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B Equivalent gravitational forces in other galaxies

  1. Jun 27, 2018 #1
    Could it be imagined that due to a particular stars' distribution in a galaxy the gravitational force felt would be like $$f(\vec{e}_r)/r^{\alpha}$$ where $$\alpha\neq 2$$ but near 2 and f a non spherically symmetric function (like a comet around a flat galaxy) ?
     
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  3. Jun 27, 2018 #2

    mfb

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    What do you mean by "like a comet around a flat galaxy"?

    The force has to be conservative, ##\displaystyle \int_0^\infty \frac{f(\vec e_r)}{r^\alpha} dr## has to be the same for all directions. If you have a weaker gravitational attraction in some direction at some point you need a stronger one in this direction at a different point.
     
  4. Jun 29, 2018 #3

    ohwilleke

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    So basically, could the strength of the gravitational pull of an asymmetric galaxy e different in different directions at distances that are near to the galaxy relative to infinity?

    If I understand you correctly, yes. It could. But, the formula you use is a point particle formula like the one in your post, and you'd need instead a formula that captures the sum of gravitational pulls from a whole host of different point-like masses. ∑ (f(i)/ri2 for i=1 . . . . billions of stars, which would create an effective force with r != 2 for an arbitrary point in space used to represent the galaxy as a whole.
     
  5. Jun 29, 2018 #4

    stefan r

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    If r is measured from the center of the Earth's then gravity does vary. The Moon and Sun cause tides everyday.

    Juno's gravity science probe flips your question. It is using differences in gravity to detect the density distribution of mass inside Jupiter.
     
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