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Minimum Satellite Period problem

  1. Sep 18, 2012 #1
    Clever guy that I am, I am writing a small paper on Celestial Mechanics...simplified...really really simplified. However, if I bore a tunnel through the Earth and drop a watermelon through it, the watermelon returns to my hand in 76 minutes. This is an example of an elliptical orbit of e=1

    But if I use the same figures and shoot the watermelon around the Earth in a grazing orbit, with eccentricity of zero, the orbital period is 6 minutes longer.


    BTW: I suspect any minimum energy subway tunnel of hypocycloidal shape has 1/2 the transit time of the watermelon, or 38 minutes, or 76 minutes round trip. A hypocycloid where the generating circle diameter is the same as the base circle radius generates a straight line through the center.

  2. jcsd
  3. Sep 18, 2012 #2
    How did you get these figures? The formula for the period should be the same for both cases. And the value around 84 minutes.
  4. Sep 18, 2012 #3
    Mean Earth Radius=6.37 x10E6 meters
    a=9.8 m/sec/sec (-)s=1/2at^2
    solve for t: t=19.00 minutes. (time for round trip...76 minutes)
    So what happened to the 8 minutes....(or the 6)?

  5. Sep 18, 2012 #4
    I'm only guessing here, but is it possible that the 84 minutes minimum orbit time includes the Earth's rotation? I.E. Cape Canaveral has moved eastward by 8 minutes of rotation during the orbit time? If you shot up a satellite (to the east to gain velocity), then it would be 76 minutes, +8 minutes (e.g.) before it was overhead.

  6. Sep 18, 2012 #5
    This does not make sense.
    The motion through the tunnel is not uniform accelerated. The motion in the tunnel is harmonic oscillation. Write the force acting on the falling body as a function of distance from the center, to see this. Then you can use the equation from that motion.
    And for the orbital motion you should use the equations for circular motion.
    In the end you'll get the same formula for both periods.
    Good luck.
  7. Sep 18, 2012 #6
    >>The motion through the tunnel is not uniform accelerated.

    Isaac Newton said it was.

    >>Write the force acting on the falling body as a function of distance from the center, to see this.

    Okay F=ma. Gravity is considered to be simply a point in the center of the sphere.

    >>And for the orbital motion you should use the equations for circular motion.

    The circular motion or the elliptical motion have the same formula, and the same orbital time

  8. Sep 18, 2012 #7


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    Staff: Mentor

    Careful - that point source rule is only valid above the surface of the earth.

    When the falling object is below the surface of the earth (requires a well or tunnel or some such) at a distance R < RE where RE is the radius of the earth, then all of the earth that lies between R and RE forms a spherical shell with the object inside it. The gravitational field in the interior of such a shell is zero. So when you're calculating the force on the object, it's as if the mass concentrated at the point in the center of the sphere is only the mass between radius 0 and R.
  9. Sep 18, 2012 #8
    Thanks. The bubble of that thought came to the surface an hour ago.
  10. Sep 18, 2012 #9
    He did not. Maybe you misunderstood him.

    Yes, but F is not a constant for the motion through the tunnel.

    So why don't you use it then and get the correct value for the period?
    I wrote you that the motion through the tunnel and around the circular path have the same period. Unfortunately, none of them is 76 minutes.
    I don't know what formula are you using.

    Albert Einstein is widely quoted as saying:
    "Everything should be made as simple as possible, but not simpler." (or variants of the same idea)
    You may have simplified a little too much.:biggrin:
    Last edited: Sep 18, 2012
  11. Sep 19, 2012 #10


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    2017 Award

    Staff: Mentor

    With the approximation of a uniform density of earth, the time would be the same.
    However, real earth has a variable density - the acceleration inside is higher than a linear formula would give, and objects falling through earth would be a bit quicker than objects in an orbit.

    No - with the approximation of a uniform density of earth, every straight line has the same transit time (about 42 minutes). Other shapes (especially the shapes for the shortest transit time) can have shorter transit times.
  12. Sep 19, 2012 #11
    Thanks all for making me smarter.

  13. Sep 19, 2012 #12
    probably not very scientific here but, if you consider that gravity's force is centered on the center of the Earth, then the closer to center the faster you accelerate. you would have to accelerate fast enough to overcome the force that you obtained in that process. But when you start to reach the other side (away from the source) you are gravitating away from the source, which would mean that whatever energy you acquired, you will now be using to produce an "escape velocity". So, the equation from the entrance side, should measure enough energy to balance the exit side. Should equal zero, unless an outside force acts on the object. Unless the force of gravity is different depending on where you are on the Earth. Or altitude, depth, things like that. I'm just a student in this at this time, just a thought.
  14. Sep 19, 2012 #13
    Not if you are inside the Earth. For homogenous Earth the acceleration will decrease as you approach the center.
    Even for actual (non-homogenous) Earth, as you go deep enough the acceleration deceases.
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