GSO: what does it depend on?

[BoilingOil]

Another general question for you guys.

I've brought several probes (typically just empty cans or random hardware stacks) into orbits around Kerbin, the Mun, Minmus and a few other places. Some went onto massive orbits around Kerbol (the Sun, for those who didn't know), mostly because I shot them way out of Kerbin's SoI without really targeting anything. So that's no longer really a challenge.

So now I though: hmm, a GSO around Kerbin would be nice. Just a probe that eternally sits right above a single spot on the surface.

Now that's all fine and dandy, but how do I do that? How do I calculate where it needs to be - and for this I'm merely interested in the height of the orbit. Figuring out how to target a specific spot in the orbit is for another day, as far as I'm concerned.

Thanks in advance for any insights you guys might have for me.

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To add: I've even had a probe in a highly excentric tilted orbit: periapsis at 150 KM over Kerbin, apoapsis well beyond the Mun, and tilted such that on its way to apoapsis it passed a long way Normal of Mun's orbit, but on the way back to Kerbin, it would pass right through Mun's orbit. This was not going to last, though: three times, on its way back, it encountered Mun closely enough for its trajectory to be radically altered. At the fourth encounter, though, it was embedded deeply inside Mun!

Edited November 6, 2015 by BoilingOil

[ElWanderer]

2869km (circular and Equitorial) will get you a period of 6 hours (edit: getting the time difference between periapsis and apoapsis to be three hours is more important than getting the altitude perfectly right), so the orbit will be stationary. The wiki has this information: http://wiki.kerbalspaceprogram.com/wiki/Kerbin#Orbits

Edited November 6, 2015 by ElWanderer

[Gaarst]

The semi-major axis a for a orbit of period T around a body of mass M is given by Kepler's 3^rd law:

T² / a³ = 4À² / GM

Rearranging for a gives:

a = (GMT² / 4À²)^1/3

To get the semi-major axis of a geo-synchronous orbit, just use the body's period of rotation for T (6h for Kerbin). Note that the heights used here are distances to the body's centre of mass, not its surface.

The semi-major axis is a property of elliptical orbits, and is not given as it is in KSP. You can calculate it from periapsis and apoapsis heights:

a = (Pe + Ap) / 2

For circular orbits, the semi-major axis is equal to the radius of the orbit, so a can be replaced by R.

Edited November 6, 2015 by Gaarst
Extended to elliptical orbits

[BoilingOil]

2869km (circular and Equitorial) will get you a period of 6 hours (edit: getting the time difference between periapsis and apoapsis to be three hours is more important than getting the altitude perfectly right), so the orbit will be stationary.

Ok, that would do it, I suppose. Thanks!

The wiki has this information: http://wiki.kerbalspaceprogram.com/wiki/Kerbin#Orbits

I'll check that out as well

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The height R for a orbit of period T around a body of mass M is given by Kepler's 3^rd law:

T² / R³ = 4À² / GM

Rearranging for R gives:

R = (GMT² / 4À²)^1/3

To get the height of a geo-synchronous orbit, just use the body's period of rotation for T (6h for Kerbin). Not that the height of the orbit here is a distance to the body's centre of mass, not its surface.

Good, a nice clean general formula! I like that... Thanks!

And by 'G', I assume that you actually meant 'g', the gravitational constant, 9.81m/s/s ??

Also, I notice something... the mass M of the planet is mentioned. But I do not see the mass of the probe represented. Doesn't that factor in at all?

Edited November 6, 2015 by BoilingOil

[Gaarst]

Good, a nice clean general formula! I like that... Thanks!

And by 'G', I assume that you actually meant 'g', the gravitational constant, 9.81m/s/s ??

G is indeed the gravitational constant, but it is not the same as g !

G is a constant and equal to 6.67x10^-11 N.m².kg^-2 for all bodies.

g is the gravitational acceleration at Earth's surface and is defined by: g = GM_E / R_E² = 9.81 m.s^-2

Where G is the gravitational constant, M_E is the mass of the Earth and R_E is the Earth's radius.

