Time of darkness period for Kerosync orbit

[legoclone09]

I just need to know how long the period of darkness is on a Keosync orbit (2868.75 x 2868.75).

Edited August 30, 2015 by legoclone09

[Yasmy]

Let's look at the point where you first go into eclipse, and draw a right triangle:

The hypotenuse goes from the center of Kerbin to the spacecraft: c = 3468.75 km = orbital altitude plus Kerbin's 600 km radius.

Leg 1 goes from the center of Kerbin to the limb of Kerbin: a = 600 km.

Leg 2 goes from the spacecraft to the limb of Kerbin (the point where you see the sun getting eclipsed): b = sqrt(c^2 - a^2).

Find the angle between the center of Kerbin and the limb of Kerbin: sin(theta) = a/c = 600 / 3468.75.

theta = asin(600/3468.75) = 9.96 degrees.

Now the angle eclipsed by Kerbin is twice this angle, which you can see by drawing similar triangles behind Kerbin.

Kerbin has a 6 hour day, so it rotates 360 degrees in 6 hours. So how long does it take to rotate 20 degrees?

6 hours * 20 / 360 = 1/3 of an hour. 20 minutes.

Edited August 30, 2015 by Yasmy

[legoclone09]

Thank you, Yamsy.

[Sharpy]

You should change the title. The orbital period is trivial, 6 hours, same as Kerbin's rotation.

[Pecan]

Yep, it can all get complicated for darkness.

i) Kerbin-synchronous is 6 hours, as Sharpy says.

ii) You can do that on a polar orbit (it's different to Kerbin-stationary, which has to be equatorial).

iii) If your polar orbit is parallel to the sun, darktime = zero.

iv) Fairly academic, since 1/4 of a year later that orbit will give you maximum darktime, same as equatorial.

Edited August 30, 2015 by Pecan

[legoclone09]

I just needed to know the rime my craft would experience darkness, gotta prep my batteries!

[ajburges]

Add a minute or two to account for a Munar eclipse on either side of the shadow.

... or just assume the vessel will never be active in that situation.

[Sharpy]

Yep, it can all get complicated for darkness.

iii) If your polar orbit is parallel to the sun, darktime = zero.

Polar keosynchronous (passing over the same meridian at all times) orbit will give night time through 1/18th of the year (time proportional to sine, peak 20min/rotation).

Earth has a sun-synchronous orbit which is slightly incluned from polar and follows the terminator line over the year, but that precession is a result of non-spherical shape of Earth; achieving such orbit in KSP is impossible.

[Pecan]

... that precession is a result of non-spherical shape of Earth; achieving such orbit in KSP is impossible.

Hence "iv) Fairly academic, since 1/4 of a year later that orbit will give you maximum darktime, same as equatorial." in my post. There was a thread about this very subject some months ago.