# Orbiting around an ellipse

## 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?

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.

What you are saying is same as the Kepler's 2nd law. Find the proof of it.

ehild
Homework Helper

## 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?

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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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?

ehild
Homework Helper
Is it torque?

No. Torque is the cross product of the radial vector with the force.

Andrew Mason
Homework Helper
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

Andrew Mason
Homework Helper
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

ehild
Homework Helper
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.

Andrew Mason