How do I calculate orbital height from orbital velocity?

AI Thread Summary
The discussion centers on calculating orbital height from orbital velocity in the context of the game Kerbal Space Program. Users explain that for circular orbits, the relationship between centripetal and gravitational forces can be used to derive the necessary velocity. The conversation highlights the complexity of delta V calculations, particularly when atmospheric drag is involved. There is confusion regarding the formulas for potential and kinetic energy, with clarifications provided about the correct relationships and constants involved. Ultimately, the original question about calculating orbital height has been resolved.
Bluelogic32
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I'm sorry if this isn't astrophysics but I didn't know what else to post on.

So I play this game called Kerbal Space Program, and I've recently been trying to calculate the delta v needed to launch my rockets and spaceships from all the planets and moons in the game. I learned how to calculate the orbital velocity needed to maintain a stable orbit around a body, and how much delta v I need to add my vehicle to account for atmospheric and gravitational drag. I could just use those equations for bodies with atmospheres because I know that scale heights of the atmospheres, so I can guess at what height my orbit will need to be. However for planets without atmospheres, I can't guess at what the height will be for the orbital velocity needed. So my question is how do I calculate the orbital height from my velocity? I should also mention that this is for a circular orbit.

As I mentioned above, I'm not sure if this is astrophysics, if it isn't could you please tell me what it is?

Thanks for your help :)
 
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Bluelogic32 said:
and how much delta v I need to add my vehicle to account for atmospheric and gravitational drag
That is not easy, and will depend on the launch profile.

I don't understand why you can calculate something with an atmospere, but not in the much easier case of no atmosphere.

For a circular orbit, the velocity is given by the equality of centripetal force mv^2/r and gravitational force mMG/r^2.
deltaV calculations are a bit more tricky, as they involve an elliptical orbit in between. I'm sure there is some online calculator or some page with formulas in the internet.
 
mfb said:
That is not easy, and will depend on the launch profile.

I don't understand why you can calculate something with an atmospere, but not in the much easier case of no atmosphere.

For a circular orbit, the velocity is given by the equality of centripetal force mv^2/r and gravitational force mMG/r^2.
deltaV calculations are a bit more tricky, as they involve an elliptical orbit in between. I'm sure there is some online calculator or some page with formulas in the internet.
I misunderstood what I was asking, I've got what I was looking for now though. Thank you for your answer.
 
To Calculate Circular Orbital Velocity, let P.E. at surface = K.E. at orbit. Thus mgh = 1/2 mvv. You see here that mass is not relevant! Thus v = square root of 2gh = approx. 17,500 mph. At this speed centripetal force = centrifugal force!
Note: H = distance from center of Earth to orbit height! or about 4200 miles for a 200 mile orbit. (Since gravity at surface is due to total Earth mass concentrated at a single point of ref, e.g.,the center of Earth's mass)

For moon just adjust H accordingly.

Isn't this less complicated to get the same answer?
 
Boemmd said:
P.E. at surface = K.E. at orbit
What does the surface have to do with the orbit?
 
Boemmd said:
To Calculate Circular Orbital Velocity, let P.E. at surface = K.E. at orbit. Thus mgh = 1/2 mvv.

This doesn't look quit right, but more like the orbital velocity at zero altitude above surface.
 
stedwards said:
This doesn't look quit right, but more like the orbital velocity at zero altitude above surface.
Not even that (the given number looks approximately right but the formula would give sqrt(2) times this value, which corresponds to the escape velocity).

- the potential energy is not mgh in any reasonable system. It is -mgR2/r where R is the radius of Earth (better: the distance where g is evaluated) and r is the distance to the center of Earth. Note: g is not constant. A better formula is -mMG/r with the gravitational constant G and the mass of Earth M. It gets rid of the detour via the radius and g at the surface.
- the necessary kinetic energy for an orbit is not equal to the potential energy (even ignoring the opposite sign) - that would give an escape trajectory. It is equal to minus one half this value.

I closed this thread, the original question has been answered in October 2014.
 
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