Rtyh-12
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At the risk of turning this thread the wrong way, exactly how did it look like?
Largest planet in the solar system could be about to be discovered
- Thread starter Kyle
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I get about half the mass of Venus, at 15000 a.u. to mimic the barycentre wobble caused by Jupiter. Then if we increase that mass, in theory, we could detect changes in barycentre accelerations. (It's easier to think of the barycentre moving for the super duper hard sums)
But, we cannot use parallax of nearby stars, we would have to use pulsar frequency variations. Those measurement are getting better but they're not good enough just yet, you can hide planets in the error bars.
This planet way out in the boonies, a rough Venus mass, will have a period of about 1.7 million years. This would mean that the Sun would rotate about 0.7 seconds of arc per year. Right so let's use that model again, where the sun is a billiard ball, the Earth is at 7.6 metres, Jupiter at 40 metres and Ignorethisbarrel's planet, "Pepsi" is at 144 km. (To give a better idea of the scale, the Earth has a diameter of 0.5 millimetres) Let's have this absolutely on a flat plain, with circular orbits and we want to get the parallax of a star at one parsec due north. When the Sun, Earth and Pepsi are in a straight line.
We take a picture wait six months, take another picture. We get one second of arc, and of course we can measure up to about 0.4 seconds of arc. Pepsi though has moved the whole Solar system round by half of 0.7 seconds of arc. The upshot is, that we are not taking the two photos at 180 degrees apart but slightly more. That "slightly more" is another base line, and it's too tiny to be of any use.
Righto, The italian guy's argument, is that we should compile a relativistic ephemera and then using that we can improve on those error bars for pulsars. He goes on to say that using that model there's a ninety five percent probability that there's no planet x out there.
You might think this is done already but as far as I know it's only done for Mercury. It might happen but Newton is good enough for any orbiter missions. Tiny relativistic effect we can leave to the geeks who can't dance, and think slide rules are Freudian.
But, we cannot use parallax of nearby stars, we would have to use pulsar frequency variations. Those measurement are getting better but they're not good enough just yet, you can hide planets in the error bars.
This planet way out in the boonies, a rough Venus mass, will have a period of about 1.7 million years. This would mean that the Sun would rotate about 0.7 seconds of arc per year. Right so let's use that model again, where the sun is a billiard ball, the Earth is at 7.6 metres, Jupiter at 40 metres and Ignorethisbarrel's planet, "Pepsi" is at 144 km. (To give a better idea of the scale, the Earth has a diameter of 0.5 millimetres) Let's have this absolutely on a flat plain, with circular orbits and we want to get the parallax of a star at one parsec due north. When the Sun, Earth and Pepsi are in a straight line.
We take a picture wait six months, take another picture. We get one second of arc, and of course we can measure up to about 0.4 seconds of arc. Pepsi though has moved the whole Solar system round by half of 0.7 seconds of arc. The upshot is, that we are not taking the two photos at 180 degrees apart but slightly more. That "slightly more" is another base line, and it's too tiny to be of any use.
Righto, The italian guy's argument, is that we should compile a relativistic ephemera and then using that we can improve on those error bars for pulsars. He goes on to say that using that model there's a ninety five percent probability that there's no planet x out there.
You might think this is done already but as far as I know it's only done for Mercury. It might happen but Newton is good enough for any orbiter missions. Tiny relativistic effect we can leave to the geeks who can't dance, and think slide rules are Freudian.
Urwumpe
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Stoat: Did you include the inverse square law in the calculations or did you just assume a rigid connection between sun and planet? because I have the strong feeling that your model is opposing real physics.
You essentially say: A probe that does a swingby far away from the planet would steal more impulse from the planet. In reality, the opposite is the case - the closer and heavier something is, the stronger the effect on the other body.
You essentially say: A probe that does a swingby far away from the planet would steal more impulse from the planet. In reality, the opposite is the case - the closer and heavier something is, the stronger the effect on the other body.
I used the cube of the semi major axis and the square of the period. i then replaced all of the planets in the Solar System with Barrel's half Venus mass planet. This to mimic the effect that Jupiter has on the Solar System. That is to shift the barycentre out to aout 107% of the Sun's diameter. Barrel's planet shifts the barycentre out to about 660 a.u. if I recall correctly, I do tend to write stuff down on bits of beer mat.
