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Celestial Mechanics

  1. Dec 30, 2003 #1
    How do I find out the minimum speed required for a satellite to place it in a geosynchronous orbit(24 h per orbit)?
     
  2. jcsd
  3. Dec 30, 2003 #2

    enigma

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    Find the acceleration by setting the acceleration of gravity equal to the rotational velocity and distance needed to match that acceleration.

    [tex]

    \frac{\mu}{r^2}=\omega^2*r

    [/tex]

    Where mu is the gravitational parameter: G*M or 398600 km^3/sec^2

    and omega is the angular rotation of the Earth or 7.29*10^-5 rad/sec

    That will give you the orbital distance.

    From there you use the Vis-Viva equation

    [tex]

    V=\sqrt{\mu*(\frac{2}{r}-\frac{1}{a})}

    [/tex]

    with both the distance, r, and the semi-major axis, a, as the value you just determined. That will give you the velocity in km/sec.
     
  4. Dec 30, 2003 #3
    Thanks for the reply but our class didnt do angular rotation or anything similiar to that, so I dont quite understand this answer
     
  5. Dec 30, 2003 #4

    NateTG

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    What equations do you have, that you think apply?
     
  6. Dec 30, 2003 #5
    I know the height of the satellites orbit. I also know the period. This object is also bound to Earth. So the binding energy of the satellite is equal to the kinetic energy of the object right? So I think I should equate the the kinetic energy and binding energy and solve for v, but I dont get the right answer
     
  7. Dec 30, 2003 #6

    NateTG

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    Then you can calculate the speed that the satelite is traveling at. The orbit is circular.
     
  8. Dec 30, 2003 #7
    Yes but how can I calculate the speed at which the satellite was launched at to reach this height?
     
  9. Dec 30, 2003 #8

    NateTG

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    It's impossible to launch a balistic projectile into Geosynchronous orbit. Getting into orbit requires accelerating in midair.
     
  10. Dec 30, 2003 #9
    The books answer is 11.2 km/s which is the escape speed. Im guessing that it was probably required to find the speed at which the satellite should be launched to just escape the Earths gravitational field. Then after that the satellite can be steered to its geosynchronous orbit. Do you think so?
     
  11. Dec 30, 2003 #10

    NateTG

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    No. The satelite won't need to get going that fast if you get to escape velocity, you'd have to slow down to get into orbit. Are you sure the question was about geosynchronous orbit, and not about escape velocity?
     
  12. Dec 31, 2003 #11

    enigma

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    Every satellite orbiting the Earth is still in the Earth's gravitational field. It's just part of the definition.

    If that is the book's answer, then the book is wrong (assuming we all understand the problem correctly).

    Can you list the exact wording of the problem?
     
  13. Dec 31, 2003 #12
    The problem is listed like so:

    Consider a geosynchronous satellite with an orbital period of 24h.

    a)What is the satellite's speed in orbit?
    b)What speed must the satellite reach during launch to attain the geosynchronous orbit?(Assume all fuel is burned in a short period and neglect air resistance)


    The answers are :

    a)3.1 x 10^3 m/s
    b)1.1 x 10^4 m/s (Which I assumed to be 11200 m/s)
     
  14. Dec 31, 2003 #13

    NateTG

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    1.1 x 104 m/s is not even in the right neighborhood.

    Let's say that a projectile is launched from a gun at 1.1x104 m/s, and let's approximate that the acceleration of gravity is constant. Since the acceleration of gravity is decreasing as the projectile moves away from earth, this will give us a lower bound on the velocity of the projectile.

    Now, the distance between the surface of the earth and Geosynchronous orbit is approximately 3.5x104 km. If the projectile is launched straight up with a velocyty of 1.1x104 m/s second, then it will reach 3.5x104 m in altitude in about 3.2 seconds, and be traveling at about 1.1x104 m/s relative to the Earth which is clearly a good bit faster than the orbital velocity of the satelite.

    That said, an arbitrarily large number could be justified for (b) because "Assume all fuel is burned in a short period and neglect air resistance" is a pretty lousy assumption.

    If you think about it for a moment, you can see that the rocket booster will need to burn fuel when it reaches geosynchrnous orbit, since it needs radial velocity to get to the orbit, but it has zero radial velocity while it is orbiting. Clearly, the rocket booster also needs to burn fuel at launch.

    So, if we combine that with the assumption that the problem gives, we can see that the rocket would have to reach geosynchronous orbit 'in a short period', and considering that the minimum distance from launch to orbit is roughly 3.5 x 104 meters, you can justify just about any speed that you like.
     
