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Thread: gravitation

  1. #1
    Senior Member
    Jan 2009


    Given that radius of orbit of earth and Mars is 1.5\times10^{11}m and 2.3\times10^11m respectively . [1 year= 3.2\times10^7 s , radius of sun = 7.0\times10^8 , mass of helium atom= 6.6\times10^{-27} kg ]


    (1) Period of revolution of Mars

    I found this to be 1.9 year .

    (2) Mass of sun

    i am not sure bout this

    (3) Escape velocity of helium atom from the surface of the sun

    i tried using this formula


    Where M is the mass of sun and r is the radius of sun and of course G is the gravitation constant .
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  2. #2
    Sep 2009
    For (2), you know the force on Mars by the sun is given by
    where the second equality is using the fact that mars is experiencing centripetal acceleration around the sun. From there you can use the fact that
    v=d/t=2\pi R/T
    where T is the period and R is the radius of Mars' orbit.

    Plugging that expression for v into the top equation gives you an equation where you know everything except M, the mass of the sun.
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