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SixtoReport: D12 Satellite Info in Post#1 - Live!

Discussion in 'DIRECTV General Discussion' started by Sixto, Jul 27, 2008.

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  1. Apr 24, 2010 #7161 of 10270
    James Long

    James Long Ready for Uplink! Staff Member Super Moderator DBSTalk Club

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    Thanks. I had the miles side right (with a difference in rounding by doing x/360*27 instead of doing 360/27=13.3 rounded). I cannot recreate how I messed up the km calculation.
     
  2. Apr 24, 2010 #7162 of 10270
    Lt Disher

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    I have two questions for you Sixto.

    1. I'm not a rocket scientist, so am not sure what you mean "at the outer edge". All of the figures on update 163 look similar to numbers from some earlier listings. I don't know how to interpret them though.

    2. I know you have alluded to this before, but since the ground control authorizations only go until May 7, should we be looking for an extension of those before the drift. In other words, from your past experience would they start the drift without authorizations for the entire thing. This assumes a 20 day drift which would not be done by May 7.

    Thanks in advance for any response you can give.
     
  3. Apr 24, 2010 #7163 of 10270
    P Smith

    P Smith Mr. FixAnything

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    D12 does keep steady in a 'box' at 76W.
     
  4. Apr 24, 2010 #7164 of 10270
    jacmyoung

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    You know what, I think I did make a mistake. The velocity I was talking about should be the absolute speed in space, along the path of the orbit, whatever that orbit is. In other words when the sat is "drifting" to a new location to the west, its orbit is raised but not its speed in space along the path of the orbit it is traveling. The higher the orbit, the longer the circumference, i.e. the distance the sat will have to travel to complete a whole cycle of the orbit, as such from a fixed location on the surface of the earth, the sat is moving west, until such time the new location is reached, the sat orbit is lowered to the geostationary orbit again, at that point, with the same velocity in space the sat is traveling along the path of the geostationary orbit, the sat again appears stationary from that fixed earth location.

    If they want to move the sat faster, they will have to shoot the sat to a much higher orbit, then bring it back from that much higher orbit later, which will consume more fuel.
     
  5. Apr 24, 2010 #7165 of 10270
    Sixto

    Sixto Well-Known Member

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    Yep, was just referring to the 76.05W (long), .05S (lat), .07 (inc). All at their highest values, which really doesn't mean much, but figured may be worth noting. Raising the orbit will be the key change.

    Relative to whether they file to extend the drift earth station communication past 5/7 before the drift starts, they may, but not sure it's required. Or maybe they're planning a quicker drift now.

    We'll see ...
     
  6. Apr 24, 2010 #7166 of 10270
    James Long

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    I'd rather think of concentric circles painted on a parking lot.

    Draw a circle with a 20ft radius and one with a 21ft radius. Place a video camera on a motorized turntable in the center with the ability to make one revolution every 84.0338 seconds. Now get an accurate speedometer on a wheel and walk around the 20ft radius circle at precisely 2 mph. Calibrate the exact rotation of the turntable so you remain in the center of the frame.

    Your walk around the circle is 125.66 ft. At 2 mph you should complete the circle every 84.0338 seconds (the same speed you calibrated the turntable to turn).

    While walking smoothly around the circle monitoring your speedometer for 2.000 mph drift out to the outside lane ... the 21ft circle. Continue to walk the exact 2.000 mph. Your new path is 131.95 ft (rounded). or 6.28 ft longer.

    Depending on the angle of your camera lens you'll be out of frame within a few of revolutions. You are still traveling the same speed (the exact 2 mph), a person on the ground may think you slowed down because you're falling behind the camera.

    There is the logic. What makes it different in space? Will a satellite traveling 11,066 km/hr (6,876 miles/hr) at a higher orbit spin out into space or fall to Earth because of some physical reason that it needs more speed to maintain any orbit (although not geostationary)? Or are we just using different points of reference for our relative speeds.

    (All things are relative.)
     
