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    Mrjnicoll's Avatar
    Mrjnicoll Posts: 1, Reputation: 1
    New Member
     
    #1

    May 12, 2010, 07:14 PM
    At one revolution per day, how far away would a ball tied to a string stay in orbit?
    If I tied a ball to the end of an indistructible string and tethered the string to the earth, how long would the string have to be to keep the ball from falling to the earth?

    Would the mass of the string or the ball be a factor?

    Could I climb up the string?
    cdad's Avatar
    cdad Posts: 12,700, Reputation: 1438
    Internet Research Expert
     
    #2

    May 12, 2010, 07:31 PM

    Ref:

    Geostationary orbit - Wikipedia, the free encyclopedia
    ballengerb1's Avatar
    ballengerb1 Posts: 27,378, Reputation: 2280
    Home Repair & Remodeling Expert
     
    #3

    May 12, 2010, 07:33 PM

    What is the purpose of the string if the ball is in a low earth orbit? I sthis an actual question that's been presented to you?
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #4

    May 13, 2010, 06:55 AM

    Geostationary orbit is a good answer; however, it assumes that the string is weightless. If the string has mass (and hence weight), or if you tried to climb the string, the satellite would have to be a bit further away than a normal geostationary orbit.

    The math works like this: for a satelliite in circular orbit the outward force on it due to its centripetal acceleration is balanced by the inward force due to gravity:



    Where is the distance from the center of the earth to the satellite, and is the mass of the earth. So given that is set by the rotational speed of the earth, you can calculate , which is the radius of a geostationary orbit.

    Now, if you add the weight of the string or a person climbing on it, there's an additional force component that you have to include:



    This means the must be a bit larger to counteract the extra weight.

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