Potential for particle circular motion

In summary, the conversation is about a question concerning a test particle in Earth's gravitational field on a circular orbit. The specific effective potential for this situation is given by two different equations, with one being incorrect due to the incorrect use of the constant $\Omega$. The correct form for the second potential is found through the use of the constant $l$, which is also dependent on $r$. The question is why there is a difference in the second derivative terms of the two potential equations. With the help of friends, the answer is found.
  • #1
2
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Hi everyone!

I have a small question that bothers me. Consider a test particle in the Earths gravitational field on circullar obrit. Speciffic effective potential is
[tex]
\begin{equation}
\widetilde{V}_{\rm eff(1)} = \frac{1}{2} \frac{l^2}{m^2} \frac{1}{r^2} - \frac{GM}{r},
\nonumber
\end{equation}
[/tex]
where
[tex]
\begin{equation}
l = m r^2 \Omega = {\rm const.}, \nonumber
\end{equation}
[/tex]
because [tex]$\theta$[/tex] is cyclic coordinate. But we have partice in circular orbit, so [tex]$\Omega = $[/tex]const. too, and we can write
[tex]
\begin{equation}
\widetilde{V}_{\rm eff(2)} = \frac{1}{2} \Omega^2 r^2 - \frac{GM}{r}.\nonumber
\end{equation}
[/tex]
Now we have correct form:
[tex]
\begin{equation}
\frac{ \partial^2 \widetilde{V}_{\rm eff(1)}}{\partial r^2} = \frac{3 l^2}{m^2 r^4} - \frac{GM}{r} = 3 \Omega^2 - \frac{2GM}{r^3}.\nonumber
\end{equation}
[/tex]
and incoorect form
[tex]
\begin{equation}
\frac{ \partial^2 \widetilde{V}_{\rm eff(2)}}{\partial r^2} = \Omega^2 - \frac{2GM}{r^3}.\nonumber
\end{equation}
[/tex]

Why is there a difference between [tex]
$ \partial^2 \widetilde{V}_{\rm eff(1)} / \partial r^2, $ [/tex] and [tex] $\partial^2 \widetilde{V}_{\rm eff(2)} / \partial r^2, $
[/tex]?

Thanks for any replies, have a nice day with physics :-)
Martin
 
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  • #2
I found the answer with the help of my friends.
Second potential is wrong, because $\Omega$ is constant only with time - it depends on $r$.
 

1. What is particle circular motion?

Particle circular motion is the movement of a particle in a circular path around a fixed point. This type of motion is characterized by a constant speed and a changing direction, as the particle continuously moves around the center point.

2. What factors affect the potential for particle circular motion?

The potential for particle circular motion is affected by the speed of the particle, the radius of the circular path, and the strength of the force acting on the particle. These factors determine the centripetal force required to keep the particle in circular motion.

3. How is the potential for particle circular motion calculated?

The potential for particle circular motion is calculated using the equation F = mv²/r, where F is the centripetal force, m is the mass of the particle, v is the velocity, and r is the radius of the circular path. This equation relates the force required for circular motion to the speed and radius of the particle's path.

4. What happens to the potential for particle circular motion if the speed is increased?

If the speed of the particle is increased, the potential for particle circular motion also increases. This is because the centripetal force required to keep the particle in circular motion is directly proportional to the square of the speed. Therefore, a higher speed will require a stronger force to maintain circular motion.

5. Can a particle maintain circular motion without a force acting on it?

No, a particle cannot maintain circular motion without a force acting on it. This is because circular motion requires a centripetal force to continuously change the direction of the particle's velocity. Without this force, the particle would continue moving in a straight line rather than a circular path.

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