Telescope to see a person on another planet

[kos]

My question is about the topic on the title. See first the original article.

So My speculative question on the subject is in the sphere of Einstein General relativity applying to the problem.
So we all now know that when light passes close to some massive objects its path is slightly changed
from its original trajectory.That's called deflection. So How will this phenomena of light deflection due to ceaselessly
moving different kinds of masses of different objects like stars planets galaxies clusters et.c . THROUGH
spacetime continium constantly wraping its curvature in different manners eventually obscure our observation of aliens on their
remote distant planet via our future telescope networks as the paths of photons constructing the image
will follow the lines of this constantly changing wraped continium.My critical thinking shows that the result should be some kind of total mess rest from the original
image we pretend to observe shouldnt it ?

[Strange]

My question is about the topic on the title. See first the original article.

 

So My speculative question on the subject is in the sphere of Einstein General relativity applying to the problem.

So we all now know that when light passes close to some massive objects its path is slightly changed

from its original trajectory.That's called deflection. So How will this phenomena of light deflection due to ceaselessly

moving different kinds of masses of different objects like stars planets galaxies clusters et.c . THROUGH

spacetime continium constantly wraping its curvature in different manners eventually obscure our observation of aliens on their

remote distant planet via our future telescope networks as the paths of photons constructing the image

will follow the lines of this constantly changing wraped continium.My critical thinking shows that the result should be some kind of total mess rest from the original

image we pretend to observe shouldnt it ?

 

 

The effects of gravitational lensing are very small unless light passes very close to a very large mass. Therefore we don't see any significant distortion of the view of distant stars because of it.

 

However, when things are nicely aligned, we can use gravitational lensing to get a better view of very distant objects. For example: http://hubblesite.org/newscenter/archive/releases/2003/01/image/a/

 

Also, very small amounts of gravitational lensing can be used to map the distribution of dark matter: http://www.nature.com/news/dark-matter-mapped-at-cosmic-scale-1.17311

[imatfaal]

!

Moderator Note

 

I have edited the website from the title - please give members here the salient points in a follow-up post.

 

Please check rule 7 - members should be able to participate without having to visit a different site

 

[Robittybob1]

I would imagine there will be a limit to our ability to resolve the image from a distant star. We can see the star but not the planets around it, and we definitely cannot see things on the planets. To have enough photons to create an image at such a distance would mean it would have to be glowing or reflecting extremely brightly in the first place.

[Airbrush]

How large a telecope would it take to be able to see a person standing on an exoplanet?

 

Suppose you can coodinate one hundred space telescopes that each has a lense one hundred feet across. Working together how fine a resolution would they have? Could they see an exoplanet? Could they see details on such a planet?

Edited June 27, 2015 by Airbrush

[imatfaal]

[latex]R \approx \frac{\lambda}{B}[/latex]

 

R is angular resolution in Radians

lambda is the wavelength in metres

D is the Baseline of the telescope array also in metres

 

So lambda is gonna be around 10^-7 and baseline at 100 metres - you are talking an angular resolution of the order of 10^-9 Radians.

 

The distance to the nearest star is about 10^16 metres - using small angle you can estimate that angle is close to Size of object / Distance.

 

The smallest object you could possibly resolve is about 10^7 metres - bigger than our earth

[Strange]

I would imagine there will be a limit to our ability to resolve the image from a distant star. We can see the star but not the planets around it, and we definitely cannot see things on the planets. To have enough photons to create an image at such a distance would mean it would have to be glowing or reflecting extremely brightly in the first place.

 

There are photos of exoplanets:

http://en.wikipedia.org/wiki/List_of_directly_imaged_exoplanets

http://www.eso.org/public/images/archive/category/exoplanets/

How large a telecope would it take to be able to see a person standing on an exoplanet?

 

That was worked out in the article the OP originally linked to...

[Sensei]

How large a telecope would it take to be able to see a person standing on an exoplanet?

 

Suppose you can coodinate one hundred space telescopes that each has a lense one hundred feet across. Working together how fine a resolution would they have? Could they see an exoplanet? Could they see details on such a planet?

 

Do you know inverse-square law.. ?

Insert power of star (or object that you want to see), and distance in equation, and you will receive W/m²

It can be split to quantity of photons with different energies, per area.

For Sun emitting 3.86*10²⁶ J per second energy in photons,

at distance 150 bln meters we receive 1367 W/m².

But at distance 4 light years it'll be just 2.15*10^-8 W/m².

Simply there is very few photons arriving per second per area unit at such distances..

 

If we assume average visible photon has 532 nm (green color), single photon has 3.73*10^-19 J,

Divide 2.15*10^-8 W/m² by 3.73*10^-19 J, and there is ~57 bln photons with average energies per m² per second.

That's ~64 billions time less than from our Sun here on Earth.

