Question about orbital mechanics

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Stewie_Griffin

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I'm not sure if this is the right place to post this but I have a question about orbital mechanics that I can't find the answer too.

I understand that the farther away from the thing you orbit the slower you move. I also understand the mechanics of the hohmann transfer orbit and that if you are in a perfectly circular orbit and you thrust then that position in the orbit will become the periapsis and the opposite position will become the apoapsis, hence putting you in an elliptical orbit.

My question is this: If you are in a highly elliptical orbit around something, will you still move at the same speed at the periapsis as you would if the orbit was perfectly circular and it was only as high as the periapsis of the highly elliptical orbit? I know that in the elliptical orbit you would move slower by the apoapsis than by the periapsis but if you took your speed at the periapsis of the elliptical orbit how would it compare to the circular orbit whose periapsis and apoapsis are both equal to the periapsis of the elliptical orbit?

If it would be the same then would that make it possible for something in LEO to dock with something in a highly elliptical orbit but whose periapsis was in LEO? would the object in the elliptical orbit take the object originally in LEO along for the ride?

Just to be clear I have only basic knowledge of orbital mechanics but I am very interested to get the answer to this question.

Thanks!
 
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kelvinzero

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I dont actually know what some of those words mean, but any two objects with the same position and velocity under the same forces would follow the same path. Therefore if two objects on different orbits are ever in the same location, they must have different velocities.
 
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samkent

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The two could meet at the same point in orbit at the same time but their velocities would not be the same. If the two were to magically dock at that point, the new orbit would be something between the original orbits. It would depend on the mass of each object.
 
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trailrider

Guest
"The two could meet at the same point in orbit at the same time but their velocities would not be the same. If the two were to magically dock at that point, the new orbit would be something between the original orbits. It would depend on the mass of each object."

Right! You get conservation of momentum, so that the
Mass(1) x Velocity(1) + Mass(2) x Velocity(2) = [Mass(1) + Mass(2)] x Velocity(Final) (Can't do subscripts on this "confuser"!) The final orbit would follow one of Kepler's Laws about conservation of angular momentum in orbit. (It's been a LOOONG time since I studied this stuff! :roll: )

I'm sure if I'm wrong about the equation, some nice soul out there will...jump all over me! And I'd welcome the correction!

Ad LEO! Ad Luna! Ad Ares! Ad Astra!
 
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aphh

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For a very brief moment the two objects would move at the same speed and could in theory dock. This is actually how Hubble was captured. That point, however, is not the perigee of the elliptical orbit, but before or after perigee.

Object with a elliptical orbit is in constant acceleration/decelaration. Object with circular orbit has constant speed (but not constant velocity). Hence at the right moment and at right location where the orbits intersect, for a brief moment the objects would move at the same speed (but not with same velocity) allowing capture and docking.

If objects are captured and docked, both orbits are affected. Ofcourse less the orbit with object with more mass (forexample Shuttle + Hubble).

To calculate the intersecting point with the same speed one needs vector math and integrals.
 
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centsworth_II

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Stewie_Griffin":3590fadt said:
...would that make it possible for something in LEO to dock with something in a highly elliptical orbit but whose periapsis was in LEO? would the object in the elliptical orbit take the object originally in LEO along for the ride?
I too know little about orbital mechanics, but here are my thoughts:

The more eccentric an orbit is, the farther away an object docked in LEO can be released, but the slower it will be going at release. So if it wants to get anywhere quickly after release, it will have to use energy to get up to speed. But what if the orbit is so eccentric that it intersects the Moon's orbit and the Moon is the destination? Would there be an energy advantage to docking in LEO and being carried up to the Moon. My guess is that the energy would eventually have to be paid back as periodic boosts to the station's orbit.
 
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aphh

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If the orbits are very dissimilar to each other, docking becomes a problem as the objects would be moving to different directions at intersect, even if their absolute speed was the same upon encounter.

In the case of Hubble and Shuttle the orbital parameters were never the same, but the arm handling the grabbing and docking of Hubble was flexible enough to handle the slight difference in velocities at intersect. Upon release of Hubble at the correct point of the orbit only slight push to other direction was needed to restore Hubble's orbit.

Going to moon can be achieved energy efficiently raising the apogee of the orbit gradually using ion drive. SMART-1 spent many months to travel from Earth orbit to Moon orbit. Not too much fuel was spent on that journey.
 