    Last edited: Dec 31, 2003
  15. Dec 31, 2003 #14
    I believe you are mistake a bit. The height of a geosynchronous orbit is about 35000 km. Thus it would take approximately 3125 seconds to get there at a speed of 11.2 km/s, isnt it?
     
  16. Dec 31, 2003 #15

    enigma

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    The book's answer to the first problem is OK.

    The book's answer to the second problem is dead wrong.

    Changing the semimajor axis of an orbit (making it orbit at a different altitude) is physically impossible to do with a single burn. You need at least two.

    The lowest energy method of changing semi-major axis is called a Hohmann Transfer. It involves two burns. The first puts the satellite into a transfer orbit to get it out to the right semimajor axis. Once it gets there, gravity has slowed it down significantly. It is going too slow to maintain that orbit. If you don't do a second burn, you'll continue on the eccentric orbit (B) shown in the figure which I linked to. The second burn is necessary to speed up to the necessary orbital velocity to maintain the geo-synchronous orbit.

    No, the satellite slows down the higher it goes. It's the same thing which happens if you coast up a hill in a car or bike. The reason escape velocity is escape velocity is because it reaches '0' velocity at an infinite distance from the Earth (all theoretical of course. In reality other forces come into play). In other words, it doesn't come back. Anything slower than escape velocity will eventually return to the starting point. Anything faster than escape velocity will have a positive velocity at infinite distance.

    Everything make sense? It's a tricky subject...


    PS Nate, you missed the *10^3 for km->m
     
  17. Dec 31, 2003 #16
    So if I were to fire some projectile at 11.2 km/s , it will continue through space forever assuming there are no other masses in space?
     
  18. Dec 31, 2003 #17

    enigma

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    If you fire it from the surface of the Earth, yes.

    It all comes from the Vis viva equation I gave earlier:

    [tex]

    V=\sqrt{\mu*(\frac{2}{r}-\frac{1}{a})}

    [/tex]

    the 'a' is the semi-major axis (which I've mentioned a few times in this thread). If you look at an ellipse, the major axis is the longest line you can draw within the ellipse. The semimajor axis is half that. It is the "longest radius" for ellipses.

    If you've got a circle, a=r.

    If you plug in a=r, and r=6378km (the radius of the Earth), you'll find the velocity necessary to be in orbit at 0 altitude. If you then plug in infinity for a (the 1/a part goes to zero), you'll have the escape velocity from the surface of the Earth.

    Now, you'll see if you play around with the numbers that if you're already in orbit at a certain altitude, the escape velocity is lower. That's because the Earth's gravity affects it less if the sat. is higher in the gravity well.

    It is probably the single most useful equation for celestial mechanics.
     
    Last edited: Dec 31, 2003
  19. Dec 31, 2003 #18

    Janus

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    The problem here is that answers a and b are for two different orbits.

    a) is for a circular orbit and b) is not.

    An orbit need not be circular to be geosynchronous. It only needs to be circular to be a geostationary orbit. (And not all 24 hr, circular orbits are geostationary either. )

    The orbit for question b is for an eliptical orbit that has a perigee at the surface of the Earth and an apogee of 77757 km. This gives the orbit a semi-major axis of 42,000 km, which gives it a period of 24 hrs. The satellite will return to the surface of the Earth at the same time as the launch point.

    The velocity at perigee of such an orbit is 1.068 x 104 m/s, which falls right in line with the 1.1 x 104 m/s given as the answer.

    As a side note, it is interesting to examine what such an orbit would look like to someone standing near the launch point:

    The satellite would whiz by at 10.6 klicks Eastward. As it travels East it will climb and slow. Eventually it will begin to move Westward, still climbing as it does. It will pass overhead at an altitude of 71379 km, still moving Westward. It will start to lose altitude and slow untill it stops its Westward movement and starts moving East again, gaining speed as it does so and still losing alititude until it whizzes by again.
     
  20. Dec 31, 2003 #19
    But how would you go about calculating the perigee and apogee?
     
  21. Dec 31, 2003 #20

    enigma

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    Ah. You caught it, Janus. It's still a sloppy question IMO, though...

    Pirsq,

    The semimajor axis is the same for a set orbital period. You know the perigee is at the surface of the Earth, so apogee will be ~42,000km * 2 away from perigee (through the earth, through the center of the orbital ellipse, to the other side of the ellipse). Subtracting the diameter of the Earth gives Janus's apogee value.
     
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