Make sure not to mess up the two while doing calculations, or you will end up with completely wrong results.

[BoilingOil]

G is indeed the gravitational constant, but it is not the same as g !

G is a constant and equal to 6.67x10^-11 N.m².kg^-2 for all bodies.

g is the gravitational acceleration at Earth's surface and is defined by: g = GM_E / R_E² = 9.81 m.s^-2

Where G is the gravitational constant, M_E is the mass of the Earth and R_E is the Earth's radius.

Make sure not to mess up the two while doing calculations, or you will end up with completely wrong results.

Ah! That was something I wasn't aware of... Now I'm double glad I asked about it, because I *would* have used g, and then I would have failed!

And just to satisfy my curiosity: 6.67 is actually rounded for practical use, right? It should probably have been 6.666... etc?

[Gaarst]

Also, I notice something... the mass M of the planet is mentioned. But I do not see the mass of the probe represented. Doesn't that factor in at all?

No it doesn't in KSP.

In real life it would impact the period because instead of using only M in the equation you would have to use M + m (m being the mass of the orbiting satellite) but we don't use it in KSP for 2 reasons:

1) For an artificial satellite orbiting a planet, m is so small compared to M that basically M + m = M

2) KSP uses on-rails orbits for celestial bodies. That means that even if you were considering a vessel ten times the mass of Kerbin, Kerbin would remain at a fixed point, and the other body would orbit it without taking its own mass into account.

[FancyMouse]

Also, I notice something... the mass M of the planet is mentioned. But I do not see the mass of the probe represented. Doesn't that factor in at all?

A legit concern, but nothing wrong here. The M in Kepler 3rd law is the sum of the mass of the two bodies, and R is measured not from CoM of planet but CoM of the system (aka barycenter). However, in most cases (including all KSP scenario), the mass of the smaller object is negligible so that it's ok to ignore that term, and use CoM of the big-mass object to measure R.

[Specialist290]

Also, I notice something... the mass M of the planet is mentioned. But I do not see the mass of the probe represented. Doesn't that factor in at all?

For all practical purposes, even in the real world, the difference between the mass of the planet and the mass of the probe body is going to be so immense that the latter can be essentially ignored, unless you're putting something really massive (like another moon) into orbit.

Additionally, in KSP, the planets are "on rails," so their orbits can't be meaningfully modified in-game without something like HyperEdit.

Edited November 6, 2015 by Specialist290
Word

[Snark]

And just to satisfy my curiosity: 6.67 is actually rounded for practical use, right? It should probably have been 6.666... etc?

Yes, it's rounded, but it's a physical constant, not a mathematical one, so it's not 6.66666. 6.672 is closer. (It's an oddly difficult number to measure directly, for various practical reasons-- we only know it well to a couple of decimal places, the third decimal place is under some contention. It's not like the speed of light, which we can measure to many many decimal places.)

https://en.wikipedia.org/wiki/Gravitational_constant

In practical terms, what you really care about is GM (G times the mass of the planet), which KSP conveniently tells you in the info panel in map view when you're focused on the planet.

Edited November 6, 2015 by Snark

[Gaarst]

Ah! That was something I wasn't aware of... Now I'm double glad I asked about it, because I *would* have used g, and then I would have failed!

And just to satisfy my curiosity: 6.67 is actually rounded for practical use, right? It should probably have been 6.666... etc?

Actually, the exact admitted value for G is 6.67408x10^-11 N.m².kg^-2

It is surprising that even today, with all the technology and all the satellites orbiting Earth, we can achieve only 6 significant figures for the value of G, right ?

Well, as it turns out, G is never used on its own for orbit calculations, so we defined a new value: the standard gravitational parameter µ, so that: µ = GM.

