Where is geostationary orbit ? :: Kerbal Space Program General Discussions

Just Jim

Dec 11, 2015 @ 5:43pm

2,863,324m and 2,863,344m

#1

Daniel

Dec 11, 2015 @ 5:55pm

Originally posted by Just Jim:
2,863,324m and 2,863,344m

ty

#2

Vectura

Dec 12, 2015 @ 5:11am

Why would the numbers be different? Surely you want a perfectly circular orbit?

#3

Psychic

Dec 12, 2015 @ 5:49am

Originally posted by CMDR Vectura #noobandproud:
Why would the numbers be different? Surely you want a perfectly circular orbit?

The planet is probably not perfectly circular. Like Earth isn't.

#4

Just Jim

Dec 12, 2015 @ 6:57am

Actually, I don't quite know myself, I copied these numbers from my most recient keostationary satellite contract.

Last edited by Just Jim; Dec 12, 2015 @ 6:58am

#5

tourist

Dec 12, 2015 @ 7:03am

Originally posted by Just Jim:
Actually, I don't quite know myself, I copied these numbers from my most recient keostationary satellite contract.

It's right in the middle, zero eccentricity (circular):

. Kerbisynchronous Equatorial Orbit (KEO) has a circularly uniform altitude of 2 863.33 km and a speed of 1 009.81 m/s.
http://wiki.kerbalspaceprogram.com/wiki/Kerbin#Orbits

OP seems to be asking for geosync. orbit that has vessle over same point on Kerbin four times a day, "get 6 hr orbit." 24hr day divide by four...

I'm not sure if that's right but the velocity is as important as the altitude.

#6

Vectura

Dec 12, 2015 @ 7:06am

Kerbin has 6 hour days you know.

#7

tourist

Dec 12, 2015 @ 7:08am

Originally posted by CMDR Vectura #noobandproud:
Kerbin has 6 hour days you know.

I do now.
edit: looks like typo
"I do, now."
"Now, I do."
"I did not know that."

Last edited by tourist; Dec 12, 2015 @ 7:11am

#8

MAD

Dec 12, 2015 @ 7:02pm

When in doubt just work it out yourself
1 solar day = 21,600 s (6.0 hours)
1 siderial day = 21,549.425 s (this is the period for geosync orbit, I don't think KSP worried about orbital precession)
Using Keplers 3rd law, equitorial geostationary orbit r = 3,463,334 m or an altitude of 2,863.334 km circular with an orbital velocity of 1,009.81 m/s

Last edited by MAD; Dec 12, 2015 @ 7:09pm

#9

tourist

Dec 13, 2015 @ 9:37am

Originally posted by MAD:
When in doubt just work it out yourself
1 solar day = 21,600 s (6.0 hours)
1 siderial day = 21,549.425 s (this is the period for geosync orbit, I don't think KSP worried about orbital precession)
Using Keplers 3rd law, equitorial geostationary orbit r = 3,463,334 m or an altitude of 2,863.334 km circular with an orbital velocity of 1,009.81 m/s

I'm confused how the Third Law was applied. Explain/elaborate/correct:
1 = ( 21 549.425)^2 / a^3 , where "a" is the altitude in question?
a = ( ( 21 549.425)^2 )^(1/3)
a= 774.3853428622663
(This doesn't make sense to me. Assuming you did something like below:)

Looking at the Third Law w.r.t. Kerbin and the Mun:
Mun Semi-Major Axis = 12 000 000 m (1 Mun_R)
Mun Sidereal Orbit = 138 984 s (1 Mun_T)

Now, looking for a Sidereal Orbit of 21 549.425 s
21 549.425 / 138 984 = 0.1550496819777816
So, 1 = ( 0.1550496819777816 )^2 / (r)^3
and, r = ( ( 0.1550496819777816 )^2 )^(1/3)
r = 0.2886116913133595 Mun_R
r = 0.2886116913133595 * 12 000 000 , r = 3463340.295760314 m
Or, r = 3 463.340 km
Then, minus the Equitorial Radius of Kerbin (600km) = 2863.340km
(Maybe rounding...? ...basicaly same value.)

The "r" calculated traces a circular orbit, and the circumference is the distance travled during one orbit. One day to make the orbit, making the velocity easy enough to calculate.

