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Gravitational Constant?

  1. Mar 4, 2012 #1
    Can someone explain the following formula to me? I would like to know what m, kg, and s are. At glance I believe this is the cross product of 6.67384 and the other half of the formula. However I am very unsure on how to continue into solving this for any specific value. I believe m is for the mass of a given object? A quick step by step solution of this formula would be helpful.

    G = 6.67384 × 10^-11 m^3 kg^-1 s^-2

    Thanks,
    Jamie
     
  2. jcsd
  3. Mar 4, 2012 #2
    m, kg and s are the units this constant is written in (meters, kilograms and seconds respectively).

    The constant is read as 6.67384 times 10 to the power of -11 cubic meters per kilogram second squared.
     
  4. Mar 4, 2012 #3
    That's a constant, not a formula.

    The formula is the Law of Universal Gravitation, by Sir Isaac Newton:

    F_Gravity = G * (m1 * m2) / r^2

    Where m1 and m2 are the masses and r is the distance between them.

    For a planet orbiting a star or moon orbiting a planet, in a circular orbit r is the radius of the circle of the orbit, hence the use of "r" instead of "d" for distance.

    The units for G can be found by working backwards from the formula:

    [Newtons] = G * [kg * kg] / [m^2]

    so G has units of kg^2 m^(-2) N^(-1) or m^3 kg^-1 s^-2

    G is measured from experiments, just like other constants.
     
  5. Mar 5, 2012 #4
    Henry Cavendish first measured G by suspending 2 little masses on a torsion setup that were allowed to rotate. And then set another 2 larger masses on the outside and then let the thing come to equilibrium, and then he moved the larger masses and then measured the change in the angle. Knowing the resistance in the thread he was able to calculate G.
    Later C.V. Boys modified the experiment and took 2 spheres and then set them on a balance and then put a larger sphere under one of the balance pans and the measured the deflection angle and was able to calculate G.
     
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