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Geostationary orbit altitudes


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What altitude is needed for geostationary orbit?

I have already searched the forums, but the altitudes described were different leaving me quite confused.

You need an 'as near as circular as possible' orbit -- 2,870,000m @ 1,008m/s for geo-stat. around Kerbin

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both values are stated on the KSP wiki!

If you\'re looking for something to do with the game, it\'s a value and you can\'t find it on the forums, there\'s a reasonable chance you will find it on the Wiki!

PD

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both values are stated on the KSP wiki!

If you\'re looking for something to do with the game, it\'s a value and you can\'t find it on the forums, there\'s a reasonable chance you will find it on the Wiki!

PD

Last time I went on the wiki it was trying to sell me all sorts of stuff; is the spam issue fixed?

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The spam issue is under control at the moment, so long as you stick to the pages you need, as opposed to just randomly surf around.

(Even then, you should find it hard to find an actual spam advert, but you will find LOTS of spam/Delete tags left by editors such as myself).

Just to point out (I added this note to the wiki after I tested the theory), that it isn\'t actually possible to achieve geo stationary orbit around the mun, as you get re-captured by Kerbin.

And yes, orbits rely on gravity, and in a vacuum, a feather falls at the same speed as a double decker bus. ;) Mass and size make no real odds.

PD

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Just for future reference, orbits do not depend on mass or craft size or anything except for velocity and radius.

This, using Kepler\'s 3rd law the square of the orbital period is proportional to the cube of the radius. This can also be derived from Newton\'s 3rd law and his theory of gravitation as:

Centrifugal force of a circular orbit = gravitational force

mv^2/r=GMm/r^2

where v=2*pi*r/T hence:

4*Pi^2*r/T^2=GM/r^2

4*Pi^2*r^3=GMT^2 which is Kepler\'s Law, for a geostationary orbit r can be solved with T=1 kerbin day in seconds.

where r is the distance from the centre of kerbin (not altitude above sea level), G is the gravitational constant, M is the mass of Kerbin.

As you can see T and r have no relation with m the mass of the orbiting object.

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This, using Kepler\'s 3rd law the square of the orbital period is proportional to the cube of the radius. This can also be derived from Newton\'s 3rd law and his theory of gravitation as:

Centrifugal force of a circular orbit = gravitational force

mv^2/r=GMm/r^2

where v=2*pi*r/T hence:

4*Pi^2*r/T^2=GM/r^2

4*Pi^2*r^3=GMT^2 which is Kepler\'s Law, for a geostationary orbit r can be solved with T=1 kerbin day in seconds.

where r is the distance from the centre of kerbin (not altitude above sea level), G is the gravitational constant, M is the mass of Kerbin.

As you can see T and r have no relation with m the mass of the orbiting object.

I feel pretty awesome for actually understanding this stuff...

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You should I did a TL:DR because I have no maths skills. Although the orbit would be only a little affected by the orbiting object, mainly by way of tidal forces acting on the object. Yes?

PD

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You should I did a TL:DR because I have no maths skills. Although the orbit would be only a little affected by the orbiting object, mainly by way of tidal forces acting on the object. Yes?

PD

There are no tidal forces in KSP. Planetary bodies are on rails and have spheres of influence. See any number of discussions on patched conics. There are no Lagrange points either. ;)

Unfortunately the processing cost of a perfect simulation is too high for a video game. Patched conic\'s are 99% as good though, and allow for Neat Stuff in the future.

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