Comparing Harmonics:
Kerbol System______(T^2/R^3)______Solar System_________(T^2/R^3)
Moho___________0.999999718587____Mercury____________0.98
Eve_____________0.999999865101____Venus______________1.01
Kerbin_______________1.0___________Earth_______________1.0
Duna____________0.999999900413____Mars________________1.01
Jool_____________0.999999914817____Jupiter______________0.99
Dres_____________0.999999911093____Ceres______0.9998255010589789
Eeloo____________0.999999911818____Pluto_______________1.0

T: Kerbin Year (9203545 s), R = 13599840256 m
Ceres T = 4.5997152557222645 y , Ceres R = 2.76596 AU
I pulled the Solar System data from wiki, also didn't see T^2/R^3 for Ceres and just did it.
Example calculation (to check for errors):
Moho Orbital Radius (semi-major axis, average of apoapsis and periapsis) = 5 263 138 304 m
Moho Orbital Radius w.r.t Kerbin "Astronomical Unit" :
5 263 138 304 / 13 599 840 256 = 0.387000009186
(Moho "R" variable, 0.387000009186 Kerbin AU from Kerbol to Moho)
Moho Sidereal Orbital Period = 2 215 754 s
Moho Orbital Period in Kerbin Years : 2 215 754 / 9 203 545 = 0.240750058809
Moho T^2/R^3 : (0.240750058809)^2 / (0.387000009186)^3 = 0.999999718587

#10

argiebarge has Kerbal Space Program

Dec 13, 2015 @ 10:04am

Originally posted by Nostradufus:
Originally posted by MAD:
...beyond me....

i'll stick to ♥♥♥♥♥♥♥

#11

MAD

Dec 13, 2015 @ 6:26pm

you only need to calculate it if you are into that sort of stuff, otherwise just take it from the wiki
http://steamcommunity.com/sharedfiles/filedetails/?id=575321399&fileuploadsuccess=1

u the standard gravitational parameter is different for every solar body
https://en.wikipedia.org/wiki/Standard_gravitational_parameter

ukerbol (the star) = 1.1723328×1018 m3/s2
ukerbin = 3.5316000×1012 m3/s2
http://wiki.kerbalspaceprogram.com/wiki/Kerbin

1 sideareal day is the time it take for a planet to make one complete revovution relative to the stars. 21,549.425s
1 solar day is the time it take for a planet to make one complete revovution relative to the star. 21,600s ie exactly 6 hours
they differ because the planet is in orbit around a star.
https://en.wikipedia.org/wiki/Solar_time

To calculate the radius for geostationary orbit we desire an orbit where the period of that orbit is 1 sidereal day
https://en.wikipedia.org/wiki/Geostationary_orbit

rather than calculate k based on the standard gravitational parameter I am going to calculate it based on something real that I can observe ie the Mun
Tmun = 138,984s
amun = 120,000,000m
kerbin k = (138,984 * 138,984) / (120,000,000 * 120,000,000 * 120,000,000) = 1.11786E-11

minmus should give the same result
T minmus = 1,077,311 s
a minmus = 47,000,000 m
kerbin k = (1,077,311 * 1,077,311) / (47,000,000 * 47,000,000 * 47,000,000) = 1.11786E-11
and it does

u kerbin = 4pi2/k = 3.5316E+12 which matches the ksp wiki

geostationary
a3 = t2/k = 21,549.425 * 21,549.425 / 1.11786E-11 = 4.15418E+19
a = 3,463,340.296m or an altitude of 2,863.34 km

the ksp wiki says 2 863.33 km (close enough)

Measuring the period is not a good way to determine if your satelite is in geostationary orbit
since an ellipse with the same 'semi-major axis' a would have the same period.
Better to rely on orbital velocity

for a circle of radius r it has a circumference 2 pi r therefore Vc = 2 pi r / T

or Vc = sqr root(u / r)

Last edited by MAD; Dec 13, 2015 @ 7:38pm

#12

Daniel

Dec 13, 2015 @ 7:04pm

Originally posted by Nostradufus:
Originally posted by MAD:
When in doubt just work it out yourself
1 solar day = 21,600 s (6.0 hours)
1 siderial day = 21,549.425 s (this is the period for geosync orbit, I don't think KSP worried about orbital precession)
Using Keplers 3rd law, equitorial geostationary orbit r = 3,463,334 m or an altitude of 2,863.334 km circular with an orbital velocity of 1,009.81 m/s
I'm confused how the Third Law was applied. Explain/elaborate/correct:
1 = ( 21 549.425)^2 / a^3 , where "a" is the altitude in question?
a = ( ( 21 549.425)^2 )^(1/3)
a= 774.3853428622663
(This doesn't make sense to me. Assuming you did something like below:)

