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decibel

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???? (lol)

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- Thread starter decibel
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decibel

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???? (lol)

- #2

Ambitwistor

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Phobos

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answer from an astronomer's FAQ website...

http://itss.raytheon.com/cafe/qadir/q2910.html

http://itss.raytheon.com/cafe/qadir/q2910.html

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- #4

StephenPrivitera

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The calculation of the mass depends on the fact that F_{g}=GMm/r^{2}. 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.

EDIT:

This site used Kepler's Laws to derives Newton's

http://www.physics.ubc.ca/~outreach/phys420/p420_95/tracy/universal.html

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.

EDIT:

This site used Kepler's Laws to derives Newton's

http://www.physics.ubc.ca/~outreach/phys420/p420_95/tracy/universal.html

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- #5

Ambitwistor

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Originally posted by StephenPrivitera

Is it derived from the fact that F=ma and planets travel in ellipses?

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 m

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

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/r

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Ambitwistor

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EDIT:

This site used Kepler's Laws to derives Newton's

http://www.physics.ubc.ca/~outreach/phys420/p420_95/tracy/universal.html [/B]

That works too... actually, I don't know for sure which (if any) is the method Newton used.

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- #7

StephenPrivitera

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I guess I am not aware of Galileo's work.Originally posted by Ambitwistor

Given Galileo's work, Newton knew that the gravitational force on a body had to be proportional to its mass.

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Ambitwistor

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- #9

StephenPrivitera

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Do you happen to know how to do this? I can't seem to figure it out for myself.Originally posted by Ambitwistor

He then calculated the orbits that would result from such a force law

- #10

Ambitwistor

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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.

- #11

StephenPrivitera

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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?

- #12

Ambitwistor

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Originally posted by StephenPrivitera

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.

Yes.

But what does this say about the shape of the orbit?

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.

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?

To see that, you have to work with the effective potential, with the 1/r

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.

- #13

StephenPrivitera

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