Results 1 to 2 of 2

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 .
    Follow Math Help Forum on Facebook and Google+

  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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. gravitation question
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: Mar 23rd 2011, 04:31 PM
  2. [SOLVED] Gravitation?
    Posted in the Advanced Applied Math Forum
    Replies: 4
    Last Post: May 20th 2009, 03:24 AM
  3. A problem os Newton´s law of gravitation
    Posted in the Calculus Forum
    Replies: 0
    Last Post: Feb 11th 2009, 07:30 PM
  4. Newton's Law of Gravitation
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: Feb 6th 2008, 03:45 AM
  5. Replies: 4
    Last Post: Oct 31st 2006, 10:49 AM

Search Tags

/mathhelpforum @mathhelpforum