Time taken to move from perigee to apogee

[Orbi]

https://www.orbiter-forum.com/showthread.php?t=26682
So I was studying the above article by george7378 and in the 3rd point, he mentioned that time to go perigee to apogee is T/2. How is it T/2?
Sorry it is a lame question and thank you.

 

[Urwumpe]

T is the orbit period, that you see in Orbit MFD. The time it takes for one full orbit around Earth.

T/2 is half of that time.

Orbits are symmetric, the velocity you have going downwards to perigee at a distance to Earth is the same velocity you have going towards apogee at the same distance.

 

[Orbi]

Is every orbit symmetric, like is there an asymmetric orbit
also?

 

[Face]

Is every orbit symmetric, like is there an asymmetric orbit
also?

You can have orbits where there is no apogee, because their eccentricity is greater or equal to one. In this light, you could say that they are not strictly symmetric in the context of the question. Mathematically, though, I guess you can argue that going from infinity to perigee and back again is also a symmetry. :shrug:

 

[Urwumpe]

Is every orbit symmetric, like is there an asymmetric orbit
also?

Well, now we get into the difference between reality and theory:

In theory, for Kepler Orbits, those are ALWAYS perfectly symmetric.

But Keplers theory assumes a two-body problem (only one planet and a small spacecraft with MUCH less mass than the planet in the whole universe), while in the reality, it is not. We have other planets.

That is why Orbit MFD does not display you the true real Orbit, but the osculating orbit. That is a Kepler Orbit that fits to your current velocity and position vector. The real orbit is made of many small segments of osculating orbits chained together. You wobble around Earth.

In reality, no Orbit is perfectly symmetric because there are more objects in the universe. Especially if you get into the boundary of the gravity well, like near Lagrange points or the Weak Stability Boundary. And then you also have no point gravity source (the whole mass of Earth is in one infinitely small volume), but a non-spherical gravity field.

Now the big question is: Does it matter?

If you are close to Earth, deep in the gravity well, you can assume that anything about Kepler Orbits applies to you and all the other effects are just making tiny errors to your orbit. Its not perfect, but its very close to it.

Just try it: Make a highly eccentric orbit (Ecc approaching 1) and measure the velocities at a distance to Earth (R)

Even if you are approaching the moon and the moon starts to slowly change your orbit from your initial parameters: The actual differences to the velocities predicted by Kepler orbit theory are small.

Now, if you would get to the Earth moon L1, a region where you clearly have a three body problem, your orbit will sure not be anything like what Kepler predicted. It can be completely chaotic.

---------- Post added at 14:41 ---------- Previous post was at 14:40 ----------

You can have orbits where there is no apogee, because their eccentricity is greater or equal to one. In this light, you could say that they are not strictly symmetric in the context of the question. Mathematically, though, I guess you can argue that going from infinity to perigee and back again is also a symmetry. :shrug:

And gravity assist maneuvers also work by the idea, that your speed leaving the planet is the same as the speed that you entered its gravity well - it just points in a different direction.

 

[Orbi]

Well, now we get into the difference between reality and theory:

In theory, for Kepler Orbits, those are ALWAYS perfectly symmetric.

But Keplers theory assumes a two-body problem (only one planet and a small spacecraft with MUCH less mass than the planet in the whole universe), while in the reality, it is not. We have other planets.

That is why Orbit MFD does not display you the true real Orbit, but the osculating orbit. That is a Kepler Orbit that fits to your current velocity and position vector. The real orbit is made of many small segments of osculating orbits chained together. You wobble around Earth.

In reality, no Orbit is perfectly symmetric because there are more objects in the universe. Especially if you get into the boundary of the gravity well, like near Lagrange points or the Weak Stability Boundary. And then you also have no point gravity source (the whole mass of Earth is in one infinitely small volume), but a non-spherical gravity field.

Now the big question is: Does it matter?

If you are close to Earth, deep in the gravity well, you can assume that anything about Kepler Orbits applies to you and all the other effects are just making tiny errors to your orbit. Its not perfect, but its very close to it.

Just try it: Make a highly eccentric orbit (Ecc approaching 1) and measure the velocities at a distance to Earth (R)

Even if you are approaching the moon and the moon starts to slowly change your orbit from your initial parameters: The actual differences to the velocities predicted by Kepler orbit theory are small.

Now, if you would get to the Earth moon L1, a region where you clearly have a three body problem, your orbit will sure not be anything like what Kepler predicted. It can be completely chaotic.

---------- Post added at 14:41 ---------- Previous post was at 14:40 ----------



And gravity assist maneuvers also work by the idea, that your speed leaving the planet is the same as the speed that you entered its gravity well - it just points in a different direction.

Yeah but I think that change in gravitational field will decrease non uniformly where at some point Earth's gravitational field will be minimum and Moon's gravitational will increase gradually, so I guess the velocity change is significant. Please correct me if I'm wrong.

 

[Urwumpe]

Yeah but I think that change in gravitational field will decrease non uniformly where at some point Earth's gravitational field will be minimum and Moon's gravitational will increase gradually, so I guess the velocity change is significant. Please correct me if I'm wrong.

No, goes into the right direction. But anyway, in about 99% of the cases, assuming the orbit is symmetric will not give you large penalties.

Just look here:

Where the lines of constant gravity potential are circular, you will have the lowest errors / perturbations from the simplified model of a Kepler Orbit.

 

[Orbi]

Thank you all for helping. Now my doubt is fully resolved.
It's really nice to discuss with you all awesome people.:hail::thumbup:

 

[cristiapi]

Here’s a real orbit example:

Four plots are showed because 2 are from SGP4 and 2 are from my numerical integration (just to be reasonably sure that the plots are accurate).
The 2 vertical dashed lines mark the start of the orbits; we see 1 full orbit and 1 incomplete orbit.
We see several “strange” things: the speed is not correlated to the position; there are no well-defined minimum/maximum speed; the orbits are asymmetric.

 

[Orbi]

What is a SGP4? If speed is not correlated with position does that mean it doesn't depend upon radius vector?

 

[Urwumpe]

What is a SGP4? If speed is not correlated with position does that mean it doesn't depend upon radius vector?

Its a way to propagate orbits into the future.



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


Note that the errors of SGP4 per day (1-3 km) is just one order of magnitude lower than the span of the altitude plot.



Also note that this example is pretty extreme - it shows a very lightweight satellite with a big aerodynamic surface (Humanity star) in very low Earth orbit, where aerodynamic drag is already a major factor. Its an example of a satellite in constant low acceleration. It would look equally strange, if you plot a satellite constantly accelerating by an electrostatic thruster.

 

[cristiapi]

What is a SGP4?

It’s a very fast, low accuracy, simplified analytical method to propagate the state (position and velocity) of an Earth’s satellite to an arbitrary time in the past or in the future.

If speed is not correlated with position does that mean it doesn't depend upon radius vector?

It depends on the radius vector, but it also depends on the atmosphere velocity and other perturbations. As Urwumpe said, this example is pretty extreme, but now you know how much the real orbits could be different from the perfect Keplerian orbits.

---------- Post added at 19:50 ---------- Previous post was at 19:34 ----------

Note that the errors of SGP4 per day (1-3 km) is just one order of magnitude lower than the span of the altitude plot.

This matter is much more complex and OT in this thread, but consider that the graph shows the magnitude of the position (radius vector), not the position.
Extensive analysis on TLE+SGP4 (called self-consistency analysis) and on the more accurate state vectors published by Planet Labs for their satellites shows a much lower uncertainty on the magnitude of the radius vector.
1 km is just a random value for the true accuracy; the accuracy can be much bigger or much smaller.