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# How to determine the orbital period

## Question

Hi KSP colleagues,

Perhaps I should be able to figure this out, but I cannot.  I would please like to determine the orbital period for a spacecraft orbiting a celestial body given a certain altitude.

For example, if I were orbiting Mun at 100 km given a certain eccentricity, how long would that orbit take?  From the formula used to determine that, I suppose it would be quite straightforward to determine the orbital period if the orbit were circular instead of elliptical.  I also suppose I could then go backwards and find out the altitude needed to achieve an orbital period of, say, three hours.

I went to Wikipedia under "orbital period," but I don't know how to translate the information for the Kerbol system.  Here is the link for the Wikipedia article:

Stanley

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3 hours ago, MetricKerbalist said:

I would please like to determine the orbital period for a spacecraft orbiting a celestial body given a certain altitude.

For example, if I were orbiting Mun at 100 km given a certain eccentricity, how long would that orbit take?

There are two little tricks that you need to know before you start calculating.  The first is that orbital period is dependent on the semi-major axis, not the eccentricity.

This should make sense if you think about it:  let's say that you're orbiting the Mun at 100 km at any arbitrary eccentricity.  Where is the 100 km point on that orbit?  Is it the apoapsis?  Is it the periapsis?  Is it somewhere in between?  Any orbit that has a periapsis at or below 100 km and an apoapsis at or above 100 km will pass through the 100 km altitude at least once, but without knowing more about how that orbit is arranged, the altitude and eccentricity do not help.

That being said, it is true that the eccentricity is derivable from the same information that gives you the semi-major axis in KSP:  the apoapsis and the periapsis.

This leads to the second little trick:  the altitude given in KSP is measured from the surface of the celestial body, and in order to get accurate information, you need to account for the body's radius.  This information is found in the Tracking Center.  It's in the Knowledge Base button that tells the planetary characteristics, (it's the one that looks like a planet with a ring around it; it should be between the one with an i and the one that looks like a planet with a wedge carved out of it) and it's the Eq. Radius parameter that should be at the top of the list that opens when you click on it.

... Or, I suppose, you can find the information on the wiki.

Anyway, on to translating the Wikipedia article:

Wikipedia shows you two ways to calculate what you want.  The first is the calculation of orbital period from the semi-major axis, and the second is calculation of the semi-major axis from the orbital period.

3 hours ago, MetricKerbalist said:

From the formula used to determine that, I suppose it would be quite straightforward to determine the orbital period if the orbit were circular instead of elliptical.

I will reproduce the formula here to assist with the explanation:

T = 2π √(a3 / μ)

Where:

T = the orbital period (in seconds),
π = the mathematical constant, approximately equal to 3.14159265 (I generally prefer to carry precision out farther than it will ever be necessary and round off later),
a = the semi-major axis (in metres), and
μ = the Standard Gravitational Parameter, a value which depends on the celestial body.  It is equivalent to the universal gravitational constant, G,  multiplied by the body's mass, M, and so some depictions of the formula will show GM instead of μ.  The symbol is the Greek letter μ pronounced roughly as 'myoo'.  The value of this number is also available in the Tracking Center, in the same place as the equatorial radius.  It's listed as GM and is a couple of lines down from the radius.  For the Mun, it is equal to 6.514 x 1010 m3/s2 according to the Knowledge Base button, but the more accurate value is 6.5138398 x 1010 m3/s2.

Let's start with a circular Mun orbit of 100 km.  That would be, therefore, 100,000 metres.  (Metric prefix notation is nice, but this is astronomy.  Confine yourself to base units and use scientific notation when you need it; it's less confusing that way and you'll be less prone to make mistakes.)

Any orbit of the Mun will need to add 200 km to its altitude to get its true gravitational distance; this means that your circular orbit will actually have a radial distance of 300,000 metres.  Since the orbit is circular, the radius is equivalent to the semi-major axis (had it not been circular, then you'd need to average the apoapsis and periapsis distances), so that's the value we want.

Therefore:

T = 2π √([300,000]3 / [6.5138398 x 1010])
T = 2π √(2.7 x 1016 / 6.5138398 x 1010)
T = 2π √(4.145020576 x 105)
T = 2π * 643.8183
T = 4045.23 seconds

(optional conversion)

T = 4045.23 seconds * (1 minute / 60 seconds)
T = 67.42 minutes

3 hours ago, MetricKerbalist said:

I also suppose I could then go backwards and find out the altitude needed to achieve an orbital period of, say, three hours.

The second formula in the Wikipedia article is the same as the first, but algebraically manipulated to find the semi-major axis variable when all other information is given.  It's popular for finding the synchronous altitude (using Kerbin's data to verify the stationary orbit is an exercise I leave to you), but you want a three-hour orbit of the Mun, so let's work on that.

Here's the formula:

a = 3√(μT2 / 4π2)

The symbols mean the same as in the previous formula.  I direct your attention to two things to note:  first, the 3√ means cube root, and sometimes you'll see the cube root of some x-value, 3√x, represented instead as x(1/3).  The two notations are equivalent.  Second, note that I condensed the GM given in the wiki to the μ of the first formula.  This was only for consistency.

The only preparatory work needed here is to convert three hours to seconds.  That's easy enough:

T = 3 hours * (60 minutes / 1 hour) * (60 seconds / 1 minute) = 10,800 seconds

Now we need to put everything into the formula.  The rest of the parameters we know from above, so I will save some time and copy mercilessly:

a = 3√([6.5138398 x 1010] * [10,800]2 / 4π2)
a = 3√([6.5138398 x 1010] * [1.164 x 108] / 4π2)
a = 3√([7.5978106 x 1018] / 4π2)
a = 3√(1.9245479 x 1017)
a = 577,534.97 m

This is the answer, but we are not finished.  We need to subtract off the 200,000 m of the Mun's radius.  By the same assumptions as before, the semi-major axis of a circular orbit is equivalent to the orbital radius:

Altitude = 577,534.97 - 200,000 = 377,534.97 m

Put a vessel at that altitude and you'll have your three-hour orbit.  You can round up to 377,535 m, if you like.

Edited by Zhetaan
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do also notice that you can get this information when planning manuevers, without calculations. the small window on the bottom left of the screen will show you apoapsis, periapsis and orbital period. if you plan a manuever, while you are selecting the manuever node, that window will show apoapsis, periapsis and orbital period after the manuever.

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Thank you so much, @Zhetaan.  I will go through your post line by line.

Hi @Zhetaan,

Nice algebra work.  That helped me a lot.

One thing I didn't know was whether the Universal Gravitational Constant applied to Kerbol -- but why shouldn't it after all.

Thanks again.

Stanley

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