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How they find weights of planets?

  1. Oct 27, 2003 #1
    ???? (lol)
  2. jcsd
  3. Oct 27, 2003 #2
    The masses of planets are estimated by observing their gravitational effects on moons, other planets, or space probes. The strength of a planet's gravity is proportional to its mass.
  4. Oct 29, 2003 #3


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  5. Oct 30, 2003 #4
    The calculation of the mass depends on the fact that Fg=GMm/r2. Where did Newton come up with this equation? Is this law empirical or derived? Is it derived from the fact that F=ma and planets travel in ellipses?
    Newton never knew the masses of the planets. So he couldn't have done anything like Kepler and taken tons of data and invented his law.
    This site used Kepler's Laws to derives Newton's
    Last edited: Oct 30, 2003
  6. Oct 30, 2003 #5
    I haven't read the historical development, but this is my guess:

    Given Galileo's work, Newton knew that the gravitational force on a body had to be proportional to its mass. He assumed that gravitation was universal, which led to the hypothesis that the force was also proportional to mass of the gravitating body. (i.e., it doesn't matter which body is the "gravitator" and which the "gravitee"; if the force is proportional to one, it should be proportional to the other.)

    For simplicity, he also assumed the force was central (i.e., it depended only on the distance between the bodies, and was attractive directly toward the other body).

    These assumptions lead to a force law of the form:

    F = -G m1 m2 f(r) r

    where G is an unknown constant and f(r) is some function of the distance r between the two bodies and r is the unit radial vector.

    He then calculated the orbits that would result from such a force law, and found that in order to get Kepler's closed elliptical orbits, f(r) had to be proportional to 1/r2.
    Last edited: Oct 30, 2003
  7. Oct 30, 2003 #6
  8. Oct 31, 2003 #7
    I guess I am not aware of Galileo's work.
  9. Oct 31, 2003 #8
    Galileo showed that the gravitational acceleration of an object did not depend on its mass. That means that the gravitational force on an object, which is its mass times the gravitational acceleration, must be proportional to its mass.
  10. Nov 2, 2003 #9
    Do you happen to know how to do this? I can't seem to figure it out for myself.
  11. Nov 2, 2003 #10
    It's not trivial. I'd recommend looking in a classical mechanics text, like Goldstein or Marion & Thornton. Or you could hunt around on Google for central force motion.

    By the way, now that I've thought about it more, I think the link you gave is probably closer to how Newton probably found his law.
  12. Nov 4, 2003 #11
    I picked up Classical Mechanics by Goldstein in my library. I skipped straight to chapter 3 which is on the two-body problem (although I had to flip back to chapter one to figure out what a Lagrangian is).
    I'm having trouble with some things.
    The graph of the gravitational potential U=-GMm/r looks like a hyperbola to me. If E<0, then certainly the particle is bounded in that it will not reach infinity. But what does this say about the shape of the orbit? Also, the fact that E<0 does not put a lower bound on r (except of course r>0). What about the graph indicates that the U of the particle will oscillate?
  13. Nov 4, 2003 #12

    It's hard to recover the shape of the orbit just by looking at the shape of the potential. You have to actually solve the equations of motion.

    To see that, you have to work with the effective potential, with the 1/r3 "centrifugal barrier" term with angular momentum.

    In a little more detail: by looking at the radial potential, you can see that a radially falling body will just drop straight to r=0. But that doesn't tell us what will happen if the body isn't moving radially; it's only one dimensional, after all. But we can convert the three dimensional motion into an effective one-dimensional problem and solve for its radial motion, this time implicitly taking into account the non-radial movement. That's what the effective potential is for. I don't have Goldstein, but I presume they derive it.
  14. Nov 4, 2003 #13
    Ah, that's great news. I had been wondering that for a while. Goldstein draws pictures of this effective potential and uses its shape to show that the object is bound at certain energies. Let me dig through Goldstein some more and I'll let you know if (when) I need more help.
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