Looking at the Third Law w.r.t. Kerbin and the Mun:
Mun Semi-Major Axis = 12 000 000 m (1 Mun_R)
Mun Sidereal Orbit = 138 984 s (1 Mun_T)

Now, looking for a Sidereal Orbit of 21 549.425 s
21 549.425 / 138 984 = 0.1550496819777816
So, 1 = ( 0.1550496819777816 )^2 / (r)^3
and, r = ( ( 0.1550496819777816 )^2 )^(1/3)
r = 0.2886116913133595 Mun_R
r = 0.2886116913133595 * 12 000 000 , r = 3463340.295760314 m
Or, r = 3 463.340 km
Then, minus the Equitorial Radius of Kerbin (600km) = 2863.340km
(Maybe rounding...? ...basicaly same value.)

The "r" calculated traces a circular orbit, and the circumference is the distance travled during one orbit. One day to make the orbit, making the velocity easy enough to calculate.

Comparing Harmonics:
Kerbol System______(T^2/R^3)______Solar System_________(T^2/R^3)
Moho___________0.999999718587____Mercury____________0.98
Eve_____________0.999999865101____Venus______________1.01
Kerbin_______________1.0___________Earth_______________1.0
Duna____________0.999999900413____Mars________________1.01
Jool_____________0.999999914817____Jupiter______________0.99
Dres_____________0.999999911093____Ceres______0.9998255010589789
Eeloo____________0.999999911818____Pluto_______________1.0

T: Kerbin Year (9203545 s), R = 13599840256 m
Ceres T = 4.5997152557222645 y , Ceres R = 2.76596 AU
I pulled the Solar System data from wiki, also didn't see T^2/R^3 for Ceres and just did it.
Example calculation (to check for errors):
Moho Orbital Radius (semi-major axis, average of apoapsis and periapsis) = 5 263 138 304 m
Moho Orbital Radius w.r.t Kerbin "Astronomical Unit" :
5 263 138 304 / 13 599 840 256 = 0.387000009186
(Moho "R" variable, 0.387000009186 Kerbin AU from Kerbol to Moho)
Moho Sidereal Orbital Period = 2 215 754 s
Moho Orbital Period in Kerbin Years : 2 215 754 / 9 203 545 = 0.240750058809
Moho T^2/R^3 : (0.240750058809)^2 / (0.387000009186)^3 = 0.999999718587

This is what it looks like to new players KSP...

#13

MAD

Dec 13, 2015 @ 7:27pm

Originally posted by Born To be Russian:
This is what it looks like to new players KSP...

not sure what point you trying to make?
writing equations in most word processors is a pain

#14

tourist

Dec 14, 2015 @ 1:41am

Originally posted by MAD:

u the standard gravitational parameter is different for every solar body
https://en.wikipedia.org/wiki/Standard_gravitational_parameter

ukerbol (the star) = 1.1723328×1018 m3/s2
ukerbin = 3.5316000×1012 m3/s2
http://wiki.kerbalspaceprogram.com/wiki/Kerbin

μ "mu" looks more cyrillic than greek in steam font. [The gravitational aspect is interesting. I'm not sure I completely understand all of it.]

Originally posted by MAD:
rather than calculate k based on the standard gravitational parameter I am going to calculate it based on something real that I can observe ie the Mun
Tmun = 138,984s
amun = 120,000,000m
kerbin k = (138,984 * 138,984) / (120,000,000 * 120,000,000 * 120,000,000) = 1.11786E-11

I did the same [sort of, used the Mun as the orbit to "harmonize" with]. Your Mun altitude is a factor of ten too great. 12 000 000 m in the wiki.
I get 1.1178560333333333e-14 running those numbers, probably just a typo.

Originally posted by MAD:

Measuring the period is not a good way to determine if your satelite is in geostationary orbit
since an ellipse with the same 'semi-major axis' a would have the same period.
Better to rely on orbital velocity

Not neccessarily the same period; a circle with a radius equal to the semi-major axis (half the big axis) of an ellipse does not mean that the permimeter [f]or the ellipse is equal to the circumference of the circle. Velocity is a consideration too. The period is [part of] the target [with altitude], geostationary orbit.

It's interesting because now any planet with a moon that isn't in some weird orbit should be able to give easy to calculate geosyncronous altitude and velocity over the planet. I can just look it up on wiki too but now the value has more meaning for me. Each to their own, it all doesn't add up [in the grand scheme of things].

I added "Johan Kerplar" to my roster via a quick persistent edit.

Last edited by tourist; Dec 14, 2015 @ 2:29am

#15

Kerbal Space Program

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