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Calculating gravitational pull on surface for large object

  1. Dec 20, 2015 #1
    1. The problem statement, all variables and given/known data
    A large sphere exists in space, which has a mass of 1 * 10^28 kg
    The sphere has a radius of 100,000 km
    What will be its gravitational pull (aka: "relative gravity") on its surface in terms of gs (1 "g" being equal to the gravitational pull of the Earth which is 9.807 m/s^2)?

    2. Relevant equations
    relative gravity = (gravitational constant * mass of object) / radius^2
    gravitational constant = 6.67408 * 10^-11 m^3 kg^-1 s^-2

    3. The attempt at a solution
    First I multiplied the Gravitational Constant by the object's mass:
    (6.67408 * 10^-11 m^3 kg^-1 s^-2) * (1*10^28 kg) = 6.67408 *10^17 m^3 s^-2

    Next I squared the radius:
    (100,000,000 m)^2 = 1*10^16 m^2

    Then I divided the numerator by the denominator:
    (6.67408 *10^17 m^3 s^-2) / (1*10^16 m^2) = 66.7408 m s^-2.
    This I define as the relative gravity upon the sphere's surface.

    Finally, I divided this relative gravity by the Earth's gravity:
    (66.7408 m s^-2) / (9.807 m s^-2) = 6.805 "gs"

    Thus: the gravitational pull upon this sphere's surface is about 6.805 times the gravity on the Earth's surface.

    Is my methodology and/or reasoning correct ladies & gentlemen?
     
  2. jcsd
  3. Dec 20, 2015 #2

    gneill

    User Avatar

    Staff: Mentor

    Looks fine.
     
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