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Theres something about gravitational

  1. Jan 24, 2004 #1
    With only two lectures in my back pocket, I still can't find a reasonable solution for this problem.

    Suppose that a planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average orbit radius of Mercury. The orbital period of Mercury is 88.0 days.

    What would be the orbital period of such a planet?

    and also, in general:

    Four identical masses of mass 500 kg each are placed at the corners of a square whose side lengths are 15. cm.

    What is the magnitude of the net gravitational force on one of the masses, due to the other three?
     
  2. jcsd
  3. Jan 24, 2004 #2

    jamesrc

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    Gold Member

    First one:
    You can use Kepler's Third Law (or equate the centripetal force with the gravitational attractio if you'd rather):

    [tex] \frac{T_1^2}{R_1^3} = \frac{T_2^2}{R_2^3} [/tex]

    With T1 = 88 days and R2 = 2*R1/3 to find the period of the mysterious new planet.

    Second one:

    Just do a little vector addition here. You know that the gravitational attraction between two masses separated by a distance r is given by

    [tex] \vec{F_{12}} = -\frac{{\rm G}m_1 m_2}{{\vec{|r_{12}|}}^3}\vec{r_{12}} [/tex]

    Just so we have our notation straight, F12 indicates the force on 1 due to 2, while r12 indicates the position vector from 1 to 2.

    Just sum the forces (good old law of superposition). Take the top left mass as an example: you'll have a force pointing down due to the bottom left mass, a force pointing to the right due to the top right mass, and a force pointing diagonally toward the bottom right mass due to that mass. The first two forces will have the same magnitude, while the third force will have a magnitude of half of either of the first two forces.

    Hint: the net force on each mass in this configuration should be directed radially inward to the center of the square the 4 masses make.
     
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