  7. Apr 24, 2010 #7167 of 10270
    Rob

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  8. Apr 24, 2010 #7168 of 10270
    jacmyoung

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    Good visual if he can visualize it:) Your explanation is the same as mine. One thing is certain, his rock tied on a string spun around a person, while much more easier to visualize, is a bad example. In space term, there is only one fixed length of the string which can allow the person to fix his eyes on the rock while spuning, but in his example, the person can increas or shorten his string at any length, still manage to fix his eyes on the rock, all he needs to do is increase and decrease the speed he turns himself.

    But our planet earth is stable, she does not change her turning speed, so we can survive.
     
  9. Apr 24, 2010 #7169 of 10270
    James Long

    James Long Ready for Uplink! Staff Member Super Moderator DBSTalk Club

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    I wouldn't say it is a bad example. It may contain the missing piece. Think about swinging the rock at a constant rate just slow enough that it remains in a horizontal plane. Then let another few inches out on the string. Does the rock remain in a horizontal plane or does it need to spin faster to maintain a plane?

    The speed the rock is moving is irrelevant to the center. We're only spinning the rock to make it move. Similar to running around in a perfect circle with a rock in our hands. It is just easier to spin a rock (and it introduces the idea of gravity pulling that rock to the center).
     
  10. Apr 24, 2010 #7170 of 10270
    P Smith

    P Smith Mr. FixAnything

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  11. Apr 24, 2010 #7171 of 10270
    Rob

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    Technicallity. :hurah:
     
  12. Apr 24, 2010 #7172 of 10270
    James Long

    James Long Ready for Uplink! Staff Member Super Moderator DBSTalk Club

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    Better Hubble than hobbled.
     
  13. Apr 24, 2010 #7173 of 10270
    LameLefty

    LameLefty I used to be a rocket scientist

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    The velocity through space is NOT constant if you change the altitude either way. To lower the orbit, you must decrease the velocity, to raise the orbit you must increase it, period. There IS no other explanation.
     
  14. Apr 24, 2010 #7174 of 10270
    jacmyoung

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    That was what I was saying, instead of as he said, having the rock "spun over your head", we need to think about having the rock spun on a level plane with you holding the string. To compare to our sat situation then, the question is not that if the string length is changed, whether the rock will drop down or up, but whether your eyes can still be fixed onto the rock while you spin with the rock. You can if you increase or decrease your speed of spin.

    But the earth can't, she does not change its speed of spin, so his example is bad.
     
  15. Apr 24, 2010 #7175 of 10270
    jacmyoung

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    That is because your velocity is a relative velocity, maybe relative to another sat in the geosationary orbit. But both sats are traveling at the same speed in space along the path of their own orbits. The one that is in the higher orbit will move to the west ("fall behind") from the view point of the sat on the lower orbit.
     
  16. Apr 24, 2010 #7176 of 10270
    LameLefty

    LameLefty I used to be a rocket scientist

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    Listen up: it's really very simple in concept.

    An object in orbit around a body will be traveling at a distance from the center of mass of that body based SOLELY upon the speed of the object and mass of the two bodies (but the mass of the satellite is negligible compared to the mass of the Earth and can be ignored here). Therefore, with the mass of the Earth essentially fixed, the ONLY thing that determines the altitude of the orbit is the velocity of the satellite.

    Now, as the altitude of the orbit increases, the period (that is, the length of time necessary for the object to complete one complete orbit) increases, because the length of the orbit (circumference, if the orbit is essentially circular) increases. At a certain altitude known as GSO, the period of the orbit matches the rotational rate of the Earth precisely (23 hrs 56 min 4 sec, more or less). A satellite at GSO will thus appear to remain fixed in space above the Earth, provided the orbit is coplanar with the equator (e.g., inclination at or very close to 0ยบ).

    Now, if one increases the velocity of that satellite, the altitude will also increase. The period will increase to something greater than the length of one day. The satellite will thus appear to drift westwards in the sky as the Earth rotates a bit faster than the period of the satellite. Conversely, if a satellite is slowed down a bit, the altitude will decrease as will the length of the orbital period, thus making the satellite appear to drift eastward relative to the surface of the Earth.
     
  17. Apr 24, 2010 #7177 of 10270
    wmb

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    No, not the same... the reason the rock drops it that gravity is acting perpendicular to the plane of the orbit of the rock. A satellite does not have similar significant external force acting on it. If you construct a free body diagram for a rock being swung on a string, the force that the string applies to the rock needs to balance the centripital force and offset gravity.

    Slowing the rock down will decrease the applied centripital force, and it will drop the rock to a lower height. The tangent of the angle of the string to the ground will be the ratio of the force of gravity to the applied centripital force.

    So, as we spin the rock faster, we are increasing the centripal force, reducing the ratio of the force of gravity (a constant) to the centripital force. This reduces the tangent of the angle between the string and the ground, which raises the height of the rock.

    So, whatever you do, the rock will move in a horizontal plane. It will just be a lower horizontal plane as you slow it down.
     
  18. Apr 24, 2010 #7178 of 10270
    wmb

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    Time to hopefully clear up a couple of things... Tryng to keep this more conceptual than quantitative...

    Angular velocity is the angle that the satellite moves during a period of times. So, since the orbit is circular, the angular velocity is degrees per time. In the case of a satellite at GSO, the angular velocity would be 2*pi radians (or 360 degrees) per sidereal day.

    A sidereal day is the amount of time it takes for the earth to revolve once relative to a fixed location, 23 hours, 56 minutes, 4.091 seconds, see wikipedia for more details... http://en.wikipedia.org/wiki/Sideral_day

    Now, to convert this to how much distance is covered per time... First, its not linear velocity since the satellite is moving along a curve. Velocity is a vector quantity, which means it tells both how fast and which way an object is moving. Speed would be a better term since speed is the magnitude of the velocity vector. Since the satellite is moving in a circle, its constantly changing direction, hence velocity. The change in direction is an acceleration caused by the force of gravity between earth and the satellite... thats why some say its falling... its constantly accelerating towards the center of the earth based on gravity, just like an object falling off a shelf.

    So, how much distance is covered per time... the length of the orbit per day. The length is the circumference of the circle that the orbit represents. In this case, the radius of the circle (distance from the center of the earth to the satellte) is 26,199 miles. So the distance covered each sidereal day is 2*pi*26,199 miles = 164,614 miles per day, or about 6859 miles per hour... a little over 110 miles per minute.

    As far as the relationship between the speed and altitude... there is a functional relationship between the two. If you want a specific velocity, such as one revolution per day, you can calculate the height you would need. Likewise, if you know how high you want to go, you can calculate the speed you need to go to get there.

    Last thing, Kepler's second law states that "The line joining a planet and the Sun sweeps out equal areas during equal intervals of time" http://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion#Second_law One way to think about this is that the orbit is a pie (pizza?). Over each equal time interval, the area of the piece of pizza pie that is cut out is the same. So, at perigee, when it is closer to earth, the angular velocity (number of degrees per time) is greater than at apogee, where it is furthest from earth. In other words, the piece of pie is short and wide when a satellite is closer to earth, and long and skinny when it is far from earth. The area of the wedge is the same in both cases.

    So, in order to move D12 west, you have to slow it down, which means making the orbit higher. But, since the velocity is being changed, the satellite is being accelerated, even though the acceleration is negative, speed decreased.
     
  19. Apr 24, 2010 #7179 of 10270
    LameLefty

    LameLefty I used to be a rocket scientist

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    You have to slow it down RELATIVE TO THE SURFACE OF THE EARTH, not in an absolute sense. The velocity of the orbit is actually increased; the orbital altitude increases; the orbital period increases; the satellite drifts westward.
     
  20. Apr 24, 2010 #7180 of 10270
    wmb

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    As long as we agree it goes up, I'm fine with whatever coordinate system you would like to use! I'm not a scientist, I'm an engineer...
     
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