Minmus orbit and night time

[LN400]

It's been a very good while since I last played KSP and most of the maths needs a good brushing up. Here's the thing, I'm planning my Minmus comms network (using Remote Tech and SCANSAT) and I need to work out the time these satellites will spend in the dark. For this one, I'm sending a scanning sat to Minmus on a 70km circular orbit.

 

I did punch in the numbers but the result I got just seems off. I got a little under 18 minutes in darkness at 70km. I barely remember Minmus and orbital periods now but 18 minutes sounds too low. I was expecting (don't know why really) more around 30-45 minutes but I did the calculations twice and I got the same result, just short of 18 minutes. Can this be right or did I mess up somewhere?

[VoidSquid]

This mod can calculate the times for you:

 

[jimmymcgoochie]

That sounds about right to me. While an orbit of Minmus at 70km might take a little over an hour and half of that orbit will be over the night side of Minmus‘ surface (or any planet/moon), the time that the orbiting spacecraft is in Minmus’ shadow is quite a bit less than half of an orbit. Minmus casts a conical shadow pointing directly away from the sun: on the surface this means that sunlight and shadow are approximately equal but in space the extra altitude means that there’s a longer arc before you enter the shadow on both sides of the orbit, reducing the time that the vessel is in shadow; with an inclined orbit- which you’ll be using for SCANsat scanning- that number comes down even more as depending on the positions of Minmus and Kerbin you can have times where the shadow is perpendicular to your orbit so the vessel is in constant sunlight.

I would try to draw a diagram, but I’m not on my PC and it would probably be terrible anyway.

[LN400]

Thanks for the heads-up on that mod. Appreciated. I would like though, to know if my maths is in worse shape than I think. I depend too much on doing calculations "by hand", it's part of the fun in playing KSP. It is no fun however, to get wrong results due to not getting the maths right.

Just now, jimmymcgoochie said:

[entire post]

Thanks heaps! It has been a long time like I said but in my faulty memory, orbiting Minmus took forever and a half even at low altitude! I guess I need to be more optimistic then

[bewing]

Yeah, the diameter of Minmus is 120km. I don't remember what the orbital speed at 70km altitude is, but I'm sure it's over 100m/s. So divide, and that's 1200 seconds. Or 20 minutes. So your calculation is fine.

 

[Zhetaan]

On 9/27/2020 at 4:29 AM, LN400 said:

I did punch in the numbers but the result I got just seems off. I got a little under 18 minutes in darkness at 70km. I barely remember Minmus and orbital periods now but 18 minutes sounds too low. I was expecting (don't know why really) more around 30-45 minutes but I did the calculations twice and I got the same result, just short of 18 minutes. Can this be right or did I mess up somewhere?

Let's go through it step by step.  There are certain assumptions and simplifications that we can (and should!) apply to this problem, but the overall result will still be close enough to correct.  First, however, since your issue is more about your assumed expectations, let's look at the problem more abstractly and see what insights we get from it to correct your intuition.

For one thing, your orbit radius is a bit over twice Minmus's radius (remember that an altitude of 70,000 metres is actually 130,000 metres from Minmus's centre).  In terms of the abstraction, let's call it exactly twice and work from that.

If Minmus is at half of the orbital radius, what then?  If we assume that the orbit is a unit circle, that the sun shines from the left side of the graph, and that Minmus's shadow is a cylinder (projected to be a rectangle--the point is that it does not taper), then Minmus's bulk will cover the right side of the orbit to half of the orbit's maximum height above the x-axis.  Thus, we want the angle that leads to a point y = .5 on the unit circle.   We can solve that with trigonometry:  arcsin (.5 / 1) = 30°.  Of course, Minmus covers a portion of the shadow below the x-axis, too, so we need to double the angle.  The total included angle is therefore 60°.

We can get that value another way, courtesy of our abstraction:  if the orbit is twice Minmus's radius, then the size of the shadow, which itself is twice Minmus's radius, will equal the orbit radius.  Since the distance from the origin to the edge of the shaded part of the orbit is going to be one orbit radius for both the upper and lower edges of the shadow, it altogether makes an equilateral triangle, so the included angle is definitely 60°.  Since your actual orbit is slightly larger than double Minmus's radius, we can safely expect that the true included angle will be near to but less than this value.

In terms of the fractional coverage of the larger orbit, 60° is one sixth of the total orbit, and of course the total is the full circle at 360°.  If you expect a darkness time of 30 - 45 minutes, then you expect an orbital period of 180 - 270 minutes, or 3 to 4.5 hours (actually slightly more).  If you find your orbital period in KSP, you'll see right away that that does not make sense.  The reasons why are less intuitive, but it's simple enough to check.

Orbital period is better to use, but your orbital speed is the first parameter that KSP displays, so let's work with that, instead.  I'll show the calculation later, but for now let's assume that it is at or very near to 116.5 m/s.  The circumference of a 130,000-metre circular orbit is 816,814 metres, so going around that orbit at that rate gives a period of 116.85 minutes.  Thirty minutes of darkness is a little more than one quarter of that orbit.  Forty-five minutes is closer to forty percent--and that should immediately feel absurd, because fifty percent coverage is what you get at the surface.  Forty percent coverage requires you to be so close to surface altitude that to achieve it should have you crashing into the mountaintops.  One sixth of the orbit gives about 19.5 minutes, but again, we've already shown that the actual value will be somewhat less than that.

 

Now to the more precise calculations:

Minmus has a standard gravitational parameter, μ, of 1.7658 x 10⁹ m³/s² and a radius, r, of 60,000 metres.  A circular orbit at 70,000 m altitude has a semimajor axis, a, of 130,000 metres, which is equal to the orbital radius at all points on the orbit.  This orbit thus has an orbital speed, v, of:

v = √(μ / a)
v = √(1.7658 x 10⁹ / 130000)
v = √(13583.1)
v = 116.5 m/s

Minmus's shadow cone technically ought to vary in length over the course of its orbit about Kerbin.  Since neither Minmus's orbit nor Kerbin's have any eccentricity, the constraints on the variation, and thus also the variation in taper angle, should be relatively easy to determine.  However, for one thing, this orbit is close enough to Minmus that approximating Minmus's shadow as a cylinder with no taper will work more than adequately (it will slightly overestimate the darkness time, but not by a lot), and for another thing, I don't know how faithful KSP is to the realities of physics in this case.  I would like to assume that it is accurate but I have neither tested it nor seen anyone else do so.

Under these assumptions, the included angle of the shadowed arc of the orbit, θ, is given by twice the arcsine of the body radius, r, divided by the orbital radius, R:

θ = 2 * arcsin (r / R)
θ = 2 * arcsin (60000 / 130000)
θ = 2 * .479729 rad
θ = .95946 rad or 54.97°

The arc length of the shadowed portion of the orbit, l, is given by the included angle multiplied by the orbital radius:

l = R * θ
l = 130000 * .95946
l = 124,729.4 m

At the previously-calculated (constant) rate of speed, your vessel will cover this length in time t:

t = l / v
t = 124729.4 / 116.5
t = 1070.6 seconds
t = 17.84 minutes

Your calculations are correct.

Edited September 28, 2020 by Zhetaan

[LN400]

5 hours ago, Zhetaan said:

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That is one thorough post. Thank you a couple of bunches! With my trusty calculator out of commision for the time being, I rely on notepad and Wolfram Alpha but that combo is not the most expedient or practical there ever was and I was afraid I had made some silly errors along the way.