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I need to argue this on why but I don't understand the question.

I have to argue why telescopes need to stay farther away from some planets but no others, I have tried searching this up but have not found anything.

So I'd like to ask for some reasons why space telescopes might need to be placed far from planets, and why it might need to be farther from some planets than from other planets.

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    $\begingroup$ So far every telescope we have built either is on earth, in earth orbit or in heliocentric orbit. So they all stay away from other planets. $\endgroup$ – Polygnome Dec 1 '19 at 23:51
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    $\begingroup$ Could be a little more specific please? Which planets? Which telescopes? Why? There are way too many possibilities to give a good answer now. $\endgroup$ – Bob Jacobsen Dec 1 '19 at 23:53
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    $\begingroup$ What is the context here? Missions to solar system planets? Direct imaging of exoplanets? Without that it's impossible to answer this question. $\endgroup$ – AtmosphericPrisonEscape Dec 2 '19 at 1:07
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    $\begingroup$ @uhoh: That's good for you. It is very unclear to me, because your interpretation doesn't make sense, as other comments have pointed out. Not objective indeed, maybe you should refrain from policing this website as if it were your own. $\endgroup$ – AtmosphericPrisonEscape Dec 2 '19 at 3:01
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    $\begingroup$ Do "telescopes" in this question include cameras of space probes, like HiRIZE on martian MRO orbiter? For them it's quality vs quantity trade. $\endgroup$ – Heopps Dec 2 '19 at 8:47
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Space telescope: In orbit around a planet but looking at other objects in space

I can think of at least three reasons:

  1. You can be closer to a small planet than a larger planet if the planet must block no more than X% of the sky. This answer gives the equation of the solid angle that a spherical body would block. $OP = R + h$ where $R$ is the radius of the planet and $h$ is the height of satellite above the planet's surface. If the planet is twice as large and you make $h$ twice as high, the solid angle blocked will be the same.

$$\Omega = 2\pi \left( 1-\frac{ \sqrt{2rh + h^2}}{R + h} \right)$$

enter image description here

  1. If you are using cooled optics and image sensors to look for objects like asteroids or comets using thermal infrared light or just looking at infrared like the JWST, you might want to be farther from a hot planet than a cold planet to decrease the thermal loading from the planet's infrared radiation.

  2. If your planet is very close to the Sun then reflected glare from the planet (e.g. Earthshine) will be more of a problem than for a planet far from the Sun. Baffles on the telescope help somewhat and you can coat the inside of your telescope with Vantablack but getting farther from the source of the scattered light will also help

Planet-observing telescope: In orbit around a planet and looking down at it

  1. You can see below that a satellite in a higher orbit will move a bit slower, and so the blurring of the planet's surface due to the satellite's motion will be reduced (neglecting the planet's own rotation which you might not be able to do) but at a given image scale (meters per pixel) you'll need to increase your focal length and aperture diameter to keep the f/no. and image brightness the same. The speed of a circular orbit around a planet is given by

$$v \approx \sqrt{\frac{GM}{r}}$$

where $G$ is the gravitational constant about 6.674E−11 m^3 /kg s^2, $M$ is the mass of the planet and $r$ is the radius or semimajor axis of the circular orbit. You can also look up the product $GM$ for the Sun, Moon and planets as their standard gravitational parameter.

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