Then I stuck this planet in to Gravity Simulator to see what it did. Very little as it happens. if we push the mass of this planet up then it starts to alter the apses of the inner planets. of course mercury is seen as belonging to uncle Albert, and that's one of the reasons for this being a little acrimonious in the press.
On the vexing question of, all vectors pointing to the barycentre, they just do! There's no speed of light lag.
Then I stuck this planet in to Gravity Simulator to see what it did. Very little as it happens. if we push the mass of this planet up then it starts to alter the apses of the inner planets. of course mercury is seen as belonging to uncle Albert, and that's one of the reasons for this being a little acrimonious in the press.
On the vexing question of, all vectors pointing to the barycentre, they just do! There's no speed of light lag.
Urwumpe
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Ah....yes....
107% of the diameter? Sorry, but this is complete nonsense, its 107% of the suns radius.
If Jupiter would have the barycenter of the solar system so far outside the sun, we would have formally a binary star system.
Also, did you check your simplified calculations against the hill sphere?
I see the math, and I realize that it represents the model well - but I feel like the model is wrong, without being exactly able to formulate why I feel so. If all objects would be connected by rigid connections, it would be perfectly right. But they are no rigid. The further something is away, the lower the impact of gravity is, the distant object just has more time accelerating something in the same direction - but a much slower lateral motion of the orbited body could already create the centrifugal force to compensate.
if you would calculate the inertia tensor of such a rigid system, you would get:
[math]I = \sum_{i}{m_i r_i^2}[/math]
The object further away as a much stronger impact by distance and mass. That seems to be how the wikipedia barycentre formulas are derived. But if the object is not rigid, you have independently moving masses. or better said, the function is only reliable if "The distance between any two given points of a rigid body remains constant in time regardless of external forces exerted on it."
This is not even the case without external forces.
107% of the diameter? Sorry, but this is complete nonsense, its 107% of the suns radius.
If Jupiter would have the barycenter of the solar system so far outside the sun, we would have formally a binary star system.
Also, did you check your simplified calculations against the hill sphere?
I see the math, and I realize that it represents the model well - but I feel like the model is wrong, without being exactly able to formulate why I feel so. If all objects would be connected by rigid connections, it would be perfectly right. But they are no rigid. The further something is away, the lower the impact of gravity is, the distant object just has more time accelerating something in the same direction - but a much slower lateral motion of the orbited body could already create the centrifugal force to compensate.if you would calculate the inertia tensor of such a rigid system, you would get:
[math]I = \sum_{i}{m_i r_i^2}[/math]
The object further away as a much stronger impact by distance and mass. That seems to be how the wikipedia barycentre formulas are derived. But if the object is not rigid, you have independently moving masses. or better said, the function is only reliable if "The distance between any two given points of a rigid body remains constant in time regardless of external forces exerted on it."
This is not even the case without external forces.
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Hi Urwumpe, thanks for the correction, it was a typo, honest! I'll have a think about your other comments and get back to you. It is back of the envelope stuff, and I want to see if IgnoreThisBarrell is going to build the planet Pepsi, and then court a certain drinks manufacturer.
---------- Post added at 09:04 PM ---------- Previous post was at 05:25 PM ----------
Oh dear!:facepalm: I checked my bit of paper and I'd badly screwed up. The bit about Barrel's planet being Venus mass is okay but my estimate of the barycentre being way out in the boonies was way off. So my apologies Urwumpe:blush:
So, a roughly Venus mass body out at 15000 a.u. will be masked by the Jupiter effect. Of course now that I've showed myself to be such a klutz at maths that soft drinks company will give me nothing
Anyway, where I screwed up was r_1 = a / (1 + m_1 /m_2)
(The underlines are subscripts. Not too sure of how I came up with such a wrong answer though. I think I didn't convert the radius of the Sun into a.u.'s)
---------- Post added at 09:04 PM ---------- Previous post was at 05:25 PM ----------
Oh dear!:facepalm: I checked my bit of paper and I'd badly screwed up. The bit about Barrel's planet being Venus mass is okay but my estimate of the barycentre being way out in the boonies was way off. So my apologies Urwumpe:blush:
So, a roughly Venus mass body out at 15000 a.u. will be masked by the Jupiter effect. Of course now that I've showed myself to be such a klutz at maths that soft drinks company will give me nothing
Anyway, where I screwed up was r_1 = a / (1 + m_1 /m_2)
(The underlines are subscripts. Not too sure of how I came up with such a wrong answer though. I think I didn't convert the radius of the Sun into a.u.'s)
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