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B Accelerating a satellite in a circular orbit

  1. Aug 4, 2017 #1
    Hello,

    I will be thankful if you could explain what appears to me as a paradox.

    We know that a satellite on a circular orbit, let say around the earth, has a uniform speed given by v=√(GM/r0).

    Now I would like to accelerate the satellite by keeping it on the same circular orbit r0. The only way to achieve this should be to "simulate" a higher gravitational field. As far as the Einstein's equivalence principle (EEP) is true, a mechanical acceleration is equivalent to a gravitational acceleration (http://www.einstein-online.info/spotlights/equivalence_principle), therefore we should be able to simulate a higher gravitational field with the thrust of the satellite's engine.

    Unfortunately the experiment, and the orbital mechanics, show that any mechanical thrust applied to the satellite, whatever its intensity and its direction, will transform the circular trajectory into an elliptic one. Therefore the mechanical acceleration is not equivalent to a gravitational acceleration, and then the EEP seems to be invalid.

    Note that the space-time box needed for this experiment might be as small as one meter large, as far as it includes the center of gravity of the satellite, and this preserves the locality necessary to apply the EEP.

    Can you help with this ?
     
  2. jcsd
  3. Aug 4, 2017 #2

    Orodruin

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    You are misunderstanding the equivalence principle. A satellite under the influence of gravity only is in free fall and feels no (proper) acceleration. A person aboard the satellite would feel no acceleration. You break this as soon as you start its thrusters. You cannot make a circular orbit at the same radius and a different speed without having an added thrust.
     
  4. Aug 4, 2017 #3
    Yes I do understand your point, that is logical, but I ask something else.
    What I want is to add a thrust to the satelite, with its engine, but keeping it on its circular orbit. As far as a mechanical acceleration should be equivalent to a gravitational acceleration (EEP), an engine thrust should be able to simulate a higher gravity, and then to keep the satellite on the same circular orbit but with a higher speed.
     
  5. Aug 4, 2017 #4

    Orodruin

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    Yes, this is possible, but an observer aboard the satellite would experience it very differently.

    Are you asking how to get it from the state where it is in the circular orbit due to gravity only to the one with a higher speed? This is trivial, you just have to provide the proper tangential acceleration to provide this additional speed at the same time as you provide the necessary additional radial acceleration. In other words, in the acceleration phase your thrust will not be directed radially nor tangentially.
     
  6. Aug 4, 2017 #5
    Sorry, I have to do right now. Il will continue this very interesting discussion in some hours, and demonstrate my point with the kinematics.
    Please apologize.
     
  7. Aug 4, 2017 #6

    Bandersnatch

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    It should be perhaps noted, that this is exactly what is routinely done in the final approach before docking, where the RCS thrusters are used in translational mode to compensate for orbital drift as the ship moves along a linear trajectory towards the station.
     
  8. Aug 4, 2017 #7

    Ibix

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    Here's the Newtonian maths: to stay in an orbit of radius ##r## while moving at speed ##v## you need a radial force of ##mv^2/r##. Gravity provides a force of ##GMm/r^2##. If your engines provide a thrust of ##T## in the (inward) radial direction then for a circular orbit $$\begin{eqnarray*}
    \frac{mv^2}r&=&\frac{GMm}{r^2}+T\\
    v&=&\sqrt\frac{GM}{r}\sqrt{1+\frac{r^2T}{GMm}}
    \end{eqnarray*}$$So you can have any velocity you like under power, but you're stuck with the one option if ##T=0##, as Trelat says. You can throw relativity into this mix if you want, but it seems rather unnecessary.
     
  9. Aug 4, 2017 #8

    Orodruin

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    To add on top of this, this gives you the general expression for the necessary centripetal thrust regardless of the tangential acceleration. The tangential acceleration is just given by the tangential force divided by the mass. In general, this will of course lead to an increase in the necessary centripetal thrust as the speed increases. The total thrust is a combination of the inward thrust and the accelerating thrust during the acceleration. Once you have reached your desired speed, you only have the radial thrust (as any tangential thrust would change the speed).
     
