How they find weights of planets?

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In summary: Originally posted by Ambitwistor 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?It doesn't say anything about the shape of the orbit. The potential just describes the force that gravity exerts on a particle. 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?
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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.
 
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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.
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
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 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.
 
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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
Originally posted by Ambitwistor
Given Galileo's work, Newton knew that the gravitational force on a body had to be proportional to its mass.
I guess I am not aware of Galileo's work.
 
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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.
 
  • #9
Originally posted by Ambitwistor

He then calculated the orbits that would result from such a force law
Do you happen to know how to do this? I can't seem to figure it out for myself.
 
  • #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.
 
  • #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?
 
  • #12
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/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.
 
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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.
 

1. How do scientists measure the weight of a planet?

Scientists measure the weight of a planet by using Newton's law of universal gravitation, which states that the force of gravity between two objects is directly proportional to their masses and inversely proportional to the square of the distance between them. By measuring the gravitational force between a planet and a known object, such as a satellite, scientists can calculate the planet's mass and therefore its weight.

2. What tools and instruments are used to measure the weight of a planet?

Scientists use a variety of tools and instruments to measure the weight of a planet. These include telescopes, which are used to observe the movement of objects in the sky and determine their mass, and spacecraft, which can collect data on a planet's gravitational pull. In addition, mathematical equations and computer simulations are used to calculate the weight of planets.

3. How accurate is the measurement of a planet's weight?

The accuracy of the measurement of a planet's weight depends on the quality of the data and the methods used. In general, modern techniques can determine a planet's weight with a high degree of accuracy, usually within a few percentage points. However, factors such as the planet's composition and the presence of other objects in its orbit can affect the accuracy of the measurement.

4. Are there any challenges in measuring the weight of a planet?

Yes, there are several challenges in measuring the weight of a planet. One major challenge is the presence of other objects in the planet's orbit, such as moons or debris, which can affect the gravitational force and make it difficult to determine the planet's weight. Additionally, the composition and density of a planet can also impact its weight, making it challenging to accurately measure.

5. How do scientists use the weight of a planet in their research?

The weight of a planet is an important factor in many areas of scientific research. For example, a planet's weight can provide valuable information about its composition, history, and formation. It is also used in astrophysics to study the dynamics of planetary systems and their interactions with other objects in the universe. Additionally, the weight of a planet can be used in space exploration to determine the feasibility of a mission and the resources needed for a spacecraft to reach and orbit the planet.

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