and jokes. Apparently they still remember me. When I got a Facebook page recently, a bunch of them--including some I'm not sure I remember--invited me to be friends on the first weekend.

I am trying to make self-evident that my talent lies in language. It doesn't lie in math. Could you explain what the formulae mean, or at least what they might mean to one of David Letterman's dumb guys? Maybe one of Monty Python's upper class twits?

Thanks.

--lemit
Dumb As Mud But Able To Turn a Phrase

### #5 modest

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Posted 03 September 2009 - 04:12 AM

I am trying to make self-evident that my talent lies in language. It doesn't lie in math. Could you explain what the formulae mean, or at least what they might mean to one of David Letterman's dumb guys?

No trouble, mate. But, I should say with assuredness: math is a language and not knowing it certainly doesn’t put into question your intelligence. I don’t know German. I would hope that says little of my competence. I would suspect you are unfamiliar with math, and that is all.

If you situate yourself some distance away from a mass like the earth and you throw something away from that mass... I suppose you might think of a person in a rocket sitting stationary above earth’s atmosphere shooting a cannonball away from the earth—there is a certain velocity for that cannonball beyond which it will not fall back toward the earth nor could it find an orbit around the earth. If the cannonball is shot away from the earth with a speed greater than the earth’s escape velocity then it will just keep moving away from the earth forever. It won’t be gravitationally bound to the earth.

So, turn the situation around a little. Let’s say you are wandering about in your rocket in space not far from earth with your good buddy Walter Cruttenden. You come across a cannonball which you notice is moving away from the earth. Your buddy Walter says “I bet that cannonball is orbiting the earth. I bet the earth and the cannonball make up a binary system”.

So, you indulge your friend and measure the velocity of the cannonball relative to the earth with your high-tech onboard radar tracking sensors (why not... it’s a nice rocket). You find that it is moving 200 kilometers per second away from the earth. But, earth’s escape velocity where you are is only 11 km/s. This means anything in that particular area of space going faster than 11 km/s away from earth will not achieve an orbit around the planet, but will rather just keep going away from it forever. That would pretty-well rule out the possibility that the cannonball is orbiting the earth. It’s going too fast.

But, here is where I throw in a curveball, what if the earth were more massive? You and Walter in the rocket might pretend that earth is twice as massive as you think it is. If that were the case then the escape velocity in that area of space would be greater than 11 kilometers per second. The next logical question would be: “how massive would the earth need to be for this cannonball to be gravitationally bound to the earth?”. The cannonball is moving 200 km/s and if the earth were massive-enough it could pull the cannonball back toward it and perhaps the cannonball could achieve an orbit. So, the question at hand is: what mass does the earth need to be in order to be bound to this cannonball?

That is the question that my first post answers. The cannonball is the sun. We know that the sun is moving away from Sirius at 7,600 meters per second (direct observation tells us this) and we know the distance to Sirius so we can find the mass that Sirius would need to be in order for the sun to be bound to it. The answer is that Sirius would need to be 17,748 times as massive as our sun. If it is any less massive than that then the sun is exceeding the escape velocity necessary to break free of Sirius’ gravity forever. In actual fact, Sirius A and Sirius B combined are only 3 times as massive as the sun. This makes it completely impossible for us to be in orbit with that star. It almost needs to be as massive as a supermassive black hole to catch us at this distance and speed. Not possible.

My second post is similar, but deals with the tangent (or sideways) velocity. If you move fast-enough past a mass then it will not catch you and put you in orbit (think of a meteor almost grazing the earth as it flies by). We are moving fast enough in a tangent direction past Sirius that we cannot be in a binary orbit with it (given that it has a reasonable mass)

That ok?

~modest

### #6 lemit

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Posted 03 September 2009 - 04:29 AM

Thank you! I love physics represented in little stories.

Now, let's see if I can ask a more intelligent question. If we assume a binary system, wouldn't we have to throw out Einstein's General Theory of Relativity as it relates to Mercury's orbit? Wouldn't that additional gravity require recalculation of the theory, or does Walter Cruttenden think Einstein just didn't understand the nature of the gravity he was measuring?

Is that how he justifies what he does?

Thanks again.

--lemit

### #7 Janus

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Posted 03 September 2009 - 07:29 AM

Thank you! I love physics represented in little stories.

Now, let's see if I can ask a more intelligent question. If we assume a binary system, wouldn't we have to throw out Einstein's General Theory of Relativity as it relates to Mercury's orbit? Wouldn't that additional gravity require recalculation of the theory, or does Walter Cruttenden think Einstein just didn't understand the nature of the gravity he was measuring?

We'd have to throw out everything we know about gravity all the way back to Newton, and there is just too much observational evidence in our own Solar System that agrees with our present understanding for it to be that far off from the truth.

### #8 lemit

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Posted 03 September 2009 - 10:50 PM

We'd have to throw out everything we know about gravity all the way back to Newton, and there is just too much observational evidence in our own Solar System that agrees with our present understanding for it to be that far off from the truth.

Thanks again.

The Binary Star theory is beginning to look like the pseudoscience I love so much.

I have another question. Does the fact we can't find 90% or so of the universe give rise to this kind of theory, or is that irrelevent to this discussion?

I know I could read Cruttenden's book or watch the video, but the more I read about him, the less I like him. I'd rather get my information from people I trust, and if you can't trust people you've never met who are using fake names, who can you trust?

--lemit
(not his real name)

### #9 freeztar

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Posted 04 September 2009 - 02:19 PM

I have another question. Does the fact we can't find 90% or so of the universe give rise to this kind of theory, or is that irrelevent to this discussion?

I wouldn't say it's necessarily irrelevant to this discussion, but we are only considering 2 bodies that are relatively close to one another so any effects from universal expansion via dark energy or gravitational influences from dark matter should be negligible. Especially considering the orders of magnitude discrepancy between the observed mass of Sirius and the mass needed to be gravitationally bound to the Earth.

I'd rather get my information from people I trust, and if you can't trust people you've never met who are using fake names, who can you trust?