  10. Aug 4, 2017 #9

    PeterDonis

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    That's not what the EEP says. Mechanical acceleration can be measured with an accelerometer; objects accelerated this way feel weight. "Gravitational acceleration" cannot--objects being "accelerated" by gravity are in free fall, weightless. They are physically different states of motion, and the EEP does not say they are the same.

    What the EEP actually says is that, if you are under mechanical acceleration (i.e., feeling weight), you can't tell, from purely local measurements, whether you are feeling weight because you are accelerating in empty space, far away from all gravitating bodies, or because you are sitting at rest in the gravitational field of some body. The key is that the weight you feel is the same in both cases. (Or, alternatively, if you are in free fall, weightless, you can't tell, from purely local measurements, whether you are just floating in empty space, far from all gravitating bodies, or falling in the gravitational field of some body.)
     
  11. Aug 4, 2017 #10

    Bandersnatch

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    This is only true if the sole force acting as the centripetal force is gravity. Here, it isn't - you're adding thrust. So you can no longer use that equation.
    See Ibix's post #10.
     
  12. Aug 4, 2017 #11

    Nugatory

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    You know that there has to be some error in this demonstration, because if the conclusion were correct it would not possible to for a centrifuge to spin an object on a circular path, nor to twirl a weight on the end of a string along a circular path - and both of these things are known to be possible.
     
  13. Aug 4, 2017 #12
    Astronauts get on centrifugal machines for high G training. Is it a case you want? Space elevator might be another one. Best.
     
    Last edited: Aug 4, 2017
  14. Aug 4, 2017 #13
    ##v_R## is the incompressible velocity, the one you can not get rid of, because it is caused by the earth, around which the satellite is orbiting. Whatever you will do, this velocity will remain. So if you add an other velocity, for instance by using a thrust, you add it to ##v_R##. I called this additional velocity ##v_T##, but you could call it ##v_{thrust}## or anything you want. This is what is so interesting in the formulation of the keplerian velocity that I remind in my post, it makes a difference between the mandatory velocity and the others.
    Sorry but I do not know where the equations of Ibix are coming from, there are no demonstration nor references, so I lack information to be able to discuss them.
    and
    As far as I know the astronauts in the ISS are not feeling any centifugal force, because they are in free-fall. So there is the "usual" rotation, for instance here on the earth (but not only), and there is the gravitational rotation. The first one is associated to a centrifugal force, the second one is not, it is then of a different kind. This is a very interesting point indeed, but each day at a time, so let us first discuss this satellite problem.
     
  15. Aug 4, 2017 #14

    Bandersnatch

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    Those are just Newton's laws (2nd and gravity) and the expression for centripetal force.

    The problem is you're using your expression for ##v_R## without understanding where it came from. As a result you don't understand its limitations. It would be best if you took a step back, and tried deriving it yourself. Again, all you need is Newton's laws and the expression for centripetal force.
     
  16. Aug 4, 2017 #15

    PeterDonis

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    This way of looking at the problem seems tailor made to cause confusion. But perhaps a better way to locate the original confusion is to go back to this statement from your OP:

    That's not correct as you state it. A correct statement would be: if you want the satellite to have the same circular orbit r0 at a higher orbital velocity while remaining in free fall, then you need a higher gravitational field. But if you drop the requirement that the satellite needs to be in free fall, then there is nothing whatever preventing you from giving it the same circular orbit r0 at any orbital velocity you want (up to the velocity of light).

    Um, basic Newtonian mechanics? This thread should really be in the Classical Physics forum as there is nothing whatever requiring relativity in your formulation of the problem, except for your reference to the EEP (which, as I've already pointed out, you have an incorrect understanding of).
    [Mentors' note: this thread was started in the relativity forum, but has since been moved to Classical Physics, for exactly this reason]

    To be explicit: for an object traveling in a circular orbit with radius ##r##, the coordinate acceleration points radially inward and has magnitude ##v^2 / r##, where ##v## is the orbital velocity. This is true for any circular orbit whatsoever, free-fall or not. This can be found in any basic textbook on Newtonian mechanics.