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MeteorWayne

Guest
In actuality, the Hubble was not pushed away at all. It was released, the arm was retracted, then the shuttle made a very small maneuver to back away.
 
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centsworth_II

Guest
aphh":1ghfucj1 said:
If the orbits are very dissimilar to each other, docking becomes a problem...
I'll say! Now that you mention it, even if both objects are traveling exactly 7km per second, it wouldn't take too much of an angle of intersection to make docking, as you say, problematic.

:eek:

Once again, no free lunches.
 
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scottb50

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centsworth_II":3a5imm1v said:
aphh":3a5imm1v said:
If the orbits are very dissimilar to each other, docking becomes a problem...
I'll say! Now that you mention it, even if both objects are traveling exactly 7km per second, it wouldn't take too much of an angle of intersection to make docking, as you say, problematic.

:eek:

Once again, no free lunches.
It would be no difference then meeting a car running a red light in an intersection, the velocities might be exactly the same, but they are aimed in different directions. If a car moves up beside you at 60 mph and matches your speed it could dock to your car, or one of the Duke brothers could climb out it's window into yours.
 
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CalliArcale

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To extend your analogy, because it's really cool:

If your car is going 60MPH and the General Lee pulls up alongside, going 60MPH right next to you, in the same direction, one of the Duke brothers can indeed transfer to your vehicle reasonably safely (though Don't Try This At Home). But if you're going 60MPH one way and they're going 60MPH the other way, you'll be next to one another for only a split second as you pass at a relative velocity of 120 MPH. (Jumping from vehicle to vehicle at this point would be suicidal, even for a movie stunt artist, because he'd hit the other car at 120 MPH.)

That's worst-case scenario -- head-on. But sideways isn't much better. Say you go through an intersection at 60MPH while the General Lee approaches at 60MPH on the cross street. Assuming you time it right so neither vehicle gets t-boned, the movie stunt artist has a split second to leap across into the other car, which is going 60MPH at a right-angle to the direction he's going before he starts his jump. This might actually be harder to achieve than the head-on one, but even if the timing is worked out exactly right, the stunt artist is now effectively getting broadsided at 60MPH when he hits your car.
 
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trailrider

Guest
Not to make light or be flippant, as I think you are getting the jist of your question answered, but I can't resist...

I think the crew of Atlantis proved to be the best "Orbital Mechanics" ever sent into orbit! :D

Seriously, about the question of the differences in velocity, one has to understand and clarify the definition of "velocity". Velocity has TWO components, speed and direction. The velocities of two rendezvousing objects in orbit depend on the relative speeds as well as the direction each is moving relative to each other. Orbits that are matched as to plane, have "only" to deal with relative speeds and altitudes. Rendezvous on two objects that are co-planer, but in orbits of different eccentricities are generally set up so as to match apoapsis (apogee when referring to Earth orbitis) by conduction burns at the appropriate points where the orbits intersect. Effectively, to "catch up" with a satellite in a higher orbit, you have to accelerate. But that raises your apogee and slows you down! So you have to stay in a lower orbit until you are ahead of the higher satellite, and then fire your rockets to accelerate and raise the orbit. You then sort of "back into" the higher satellite. This is somewhat of a simplification, but hopefully it gets you the idea.
 
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ThereIWas2

Guest
Back to the original question, is an object in an elliptical orbit that intersects a circular orbit tangentially moving at the same velocity at the point of intersection as an object in the circular orbit? The answer is no, it is not. If it was, it would not be in the elliptical orbit, but in the circular one.

At the apogee of the elliptical orbit an object is moving slower than an object in a circular orbit at the same distance from the planet, and at perigee it is moving faster than an object in a circular orbit at that lower altitude would be.

This is why a "circularizing burn" is required as the second step of a Hohman transfer.
 
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SpaceTas

Guest
At every point in the elliptical orbit the speed (just the size of the velocity not direction) is the same as for the circular orbit at that distance from the central object being orbited (actually center of mass of system). You can derive this by considering the force (or acceleration) of gravity and the force (or acceleration) required to create circular orbit (centripedal force) by constantly changing the velocity direction. Changing velocity even if it is only the direction is an acceleration and so requires a force.
 
