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Is the force of gravity a function of a planet's orbit?

  • #1

Homework Statement



Calculate the formula for the force of gravity in a universe where Kepler's 2nd law holds and a planet revolves around the sun with polar eq r = 1 + sin (theta). Theta = pi/2 is where velocity and position vector are perpendicular.

Homework Equations



Universal gravitation law. Perihelion = point at which velocity and position vectors are perpendicular = 2.


The Attempt at a Solution



I don't understand something. How does the shape of the orbit of the planet around the sun affect the formula for force of gravity? Wouldn't the solution simply be Newton's Universal Gravitation formula, the radius being a function of theta?

I guess what I'm looking for is an explanation as to why my assumption isn't correct, which hopefully will give me a hint as to how best to approach this problem.
 
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Answers and Replies

  • #2
kuruman
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I think the problem is asking you to cast the law of gravitation using a new r measured from the polar center, i.e. where the ellipse major and minor axes intersect.
 
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  • #3
The problem says that the orbit isn't an ellipse, though. It's defined by r=1 + sin (theta).
 
  • #4
D H
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That is the equation for a cardoid, not an ellipse. So -- are you sure you have the problem statement correct?
 
  • #5
Here's the exact wording of the problem:

"Suppose that in a universe where Kepler's second law holds, a planet revolves around the sun on the path with polar eq r= 1 + sin(theta). When theta = pi/2, the velocity vector and position vector are perpendicular to one another. Calculate the formula for the force of gravity in this universe. Give your answer in terms of r; there should be no theta at the end."
 
  • #6
kuruman
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Look in your textbook. There is an equation that relates the force law to the shape of the orbit. I don't remember that equation off the top of my head, but that's what you have to use.
 
  • #7
Ugh, I was afraid of that. My professor is old fashioned and doesn't use textbooks for any of his classes. I've scoured the internet and my notes and haven't found mention of non-elliptical orbits.
 
  • #8
kuruman
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I don't have my textbook with me. I can post the equation tomorrow unless someone else beats me to it.
 
  • #9
D H
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Orbits are conic sections in a universe with two bodies that obeys Newton's laws. This is the equation for a cardoid (the canonical equation for a cardoid, for that matter), and a cardiod is not a conic section.
 
  • #10
That'd be great. I'd be interested in knowing how to derive it. I'm surprised I can't find anything about this online.
 
  • #11
Orbits are conic sections in a universe with two bodies that obeys Newton's laws. This is the equation for a cardoid (the canonical equation for a cardoid, for that matter), and a cardiod is not a conic section.
I don't think the professor is suggesting that this is a plausible orbit. For what it's worth, this is a problem given to me in a class called History of Mathematics -- it's not a physics class.
 
  • #12
D H
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OK then. Read the question as stated. Do not assume Newtonian gravity; it obviously does not apply in this universe. Kepler's second law does, however. What is Kepler's second law? The problem is asking you to deduce a gravitational law for this universe.
 
  • #13
Kepler's 2nd law says that the rate at which the radius vector sweeps area is constant. That is, the orbiting planet will adjust it's velocity accordingly to ensure that this is so. I'm not sure what the connection is between this and gravity is though.
 
  • #14
D H
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Forget about Newton's law of gravitation. That is not a relevant equation here. Gravity behaves differently in that universe.

That said, that orbits obey Kepler laws (not just the second) in our universe was one of the things that motivated Newton in his formulation of Newton's law of gravitation. The shape and temporal behavior of orbits says a lot about the underlying force law that generates that behavior.
 
  • #15
ehild
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Determine the vector of acceleration, by calculating the second time derivative of the position vector. The force is still ma. Using Kepler's second law, you will find that the force acts along the radius and depends only on the distance between the planet and its sun.

The position vector is

[tex]\vec r = (1+\sin{\theta})\vec e_r[/tex]

Where er is the radial unit vector.

ehild
 
  • #16
kuruman
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Here is another way to do it (as promised). Start with the equation of motion, which you can get for a general central potential U(r) in spherical coordinates. You can get that through the Lagrangian. Note that, because Kepler's Second Law is valid in this Universe, angular momentum must be conserved. Substitute u = 1/r in the equation of motion, use the chain rule of differentiation correctly and you should end up with

[tex]\frac{d^2u(\theta)}{d \theta^2}+u(\theta)=-\frac{1}{m^2L^2u(\theta)^2}F(u^{-1})[/tex]

This equation is derived in many textbooks, I cite "Analytical Mechanics" by Fowles and Cassiday, "Classical Mechanics" by Taylor and "Classical Mechanics" by Finn because these are the ones I have on my shelves. You can solve the equation for the force and put the radial dependence back in through r = 1/u.
 
  • #17
ehild
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Phantomcow2, do not let let you frighten off. This problem is not that complicated, you can do it. First see what Kepler's second law means for this problem. What is the area swept by the radius vector in unit time? By the way, you can formulate it using the definition of angular momentum: It is the vector product of the position vector r with the momentum mv.

We use polar coordinates, [itex]e_r[/itex] is the radial unit vector and [itex]e_{\theta}[/itex] is the azimuthal unit vector. You certainly know the time derivative of the unit vectors:

[tex]\dot \vec {e_r}= \dot {\theta} \vec {e_{\theta}}[/tex]

and

[tex]\dot e_{\theta}= -\dot {\theta} \vec e_{r}[/tex]

So the position vector is

[tex]\vec r = r\vec {e_r}[/tex],

the velocity is

[tex]\vec v =\dot r \vec {e_r}+r \dot{\theta}\dot e_{\theta}[/tex]

and the magnitude of their vector product is

[tex] r^2 \dot {\theta}=C[/tex] (*)

with C a constant.


The acceleration in polar coordinates is

[tex]\vec a =(\ddot r-r\dot{\theta}^2)\vec {e_r}+(2\dot r \dot{\theta}+r\ddot {\theta}) \vec{e_{\theta}} [/tex]

You have to use the expression for r(t) to get its derivatives and use the condition (*) to express theta and its derivatives with r(t).

ehild
 
  • #18
Wow, thanks for the helpful replies! :). ehild, it turns out that your derivation was one that was gone over in class during the one time I wasn't there :p (I had a medical appointment).

I found notes from a classmate and to sum it up the resulting formula is
[tex]\frac{-mr_{0}^2v_{0}^2}{r^2} (y''+y)[/tex] where y=1/r

I could probably use such a method to derive the Universal Gravitation law by substituting the polar equation of an ellipse.
 
  • #19
ehild
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Yes, for a normal world it is very easy. The world of your professor is a bit more complicated and all planets end up inside their sun before completing one period.

ehild
 
  • #20
kuruman
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Wow, thanks for the helpful replies! :). ehild, it turns out that your derivation was one that was gone over in class during the one time I wasn't there :p (I had a medical appointment).

I found notes from a classmate and to sum it up the resulting formula is
[tex]\frac{-mr_{0}^2v_{0}^2}{r^2} (y''+y)[/tex] where y=1/r

I could probably use such a method to derive the Universal Gravitation law by substituting the polar equation of an ellipse.
Which is the equation I gave you except that I used u instead of y and L instead of mv0r0. :smile:
 

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