Physics Help Forum (orbital) momentum

 Periodic and Circular Motion Periodic and Circular Motion Physics Help Forum

 Jul 5th 2017, 11:01 PM #1 Junior Member   Join Date: Jul 2017 Posts: 27 (orbital) momentum Hello everybody! I'm layman in physics and new in this forum, but recently I feel very strong interest in physics... and now I am struggling to obtain some knowledge. That's why i decided to sign up in the forum and I hope to get help from people who are versed and educated in physics. Here is the first question which I hope to solve.... I know about the notorious formulation: p = mv p - momentum; m - mass; v - velocity. So far, so good... Two weeks ago, I read a text, where the author talks about "free movement (circulation) in gravitational orbit" and he (the author) describes the momentum like this: p = miv/(2πl) = const p - momentum; iv - orbital velocity (velocity of circulation); 2πl - orbital length (perimeter of circumference). After that, he says: when we have 2πl= i (imaginary number), we get: p = mv (the notorious formulation) I can't grasp his idea... I searched all the google and and all the textbooks, which i have at home... but i can't find information about the author's primary formulation: p = miv/(2πl) = const I can't post the original text, because it is written in bulgarian. the literal translation will be difficult for me. English is not my native language, but i hope I asked my question clearly... So I am looking for some comments, which will explain me the formulation... I will be very thankful for every explanation.... Have a nice day!
 Jul 6th 2017, 09:19 PM #2 Senior Member   Join Date: Apr 2017 Posts: 156 Hi ..And welcome I don't understand where this formula comes from ,or why an 'imaginary' number should be there ... I wouldn't worry too much about it ...this could be an error ! Momentum is very easy , just mv , whether the mass is in orbit or travelling in a strait line . When in an elliptical orbit the momentum is continually changing ... the further apart the two bodies the slower they move ... the lose in kinetic energy is converted into increased potential energy as the two bodies are further apart.
Jul 6th 2017, 10:12 PM   #3
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Join Date: Nov 2013
Location: New Zealand
Posts: 534
 Originally Posted by DesertFox Two weeks ago, I read a text, where the author talks about "free movement (circulation) in gravitational orbit" and he (the author) describes the momentum like this: p = miv/(2πl) = const p - momentum; iv - orbital velocity (velocity of circulation); 2πl - orbital length (perimeter of circumference). After that, he says: when we have 2πl= i (imaginary number), we get: p = mv (the notorious formulation) I can't grasp his idea... I searched all the google and and all the textbooks, which i have at home... but i can't find information about the author's primary formulation: p = miv/(2πl) = const I can't post the original text, because it is written in bulgarian. the literal translation will be difficult for me. English is not my native language, but i hope I asked my question clearly... So I am looking for some comments, which will explain me the formulation... I will be very thankful for every explanation.... Have a nice day!
Hi DesertFox,

Welcome to the forum. Your English is alright so don't worry about that. I am practically a layperson also but have managed to pick up some advanced maths skills here and there and have learned a lot just doing what you are doing.

I looks like p = miv/(2πl) = const might be a statement about conservation of angular momentum as $\displaystyle \mathbf{r} \times \mathbf{P} = \mathbf{L}$ or perhaps more understandable as $\displaystyle r P sin(\theta) = L$ which, using circular motion as an example, is the radius of motion and P is the momentum and theta is the angle between the radius of motion and the direction of P and L is angular momentum which is generally conserved. Here is basic explanation of angular momentum This Url Link. Others on this forum may have better sources.

Last edited by kiwiheretic; Jul 6th 2017 at 10:19 PM.

 Jul 7th 2017, 08:27 AM #4 Physics Team     Join Date: Jun 2010 Location: Naperville, IL USA Posts: 2,271 The author's assertion that: $\displaystyle p = \frac {miv}{2 \pi l}$ is clearly incorrect, as the units of momentum are Kg-m/s, but the right hand side of this equation has units of Kg/s.
 Jul 7th 2017, 09:32 AM #5 Junior Member   Join Date: Jul 2017 Posts: 27 Further in the text, the author talks about photon, which is circulating around kernel. The author represents it like this: p = mic/(2πl) (c - the speed of light)... He draws a parallel between "free fall" and "free moving (circulating) in an orbit" But i can't really realize how does he come up with these mathematical formulations.....
 Jul 7th 2017, 09:35 AM #6 Junior Member   Join Date: Jul 2017 Posts: 27 Thank you for all the replies and your efforts! I'm still trying to understand whether this is mistake or the author tries to devise some kine of new mathematical representation (formulation).....
Jul 7th 2017, 01:44 PM   #7
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Join Date: Nov 2013
Location: New Zealand
Posts: 534
 Originally Posted by DesertFox Further in the text, the author talks about photon, which is circulating around kernel. The author represents it like this: p = mic/(2πl) (c - the speed of light)...
Then what does "m" stand for? (A photon has no (rest) mass.)

 Jul 7th 2017, 02:27 PM #8 Senior Member   Join Date: Nov 2013 Location: New Zealand Posts: 534 Here is a Khytoransky video with captions explaining momentum and angular momentum topsquark and DesertFox like this.

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