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MeteorWayne

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I don't believe that's correct Tas.
If the orbital speed was the same for an elliptical and circular orbit, the orbit would be the same; it would have to be circular.
For example, it's it's near circular orbit, the earth does about 30 kps around the sun. A comet on an highly elliptical orbit is moving about 41 km/sec at 1 AU. It is this speed which allows it to continue it's elliptical orbit.

MW
 
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enzoenzo

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I think i need to get some physics lessons, this is way over my head.
 
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MeteorWayne

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LOL, believe me, these formula are not just floating around in my brain. Good reference books are essential :)
 
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aphh

Guest
Object on circular orbit has constant speed and constant velocity (=constant centripetal acceleration)

Object on elliptical orbit has constantly varying speed and constantly varying velocity.

At some point both objects would have the same speed, but still slightly different velocity (different direction).

According to Kepler, a line sweeps out an equal area of the orbit at equal period of time. Hence on a elliptical orbit the object must be constantly accelerating/decelerating (besides the apogee and perigee points, where for a short while the speed is constant). At some point of the elliptical orbit the speed matches the speed of an object on a circular orbit.
 
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aphh

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MeteorWayne":2w1vg6zn said:
For example, it's it's near circular orbit, the earth does about 30 kps around the sun. A comet on an highly elliptical orbit is moving about 41 km/sec at 1 AU. It is this speed which allows it to continue it's elliptical orbit.
Yes, it is that speed that swings the comet past sun back towards a higher apogee. Think it as a half a loop on a roller coaster, the higher and thus faster you enter the loop, the higher you would get slinged back up.

Enter the loop too fast and you would escape the loop system altogether and not land back on the tracks again ;)
 
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shuttle_guy

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aphh said:
For a very brief moment the two objects would move at the same speed.......
No that is not correct. The two objects MUST be in the same orbit to match velocities. Inorder to dock the 2 spacecraft must adjust their orbits to get into the same orbit. When the Shuttle Orbiter docked with Hubble it slowly adjusted it's orbit to match the orbit of the Hubble.
 
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shuttle_guy

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aphh":3eigep9s said:
.....In the case of Hubble and Shuttle the orbital parameters were never the same, but the arm handling the grabbing and docking of Hubble was flexible enough to handle the slight difference in velocities at intersect. Upon release of Hubble at the correct point of the orbit only slight push to other direction was needed to restore Hubble's orbit........
No that is not correct. The two objects MUST be in the same orbit to match velocities. Inorder to dock the 2 spacecraft must adjust their orbits to get into the same orbit. When the Shuttle Orbiter arrived to "fly" along in the same orbit as Hubble the Shuttle Orbiter had slowly adjusted it's orbit to match the orbit of the Hubble. The RMS "arm" can NOT grapple anything if the Orbiter and the object to be grappled are not at the exact same velocities. The arm can not take the the loads otherwise.
 
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shuttle_guy

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MeteorWayne":ku1p2ye4 said:
In actuality, the Hubble was not pushed away at all. It was released, the arm was retracted, then the shuttle made a very small maneuver to back away.

Exactly...the "arm" (RMS) structure can not take any significant loads except VERY small movements when it is attached to an object. It is NEVER used to push an object away.
 
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scottb50

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shuttle_guy":1y76tiwd said:
aphh":1y76tiwd said:
For a very brief moment the two objects would move at the same speed.......
No that is not correct. The two objects MUST be in the same orbit to match velocities. Inorder to dock the 2 spacecraft must adjust their orbits to get into the same orbit. When the Shuttle Orbiter docked with Hubble it slowly adjusted it's orbit to match the orbit of the Hubble.
If the result of their impact kept them together they would move at the same speed. Except it would not be in the same orbit as either was to begin with. It would be like a full motion pool table, the cue ball hitting the other ball changes the path of both balls. If the path and velocity is identical, or at least close, they may end up together.
 
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shuttle_guy

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MeteorWayne":25amn1ns said:
Hey s_g. Good to see you again!!!!!!!!!!!!

Hello MW. I will get back here when i can make time.

One thing I neglected to mention earlier. The poster that said that 2 objects in different orbits could dock when their velocity was almost the same was forgetting that velocity is speed and direction. The direction (velocity vector) of objects in different orbits are not pointing in the same direction thus the speed may be the same however their velocity is not the same.
 
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