µ is unique for each body, as it depends on the body's mass, and it can be determined directly using orbital measurements, with no need for the body's exact mass. Therefore, µ values are often known to 10 significant figures: for example, µ for Earth is 3.986004418x10⁸ m³.s^-2

Edited November 6, 2015 by Gaarst

[BoilingOil]

Actually, the exact admitted value for G is 6.67408x10^-11 N.m².kg^-2

I should have learned by now, not to assume so much! Thanks again!

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No it doesn't in KSP.

In real life it would impact the period because instead of using only M in the equation you would have to use M + m (m being the mass of the orbiting satellite) but we don't use it in KSP for 2 reasons:

1) For an artificial satellite orbiting a planet, m is so small compared to M that basically M + m = M

2) KSP uses on-rails orbits for celestial bodies. That means that even if you were considering a vessel ten times the mass of Kerbin, Kerbin would remain at a fixed point, and the other body would orbit it without taking its own mass into account.

Ah, got it! And IRL, the same would mostly be true as well: the mass of any probe or satellite in Earth's orbit is inconsequential in comparison to the shere mass of Earth itself!

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A legit concern, but nothing wrong here. The M in Kepler 3rd law is the sum of the mass of the two bodies, and R is measured not from CoM of planet but CoM of the system (aka barycenter). However, in most cases (including all KSP scenario), the mass of the smaller object is negligible so that it's ok to ignore that term, and use CoM of the big-mass object to measure R.

Precisely. Now that makes sense as well!

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For all practical purposes, even in the real world, the difference between the mass of the planet and the mass of the probe body is going to be so negligible that the latter can be essentially ignored, unless you're putting something really massive (like another moon) into orbit.

Additionally, in KSP, the planets are "on rails," so their orbits can't be meaningfully modified in-game without something like HyperEdit.

So many people knowing what I *should* have known... I'm beginning to experience some shame now

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Yes, it's rounded, but it's a physical constant, not a mathematical one, so it's not 6.66666. 6.672 is closer. (It's an oddly difficult number to measure directly, for various practical reasons-- we only know it well to a couple of decimal places, the third decimal place is under some contention. It's not like the speed of light, which we can measure to many many decimal places.)

https://en.wikipedia.org/wiki/Gravitational_constant

In practical terms, what you really care about is GM (G times the mass of the planet), which KSP conveniently tells you in the info panel in map view when you're focused on the planet.

Ah, so that makes it even simpler to find the right orbit myself! Thanks again, Snark. Most helpful!

[FancyMouse]

It is surprising that even today, with all the technology and all the satellites orbiting Earth, we can achieve only 6 significant figures for the value of G, right ?

And appreciate the fascinating experiment carried out by Cavendish over 200 years ago, which gives an error of just about 1%

[GoSlash27]

BoilingOil,

You've got the formula for the kerbosynchronous orbit, which my spreadsheet says is an altitude of 2,868,723m.

The next step is to figure out how much DV you need to place an object there.

The vis-viva will tell you how much additional DV is required to execute a Hohmann transfer from LKO to KSO. There's an excellent write-up of the procedure here:

http://www.braeunig.us/space/orbmech.htm

^ Bookmark or download a copy of this!

for a Hohmann transfer from an altitude of 101,305m to LKO, the first burn will be 650 m/sec and the circularization burn will be 424 m/sec; a total budget of 1,074 m/sec. Why such an odd LKO altitude? well...

Where it really gets fun is figuring out the longitude your vehicle needs to be at in order to place the sat over a specific longitude on the globe in KSO.

In order to do this, you need to look back at your transfer orbit from the previous problem. You need to convert it's SMA into an orbital period (also in the guide I linked) and then divide by 2. This is how much time it will take you to go from Pe to Ap; 5,033 seconds.

You need to know this because Kerbin is going to rotate during the transfer and you need to lead it so that you will arrive overhead the target. Kerbin has a rotation period of 6 hours (21,600 seconds), so it will rotate (5,033/21,600) *360° = 83.88° in that time.

Since your Pe must be on the opposite side of the planet, you would start the transfer when the vehicle's longitude is 180°-83.9°= 96.1° west of the target longitude.