[Robittybob1]

 

There are photos of exoplanets:

http://en.wikipedia.org/wiki/List_of_directly_imaged_exoplanets

http://www.eso.org/public/images/archive/category/exoplanets/

...

Photos of a sort.

 

Do you know inverse-square law.. ?

Insert power of star (or object that you want to see), and distance in equation, and you will receive W/m²

It can be split to quantity of photons with different energies, per area.

For Sun emitting 3.86*10²⁶ J per second energy in photons,

at distance 150 bln meters we receive 1367 W/m².

But at distance 4 light years it'll be just 2.15*10^-8 W/m².

Simply there is very few photons arriving per second per area unit at such distances..

 

If we assume average visible photon has 532 nm (green color), single photon has 3.73*10^-19 J,

Divide 2.15*10^-8 W/m² by 3.73*10^-19 J, and there is ~57 bln photons with average energies per m² per second.

That's ~64 billions time less than from our Sun here on Earth.

Great bit of work. Shows the sort of problem we are up against when looking at another planet in detail.

[kos]

Any time we talk about some subject its always the same . The discussion goes to the wrong direction. Limitations about telescopes , about networks , about computing power , about refraction, atmospheric conditions e t.c . .. I don't care them. The question is how does every gravitational pool of any object between the surveillancing aliens and our technique distort the original image coming from their distant planet for instance situated at the other end of the Milky way galaxy or in the begining of the open cluster of some other galaxies ? If that will help you I show you this interactive map . https://in-the-sky.org/ngc3d.php

[John Cuthber]

Any time we talk about some subject its always the same . The discussion goes to the wrong direction. Limitations about telescopes , about networks , about computing power , about refraction, atmospheric conditions e t.c . .. I don't care them. The question is how does every gravitational pool of any object between the surveillancing aliens and our technique distort the original image coming from their distant planet for instance situated at the other end of the Milky way galaxy or in the begining of the open cluster of some other galaxies ? If that will help you I show you this interactive map . https://in-the-sky.org/ngc3d.php

The discussions you describe only started after the question had been answered.

 

The effects of gravitational lensing are very small unless light passes very close to a very large mass. Therefore we don't see any significant distortion of the view of distant stars because of it.

 

[Strange]

Any time we talk about some subject its always the same . The discussion goes to the wrong direction.

 

You said: My question is about the topic on the title.

 

That is what people have been discussing. Don't blame them. Who do you think is to blame for your lack of clarity?

[kos]

Ok it's nice answer but can you make it a little bit more exact by some equations or some calculus ? What does it mean this "very small" and "very close " "and very large mass" . I don't expect some very exact values cause I know task seems to be quite impossible even for NASA engeneers so I need nothing more than an approximation.Please excuse me and please don't be angry I just study engineering physics and I need a little bit more numbers.And please excuse my bad english but it's my third language ! Thank you in advance

[John Cuthber]

 

You said: My question is about the topic on the title.

 

That is what people have been discussing. Don't blame them. Who do you think is to blame for your lack of clarity?

To be fair, the title was changed.

[Strange]

Ok it's nice answer but can you make it a little bit more exact by some equations or some calculus ? What does it mean this "very small" and "very close " "and very large mass" . I don't expect some very exact values cause I know task seems to be quite impossible even for NASA engeneers so I need nothing more than an approximation.

 

Not sure why you say the task seems to be impossible.

 

The angle of deflection is:

[math]\theta = \frac{4GM}{rc^2}[/math]

Where M is the mass of the object, r is the distance at which the light passes it, c is the speed of light and G is the gravitational constant.

https://en.wikipedia.org/wiki/Gravitational_lens#Explanation_in_terms_of_space.E2.80.93time_curvature

 

Practically, if you want to use gravitational lensing to observe a planet then you would need a very massive, compact body that emits no light. In other words, a black hole (with no accretion disk). You can use the formula to work out what the effect of a black hole would be. Note that the light would have pass outside the photon sphere: https://en.wikipedia.org/wiki/Photon_sphere

 

Because the light passes round the outside of the lensing object (rather than through a normal lens) the image is highly distorted and needs to be corrected.

 

Also this black hole would have to be precisely aligned to lens something like a distant solar system; so obviously it would only be a transient effect.

 

Some more info here:

http://news.discovery.com/space/hunting-black-holes-with-gravitational-lenses-120210.htm

[kos]

That is something nice thank you Scientist )))))

[LaurieAG]

Just for comparison.

 

 

Hubble’s resolution in visible light is about 0.05 arcseconds (where 1 arcsecond = 1/60 arcminute = 1/3600 degree). To give some idea of what this means, if the Hubble Space Telescope were in Washington DC, it could distinguish two objects in New York City if they were separated by a distance of just 3 inches: