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Releasing the object from Space Elevator

  1. Jul 1, 2011 #1
    Let’s imagine that the Space Elevator is built and we ascend to space with its cabin. The cabin stops temporarily at the altitude of 100 km (or a bit more) and then we release some object (sphere, hemisphere) downwards, the question is: what trajectory will it (object) follow when falling down?
    In Wikipedia is written that:
    “As the car climbs, the elevator takes on a 1 degree lean, due to the top of the elevator traveling faster than the bottom around the Earth (Coriolis force). This diagram is not to scale” (under the picture).
    http://en.wikipedia.org/wiki/Space_Elevator#Climbers
    So, when the cabin ascends the cable will be bent due to Coriolis force as shown on the picture.
    And I would like to know how exactly the object will fall that we will release from the cabin? Will it fall vertically (straight line) and hence continuously recede from east-going cable or will it fall in a curved line due to the same Coriolis force? I have read that if you release the object (never mind from Space Elevator or other) it will fall exactly vertically on the poles but not on the equator (the Space Elevator will be built on the equator). So, which trajectory will the object “choose”? There are two options:
    1. Object falls vertically (red spheres)
    2. Object fall along the cable (green spheres) and therefore the distance between cable and object remains the same (let’s forget about atmosphere’s resistance for a while).
    w6uly0.jpg
    I want to find the solution for the following problem: how to release some object from Space Elevator so that it not to cross/cut Space Elevator’s cable during falling down :smile:
     
  2. jcsd
  3. Jul 3, 2011 #2

    Drakkith

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    Staff: Mentor

    You should throw this into the Classical or General physics forums. I bet you'd get more help there.
     
  4. Jul 4, 2011 #3
    From a reference frame attached to the surface of the earth (i.e. a non-inertial rotating reference frame), the object will not fall 'straight' down, but will instead appear to curve due to the coriolis and centrifugal forces. Depending on the nature of the object (e.g. density) the effects of wind could be highly non-negligible.

    To avoid having the falling object hit the tether, one would simply drop the object on the appropriate side so that it curves away.
     
  5. Jul 4, 2011 #4
    Even though there is a 1 degree lean, I think the object will have a linear velocity greater than 25000 miles per 24 hours.
    The cable anchor at earth's surface will have a velocity of 25000 miles per 24 hours.
    Therefore, I think the object will strike the suface to the east of the cable anchor.
    I think one would arrive at the same conclusion when considering the problem from the standpoint that the earth is not flat and using angular momentum, energy, etc.
     
  6. Jul 7, 2011 #5
    zhermes
    So, as I understand from your post and my little knowledge from celestial mechanics the objects released from altitude below Geostationary orbit will fall downwards and forward (to East), at Geostationary orbit-stay there motionlessly, above Geostationary orbit-will go upwards and backwards (to West), right? :uhh:
    http://img39.imageshack.us/img39/3576/graphic1oa.jpg [Broken]
    That is the object should be pushed northwards or southwards?

    Fun Value
    By the way I would like to clarify this issue: when the cabin ascends along the the elevator takes on a 1 degree lean. Does this angle changes when the cabin ascends/descends? :smile:
     
    Last edited by a moderator: May 5, 2017
  7. Jul 19, 2011 #6
    Roger, Eagle9. I have received and copy your transmission sent on July 7.
    I think I am trying to catch up to you on this. Here is the way I see it so far, and I am trying to see it in terms of linear velocity.
    At 62000 miles the counterweight is stationary above a point on the earth's surface. Its angular velocity is 2pi radians per 24 hours which translates to a linear velocity of about 389360 miles per 24 hours or about 16,233 miles per hour (ignoring sig.figs). This velocity, by the way, is greater than the normal free satelite at geosynchronous orbit having a 26000 mile radius.
    As the elevator ascends I was thinking the counterweight would be pulled down toward earth - pulled down to an orbit where the tethered geostationary velocity is a bit less than 16,233 mph. Yet the counterweight might not lose any linear velocity (well, no loss if rocket engines could keep it going at the same speed - which I know is not part of the plan). Anyway, not counting loses to the ascending cabin, the counterweight would take up a lower orbiting altitude and somehow not lose linear velocity (or angular momentum). For a short time, it would try to keep going 16,233 mph, but at a lower altitude. For a short adjustment time it would go faster than the geosynchronous linear velocity required for the lower altitude. Due to this greater linear velocity or greater angular velocity, I thought it would move ahead (east) of its normal resting (unloaded) geostationary point above the earth.
    To answer your question, I think the angle will change as the cabin ascends and gains linear velocity. But I think rocket engines are needed either at the counterweight or on the cabin to push the cabin from a linear velocity of about 25000 miles / 24 hours or about 1000 mph to about 16000 miles per hour - in order to recover the angular loss or lean.
    Finally, as I said on the other space elevator thread, even though cabin or counterweight rockets seem to defeat the purpose or intent of the space elevator, I think we need as many ways to access the opportunities of space travel as we can find.
     
    Last edited: Jul 19, 2011
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