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Thread: gravitation

  1. #1
    Senior Member
    Jan 2009


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

    $\displaystyle v=\sqrt{\frac{2GM}{r}}$

    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
    $\displaystyle F=GmM/R=mv^2/2$
    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
    $\displaystyle v=d/t=2\pi R/T$
    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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