And before you can make use of that, you have to answer a more fundamental problem: How do you know your vehicle's longitude at any given moment?

On a bone-stock installation, there's no direct readout of the vehicle's longitude. This requires some inventiveness to work around and this is how I solve it:

The reason I chose such an oddball LKO altitude is because it has an orbital period that orbits Kerbin exactly 11 times per day. Since Kerbin rotates once per day, that means that it passes over KSC 10 times per day, or once every 36 minutes.

By placing a seeker pointed up on KSC grounds and tracking it, I can mark the time of it's passage over KSC down to the second.

Since it passes over KSC once every 36 minutes, it covers 10° per minute and 1° every 6 seconds. I therefore know what it's longitude is simply by the time of day.

It's then a simple matter to calculate a transfer window from the time it passes over KSC. If I wanted to place a sat directly over KSC, I would fire at 360°-96.1°= 263.9° after passage. That's 26 minutes, 23.4 seconds.

For ultimate accuracy, I would want to know how long my transfer burn will take so I can split the difference.

Orbital mechanics is fun math once you get the hang of it, and it opens up all sorts of interesting possibilities.

Good luck!

-Slashy

Edited November 6, 2015 by GoSlash27

[BoilingOil]

BoilingOil,

You've got the formula for the kerbosynchronous orbit, which my spreadsheet says is an altitude of 2,868,723m.

The next step is to figure out how much DV you need to place an object there.

Thanks, Slashy. I will know to use this when the time comes that I want to get this deep into it. For now, I haven't even figured out yet, how to get something into an LKO at a specific hight efficiently from launch, yet. And to make it circular at that orbit is yet another issue. Once I've got my space legs, I'll get there eventually. And then I'll NEED this stuff.

Edited November 6, 2015 by BoilingOil

[Highguard]

I should have learned by now ... Ah, got it! ... Now that makes sense as well! ...

So many people knowing what I *should* have known... I'm beginning to experience some shame now

Your first comments make the last one unnecessary, we've all been there and if you're starting to learn all the other stuff that's new to you then you know why KSP is so cool :-)

[BoilingOil]

You are so right about that, Highguard. KSP is awesome! And some of you guys here are awesome as well! I've already discarded a number of highly inefficient rockets that I had created, in favor of some cheaper, lighter ones made with the knowledge that I've collected here in only two days!

[Curveball Anders]

fun math

That's two words I'd never believe that I'd see together before trying KSP

[BoilingOil]

That's two words I'd never believe that I'd see together before trying KSP

For me, those two words went together quite well from the moment I learned about calculus. But I understand how for most people that would be different. Sometimes, these numbers can just make your head spin!

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The semi-major axis a for a orbit of period T around a body of mass M is given by Kepler's 3^rd law:

T² / a³ = 4À² / GM

Rearranging for a gives:

a = (GMT² / 4À²)^1/3

To get the semi-major axis of a geo-synchronous orbit, just use the body's period of rotation for T (6h for Kerbin). Note that the heights used here are distances to the body's centre of mass, not its surface.

The semi-major axis is a property of elliptical orbits, and is not given as it is in KSP. You can calculate it from periapsis and apoapsis heights:

a = (Pe + Ap) / 2

For circular orbits, the semi-major axis is equal to the radius of the orbit, so a can be replaced by R.

Nice edit to include eliptical orbits, Gaarst.

Also, I must point out that it took me a bit to actually see what I was looking at. At fist, I was like "Hmm, the speed of the vessel must be sufficient to overcome the gravitational force working on your vessel, which is dictated by the height of your orbit. That's complicated, if this height (R or (Ap+Pe)/2) is what you want to calculate in order to get the required speed (v) that gets you around the planet in a specific time (T)!"

But it is not at all complicated, if you ignore v altogether, and simply use the aforementioned time (T) directly, which the above formula does! Thank you again for this most insightful response!

Edited November 7, 2015 by BoilingOil