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I Gravitational constant for normalized masses

  1. Apr 20, 2016 #1
    hi,
    i'm totally confused right now. i'm playing around with a simple orbital model of the sun and the earth. since my rigid body solver doesn't like the huge masses of those bodies, i just normalized the masses to those of the sun. so i have m_sun = 1 and m_earth=3.0044e-6 as dimensionless numbers. also i reduced the distances to dimensionless numbers by dividing the real ones by 1 million km. so e.g. i got about r_earth = 149. accordingly i've divided the velocity by 1 million km to match it and multiplied it again by 1 million to speed things up. so i got a speed of about v=29/s (which comes from 29km/s). of course all this stuff are actually 3d vectors, i just gave you the magnitudes here for simplification.

    stuff works fine so far. but of course my gravitational constant now got changed. by trying around i found that it is somewhere near G_mine = 132200. this keeps the radius of the earth between 147 and 151.
    so my gravitational law looks like: F = G_mine * m1 * m2 / r², which in my case is about F = 132200 * 1 * 3.0044e-6 / (149)²

    now i wanted to calculate the exact value for G_mine. and that's where i got stucked. i thought i should get the right number (maybe not the magnitude) by just using m_earth=3.0044e-006 and maybe the actual value for G (6.674e-11). i ignored my changes to the distances, since they are multiples of 10 and even if squared they should just change the magnitude of G_mine (e.g. if i enhance the speed even more by 10, i gotta multiply G_mine by 100 because the energy is proportional to the square of the speed, which works nicely). anyway, i don't seem to be able to get where i want.

    any ideas what i'm overlooking here?
    kind regards
     
  2. jcsd
  3. Apr 20, 2016 #2

    jbriggs444

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    When you changed the mass units from kilograms to the mass of the sun, that is not a change by a factor of 3.044 x 10-6 but instead by a larger factor.

    What is the mass of the sun in kilograms?
     
  4. Apr 20, 2016 #3
    5.974 · 10^24 kg / 1.9884 · 10^30 kg = 3.044 · 10^-6
    think 3 million is still enough :)
     
  5. Apr 20, 2016 #4

    jbriggs444

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    What possible use is the number you just computed? How does it relate to the question that I asked?
     
  6. Apr 20, 2016 #5
    how do i reply to an answer?
    anyway, it's the ratio of the earth's mass to the sun's mass... it is indeed in this order of magnitude... m_earth = 5.974 · 10^24 kg / m_sun = 1.9884 · 10^30 kg = 3.044 · 10^-6 ...
    it's the normalized masses i've used...
     
  7. Apr 20, 2016 #6

    jbriggs444

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    Newton's Universal gravitational constant, "G" is normally posed in terms of kilograms, meters and seconds. That is where the figure that you quoted for G (6.674 x 10-11)" came from. That constant has that numeric value in that system of units. Change to another system of units and the numeric value for G will change. You seem to understand this.

    The system of units that you have chosen to use is one in which the the unit of mass is 1 solar mass. In order to convert from a system of units in which the unit of mass is 1 kilogram to a system of units in which the unit of mass is 1 solar mass, the conversion factor for mass is what?

    If you want to update the numeric value of G to compensate for the change in mass units, does the mass of the Earth enter in?
     
  8. Apr 21, 2016 #7
    Well the conversion factor should be 1 solar mass in kilogram... so 1.9884 · 10^30 kg ... mass of earth shouldn't enter the conversion, since it's just another mass (which needs to be converted as well of course, but it has nothing to do with G). So actually it should be G_mine = G * m_sun = 6.67408e-11m³/(kg*s²) * 1.9884 · 10^30 kg = 1.3271e+020 m³/s² ... which is the value i was looking for! :) Of course i also have to compensate for my length conversions, but this is no problem now ;)

    Thank you very much... Sometimes a little push by someone else is already enough :)
     
  9. Apr 21, 2016 #8
    ah! now i understand that first question of yours in the right way! yeah, there was already the answer in there, i just didn't get it at that time.
     
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