Kerbal Astronomy 101

[Alphard]

6 hours ago, OhioBob said:

The presumption is that Kerbal and human technology have developed along a similar timeline.  My article is written assuming a 1950s level of technology.  Adaptive optics didn't become practical until the 1990s.

Oh, I get it now. Anyways, the images you made for the book are just awesome.

[_Augustus_]

Here on Earth, Jupiter's largest moon Ganymede never reaches more than 2" in diameter, but astronomers have been able to visually detect albedo features on the surface with 8" or larger telescopes. While Mercury can get to over 4 times that size, it lacks any high-contrast markings and thus detail is very difficult to see with any telescope, especially without lucky imaging technology. 

Thus I suspect that Kerbal astronomers by the 1950s would've discovered Moho's basins, Dres' maria, the islands of Laythe (assuming it doesn't have 100% cloud cover like Venus/Titan), and maybe some features on Tylo.

Edited May 28, 2017 by _Augustus_

[Adstri]

On 5/28/2017 at 3:55 PM, _Augustus_ said:

Here on Earth, Jupiter's largest moon Ganymede never reaches more than 2" in diameter, but astronomers have been able to visually detect albedo features on the surface with 8" or larger telescopes. While Mercury can get to over 4 times that size, it lacks any high-contrast markings and thus detail is very difficult to see with any telescope, especially without lucky imaging technology. 

Thus I suspect that Kerbal astronomers by the 1950s would've discovered Moho's basins, Dres' maria, the islands of Laythe (assuming it doesn't have 100% cloud cover like Venus/Titan), and maybe some features on Tylo.

Possibly...

[FreeThinker]

 

On 8/7/2015 at 11:50 PM, OhioBob said:

By measuring the value of the solar constant at Kerbin, astronomers can compute the Sun's luminosity. The solar constant is defined in the game's configuration files as 1360 W/m², from which a luminosity of 3.161×10²⁴ watts is calculated.

Exactly how did you calculate 3.161×10²⁴  from 1360 W/m² ?

Edited October 28, 2018 by FreeThinker

[Mike Mars]

Just found this.

What an awesome piece of work.

Thank you so much.

I second the vote for including OPM bodies in this.

Thanks again for the great contribution.

MM

[OhioBob]

1 hour ago, FreeThinker said:

Exactly how did you calculate 3.161×10²⁴  from 1360 W/m² ?

If we were to build a big hollow sphere around the sun having a radius equal to that of Kerbin's orbit, then all of the sun's energy would fall on the inside surface of that sphere.  If we were to measure the amount of energy falling on that surface per unit area, we would measure 1360 W/m².  This measurement in called the solar constant, and the value of 1360 is defined in the game.  So to compute the total energy that the sun radiates, we just have to compute the total surface area of the sphere and multiply the result it by 1360 W/m².

The radius of Kerbin's orbit is 13,599,840,256 meter.  So the surface area of a sphere having that radius is,

     area = 4 * pi * radius^2 = 4 * pi * 13599840256^2 = 2.324221308E+21 m²

So the sun's luminosity is,

     Luminosity = solar constant * area = 1360 * 2.324221308E+21 = 3.161E+24 W.

[FreeThinker]

Thanks, but this number contradicts if you calculate Kerbol luminosity based on diameter and temperature

 luminocityKerbol = 4 * Math.PI * kerbolRadius^2 * PhysicsGlobals.StefanBoltzmanConstant * kerbolTemperature^4
= 4 * Math.PI * 261600000^2 *  PhysicsGlobals.StefanBoltzmanConstant * 5840^4 = 5.67215852172875E+25

 

[OhioBob]

25 minutes ago, FreeThinker said:

Thanks, but this number contradicts if you calculate Kerbol luminosity based on diameter and temperature

 luminocityKerbol = 4 * Math.PI * kerbolRadius^2 * PhysicsGlobals.StefanBoltzmanConstant * kerbolTemperature^4
= 4 * Math.PI * 261600000^2 *  PhysicsGlobals.StefanBoltzmanConstant * 5840^4 = 5.67215852172875E+25

 

Absolutely, I agree.  Kerbol is one big contradiction.  The reason the two methods yield such different results is because Squad made Kerbol so large.  If it's radius was at the same scale as the rest of the solar system, the two methods would yield matching results.  I prefer to go with the 3.161E+24 number because temperatures in KSP are computed using the solar constant (at least I think they are).
 

Edited October 28, 2018 by OhioBob

[FreeThinker]

2 hours ago, OhioBob said:

Absolutely, I agree.  Kerbol is one big contradiction.  The reason the two methods yield such different results is because Squad made Kerbol so large.  If it's radius was at the same scale as the rest of the solar system, the two methods would yield matching results.  I prefer to go with the 3.161E+24 number because temperatures in KSP are computed using the solar constant (at least I think they are).
 

it appears the kerbols diameter is 4.236 times to big for its luminosity. My guess they made it bigger to make it a more prominent feature when looking at KSC. The problem is that is becomes hard to predict expected luminosity for other stars other then Kerbol.

Edited October 28, 2018 by FreeThinker

[Pecan]

3 hours ago, OhioBob said:

*snip*

Three year old thread, no problem :-)

Applause - and waves.

[Superfluous J]

9 hours ago, FreeThinker said:

it appears the kerbols diameter is 4.236 times to big for its luminosity. My guess they made it bigger to make it a more prominent feature when looking at KSC. The problem is that is becomes hard to predict expected luminosity for other stars other then Kerbol.

I think they made it that size so you'd get eclipses from Mun, and Mun turned out that size for whatever reason (I think "because it looks about right from the surface even though it's really too big").

[LusitaniaU20]

This isn't related to Kerbol, but I haven't seen the plain angular diameter values for the planets/moons at their semi-major axes on this thread, so here are some I've computed:

Mun: 1 degree, 54 arc-minutes, 47.56 arc-seconds

Duna: 6.3 arc-seconds

Dres: 1.4 arc-seconds

Jool: 36 arc-seconds

Eeloo: 1 arc-second

I'll post more diameters later, if I get the time to. (I should, because of Winter Break, but you never know.)

 

Here's a picture of what I calculated Eeloo would look like if viewed from the resolution of the Hubble Space Telescope:

https://photos.google.com/photo/AF1QipPjP41XmIkdYFqXF5BCLiKv_yXuJgJmymBoniac

Edited December 28, 2018 by LusitaniaU20

[Walker]

@OhioBob thanks for supporting this, this is fantastic thread and I'm still using this. Any chances for OPM? 

[TheFlyingKerman]

On 5/20/2017 at 3:38 AM, Alphard said:

Also, on mountaintops of small islands like Mauna Kea or La Palma (where some big observatories are located), there can be seeing good enough to resolve a disc with a diameter of only 0.4", using lucky imaging. Also, kerbals may have invented adaptive optics, reducing the seeing limit to nearly zero (it is not so extremely perfect)  The only limiting thing then is the optical resolution, given by a telescopes aperture.

Yes. I suspect there is no jet stream in Kerbin and there are places with super good seeing, like the crater island.

[Curveball Anders]

2 hours ago, TheFlyingKerman said:

Yes. I suspect there is no jet stream in Kerbin and there are places with super good seeing, like the crater island.

And no big issues with light pollution