Orbiting around an ellipse

  • Thread starter Calpalned
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  • #1
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Homework Statement


For an ellipse, if we take the cross product of the velocity vector with the radial vector (distance from center of mass), it is equal to a constant h. Is this true? If so, what is the proof?

Homework Equations


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The Attempt at a Solution


Conceptually it makes sense. When a comet is farther from the sun, the radial vector is longer, but the velocity vector is less. Likewise, at perihelion, the comet has more velocity but less radius. So it should be a constant? I can't justify it mathematically though.
 

Answers and Replies

  • #2
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What you are saying is same as the Kepler's 2nd law. Find the proof of it.
 
  • #3
ehild
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Homework Statement


For an ellipse, if we take the cross product of the velocity vector with the radial vector (distance from center of mass), it is equal to a constant h. Is this true? If so, what is the proof?

Homework Equations


n/a

The Attempt at a Solution


Conceptually it makes sense. When a comet is farther from the sun, the radial vector is longer, but the velocity vector is less. Likewise, at perihelion, the comet has more velocity but less radius. So it should be a constant? I can't justify it mathematically though.
The cross product of the radial vector and velocity (and multiplied with the mass) is an important physical quantity - what is the name? And there is a conservation law, connected to it.
Take the time derivative of the cross product. When is it zero?
 
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  • #4
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The cross product of the radial vector and velocity (and multiplied with the mass) is an important physical quantity - what is the name? And there is a conservation law, connected to it.
Take the time derivative of the cross product. When is it zero?
Is it torque?
 
  • #5
ehild
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Is it torque?

No. Torque is the cross product of the radial vector with the force.
 
  • #6
Andrew Mason
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No. Torque is the cross product of the radial vector with the force.
?? ##\vec{p}\times\vec{r} = \vec{L}## and ##d\vec{L}/dt = d/dt(\vec{p} \times \vec{r}) = d/dt(\vec{p}) \times \vec{r} + d/dt(\vec{r}) \times \vec{p} = d/dt(\vec{p}) \times \vec{r} +0 = \vec{F} \times \vec{r} = \tau##

AM
 
  • #7
Andrew Mason
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Is it torque?
Yes. The rate of change of angular momentum is torque. Is there torque acting on a body in gravitational orbit (think central force)? So what can you say about the angular momentum?

AM
 
  • #8
ehild
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Yes. The rate of change of angular momentum is torque.

AM
The constancy of the cross product of the radial vector with the velocity was the question, and my question referred to that cross product. It is not torque.
And it would be more useful for the OP if he figures it out by himself. We do not give out full solution.
Moreover, angular momentum is ##\vec{L}=\vec{r}\times\vec{p}## and torque is ##\vec{\tau}=\vec{r}\times\vec{F}##, not the other way round.
 
  • #9
Andrew Mason
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The constancy of the cross product of the radial vector with the velocity was the question, and my question referred to that cross product. It is not torque.
You asked him to take the time derivative of the "cross product of the radial vector and velocity (and multiplied with the mass)" and asked when it was zero. He replied: "is it torque?". Your answer seemed a bit confusing because the time derivative of that cross product is torque.
And it would be more useful for the OP if he figures it out by himself. We do not give out full solution.
I did not provide the solution to the question that was posed by the OP. The solution is to explain why the cross product of velocity with the radius vector is constant. That still has not been answered.
Moreover, angular momentum is ##\vec{L}=\vec{r}\times\vec{p}## and torque is ##\vec{\tau}=\vec{r}\times\vec{F}##, not the other way round.
Quite right. That is the convention. The difference is the sign.

AM
 

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