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How to calculate altitude for a given orbital period? (& vice-versa)


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T= 2 * pi * sqrt( a^3 / mu )

T is period, a is semi-major axis, mu is gravitational parameter for the body you are orbiting (which you can get from the KSP wiki).

Don't forget to include the radius of the planet (also on KSP wiki) when calculating the semi-major axis:

a= h + R

R is the radius of the planet and h is altitude above the surface (what the altimeter says at the top of the screen).

So if you know the period that you want, solve the first equation for a:

a^3 = mu * (T/(2*pi))^2

Now you can calculate a. To get the altitude, subtract the radius of the planet:

h= a-R

And you're done.

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T= 2 * pi * sqrt( a^3 / mu )

T is period, a is semi-major axis, mu is gravitational parameter for the body you are orbiting (which you can get from the KSP wiki).

Don't forget to include the radius of the planet (also on KSP wiki) when calculating the semi-major axis:

a= h + R

R is the radius of the planet and h is altitude above the surface (what the altimeter says at the top of the screen).

So if you know the period that you want, solve the first equation for a:

a^3 = mu * (T/(2*pi))^2

Now you can calculate a. To get the altitude, subtract the radius of the planet:

h= a-R

And you're done.

I think it should be a = 2 (R + h) for a circular orbit? Or a = 2R + h1 + h2 for an elliptical orbit? Other than that spot on.

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So if you know the period that you want, solve the first equation for a:

a^3 = mu * (T/(2*pi))^2

Now you can calculate a. To get the altitude, subtract the radius of the planet:

h= a-R

And you're done.

Ok, I've managed to calculate the orbital period from altitude (first equation), which I tested using the data for a geosynchronous orbit (2868.75km), and I've understood most of what's going on with the math, but I still have one problem:

How do I solve "a^3"?

I'm trying "log((3.5316 * (10 ^ 12)) * ( 21600 / ( 2 * pi ) ) ^ 2) / log(3)" but it results in 41.1227105891 rather than the geosynchronous semi-major axis of 3468750 (600000 + 2868750, altitude of 2868.75km)

"(3.5316 * (10 ^ 12))" being Kerbin's gravitational parameter and "21600" being 6 hours in seconds.

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Ah, I see what you are asking...

T= 2 * pi * sqrt( a^3 / mu )

=> T/(2pi) = sqrt(a^3 / mu)

=> (T/2pi)^2 = a^3 / mu

=> mu(T/2pi)^2 = a^3

=> cube-root(mu(T/2pi)^2) = a

Edited by EtherDragon
Actually answered question, I think
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Oh, I was turning a simple thing into a seven-headed monster...Again...:P

"((3.5316*(10^12)) * (21600 / (2*pi)) ^2) ^(1/3)"

[EDIT]

Thanks to the community's help, with special thanks to @Horn Brain, here are the formulas for both operations (for circular orbit), which you can paste on google (without the quotes) to use the hidden calculator, needing only to substitute the texts with the appropriate values, that you can get in the wiki:

Altitude-To-Orbital Period: "2 * pi * sqrt( ( ( PLANET_RADIUS_METERS + ALTITUDE_METERS ) ^ 3 ) / GRAVITATIONAL_PARAMETER ) seconds to hours"

Orbital Period-To-Altitude: "(((GRAVITATIONAL_PARAMETER * ( PERIOD_SECONDS /(2*pi))^2)^(1/3)) - PLANET_RADIUS_METERS) meters to kilometers"

Edited by AlmightyR
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Oh, I was turning a simple thing into a seven-headed monster...Again...:P

Happens to all of us - It usually takes me a couple days thinking about how to simplify an orbital mechanical problem to work it out to my satisfaction. But, a lot of that time is spent trimming out unneeded complexity.

For instance, I just finished work on this little guy:

R = 1 / (2*sqrt (Ad^3 / Ah^3))

Where

Ad is the destination semi-major axis

Ah is the Hohmann transfer semi-major axis

Which simply tells you what portion of the target's orbit ® will be completed during your Hohmann Transfer (used to plan interplanetary encounters).

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