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TV and Newspapers in . U.k. Show new object in solar system .


Mike Smith Cosmos

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post-33514-0-02788400-1453365791_thumb.jpg

 

This picture of a planet like disc , Appeared on BBC . TV . Last night 20/01/16

 

Suggestions of an object 10 times heavier than Earth, a gas giant , identified by an American University, Californian Institute of Technology , on the fringes of , yet within , the Solar system

 

Mike

Edited by Mike Smith Cosmos
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Note that this hypothetical body has not been observed directly yet. It is based on its effects on other objects in the Kuiper belt:

 

The researchers, Konstantin Batygin and Mike Brown, discovered the planet's existence through mathematical modeling and computer simulations but have not yet observed the object directly.

https://www.caltech.edu/news/caltech-researchers-find-evidence-real-ninth-planet-49523

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I made comparison sheet between Earth, Ice-only planet, same as Earth density and Iron-only planet for object that is 10x more massive than Earth:

 

post-100882-0-75323200-1453396781_thumb.png

 

Radius/volume of planet is calculated from mass/density.

 

Somebody wants to make equation how big will be such object on photo taken from Earth.. ?

Edited by Sensei
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I made comparison sheet between Earth, Ice-only planet, same as Earth density and Iron-only planet for object that is 10x more massive than Earth:

 

attachicon.gifComparison.png

 

Radius/volume of planet is calculated from mass/density.

 

Somebody wants to make equation how big will be such object on photo taken from Earth.. ?

At its average distance from the Sun (given as 20 times that of Neptune and using the largest radius given (for ice), it would have an angular size of somewhere in the neighborhood of 0.1 arc second, which twice the angular resolution for the Hubble. In other words just resolvable.

However, at that distance, it will also get 1/400 as much light as Neptune(per square meter) or 1/363,000 as much as the Earth. Which means it would be lit by about as much light as reaches the ground under a full moon.

It is not going to reflect a lot of light back to us, and depending on its albedo, It could take a long exposure time to get any image at all of it.

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At its average distance from the Sun (given as 20 times that of Neptune and using the largest radius given (for ice), it would have an angular size of somewhere in the neighborhood of 0.1 arc second, which twice the angular resolution for the Hubble. In other words just resolvable.

Hmm.. I was thinking about something more accurate, like my sheet.

 

Date of year when there is taken photo also matter: in f.e. January Earth is closer to some objects by 300 mln km, while half-year later in June/July it's 300 mln km farther (than in previous closer case).

 

Anyway, thank you.

 

However, at that distance, it will also get 1/400 as much light as Neptune(per square meter) or 1/363,000 as much as the Earth. Which means it would be lit by about as much light as reaches the ground under a full moon.

It is not going to reflect a lot of light back to us, and depending on its albedo, It could take a long exposure time to get any image at all of it.

If there will done periodical examination of position of planet-to-be, and known stars behind it, planet should hide them, and known star should disappear from photos.

We don't need to rely on light reflected/emitted by body, but also by hiding what is behind. Of course, if there is something.

Edited by Sensei
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Hmm.. I was thinking about something more accurate, like my sheet.

Without more accuracy in initial data, you aren't going to get more accuracy. Its been said to be an average of 20 times further than Neptune, which gives it a period of ~14,000 yrs, or having a period between 10,000 and 20,000 yrs. So unless you can point to some figures more nailed down than that, a close estimate is all you are going to get.

 

Date of year when there is taken photo also matter: in f.e. January Earth is closer to some objects by 300 mln km, while half-year later in June/July it's 300 mln km farther.

Insignificant, really. At the 20 times Neptune distance it is some 600 times further from the Sun than the Earth. So it would be like noted the apparent difference in size between a grain of sand 300 miles away vs one 301 miles away. The major difference will be due to the planet's own eccentricity of orbit, and to work that out you would need all the orbital parameters for it.

Anyway, thank you.

 

 

If there will done periodical examination of position of planet-to-be, and known stars behind it, planet should hide them, and known star should disappear from photos.

We don't need to rely on light reflected/emitted by body, but also by hiding what is behind. Of course, if there is something.

If you are waiting on an object with that small an angular size to occult a star, you are going likely be in for a long wait. Much better to pin down the expected position of the planet. Do a long enough exposure to detect it, then do the same thing 6 months later and look for a parallax.

Edited by Janus
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