    The above coordinate acceleration needs to be provided by some force or forces, by ##F = ma##. So there must be a force on the object pointing radially inward, with magnitude ##m v^2 / r##, where ##m## is the object's mass. There are two possible sources for such a force: gravity, which is always present and determined by the source (in this case planet Earth), and rocket thrust, which is under the experimenter's control. So schematically, we have ##m v^2 / r## equal to gravity plus thrust. All Ibix did was write that out explicitly, using the known formula for the force of gravity at radius ##r##.

    Now, given that, let's go back to your statement about "velocity". The ##v## that appears in the above formula should, if we do our definitions properly, be what you are calling ##v_R##--the "rotational velocity". It's the velocity of the object along the circular orbit at radius ##r##. Its magnitude will be constant, but its direction will be changing as the object goes around the orbit.

    However, you have confused yourself by adopting the definition ##v_R = \sqrt{GM / r}##, which, as should now be obvious from the above, amounts to assuming that there is no rocket thrust being applied radially, i.e., applied to holding the object in its current circular orbit. So you've convinced yourself that a circular orbit under thrust is impossible by adopting a definition that forbids you from even considering the possibility. If you drop this confusing definition and just look at things using basic Newtonian mechanics, as above, there's no problem.

    This is also a confusing way to look at it if your goal is to understand how a circular orbit under thrust is possible. Basically, in the terms you are using here, a circular orbit under thrust is a combination of "usual rotation" (due to a non-gravitational force being applied radially inward) and "gravitational rotation" (due to gravity). But your separation of those two things has apparently confused you into thinking that they can't be combined this way, which is wrong.

    It certainly is from the standpoint of GR; but I think that, as you say, you need to get clear about the Newtonian formulation of the problem first.
     
    Last edited by a moderator: Aug 4, 2017
  17. Aug 4, 2017 #16

    PeterDonis

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    Yes, but the whole point of a transfer orbit is that the thrust is applied tangentially, not radially. So your whole formulation is simply not adapted to the problem you are trying to solve in this thread (how a circular orbit under thrust is possible). It is adapted to a different problem (how to transfer between two circular orbits of different radii, by means of an elliptical orbit, using minimum thrust). The solution of that problem is basically to never apply thrust radially, because that just wastes energy. But that doesn't mean applying thrust radially, to allow a circular orbit under thrust, is impossible; it just means it's using more energy.
     
  18. Aug 4, 2017 #17
    You are right, I should have been more precise.
    The gravitational acceleration is ## a=GM/r^2 = v_R^2 /r## therefore ## v_R= \sqrt {GM \over r} ## [1]
    -------
    Reference :
    [1] https://en.wikipedia.org/wiki/Orbital_speed
     
  19. Aug 4, 2017 #18

    PeterDonis

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    You're still confusing yourself. If this is your definition of ##v_R##, then your claim that ##v_R## is also the "rotational velocity" (i.e., the velocity along the circular orbit) is not correct, unless you also assume that no thrust is directed radially. But assuming that no thrust is directed radially rules out the very possibility we are discussing in this thread--of having a circular orbit under thrust.
     
  20. Aug 4, 2017 #19
    Sorry, not at all. My point is : because it is impossible to accelerate an orbiter on the same circular orbit, there is a paradox with the EEP, as far as a rooky like me in GR can understand. I am surprized that the members did know that it is impossible to accelerate an orbiter on the same circular orbit, because it is a basic of orbital mechanics, so I issued a demonstration. But my aim still remains, and concerns the GR.
     
  21. Aug 4, 2017 #20

    Bandersnatch

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    This is correct if gravity is the only force.
    You need to apply Newton's 2nd. Specifically where it says that acceleration times mass equals the sum of all forces acting on it. What happens if there's an additional force acting in the same direction as gravity? What is the expression for ##v_R## then?
     
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