Robittybob's Law - orbital issue???

[Robittybob1]

I was trying to explain an issue on another thread http://www.scienceforums.net/topic/93442-gravitational-waves-discovery-expected/page-3#entry906394

and I wasn't making much progress

 

Try and write formulas for the following. Normally to lose orbital energy you need drag or reverse thrust. Gravitational energy is equivalent to that drag, so the orbital energy is converted to G-Rad and kinetic energy, whereas in normal orbital decay orbital energy is converted to drag (heat energy) and kinetic energy. You lose that orbital energy which has a component of gravitational potential energy in it, it is losing (reducing) that potential energy factor that is the source of the heat and kinetic and in this case G-radiation. There is the factor of 2 in those equations.

 

But on a site I found the effect I was trying to explain:http://www.sparknotes.com/physics/gravitation/orbits/section2.rhtml

 

However, this equation implies that as the velocity of the satellite decreases, the orbit should increase in radius. This appears to contradict our idea that viscous drag slows the satellite down, causing it to spiral towards the earth. We would expect viscous drag to cause the satellite to spiral away from the earth. In fact, the equation is right, and our intuition about the effect of the drag was wrong. The viscous drag due to the atmosphere actually speeds up the satellite in its orbit, but causes it to move to a lower energy (lower radius) orbit. In this lower orbit, the potential energy of the satellite is reduced, but since it has increased in speed, the kinetic energy has increased. Only in this way can the total energy be conserved.

Now years ago when I had said that the reduction in the potential energy is always split, half goes into additional kinetic energy and the other half is lost as drag, someone called that the "RB Law".

 

I think it was a known fact long before I found it out so does anyone know the proper name for this relationship?

 

The Potential energy gets more negative but the energy lost to drag and the kinetic energy are positive.

(I suppose it is not that useful for it would have to be an ideal circular orbit for the relationship to work.)

 

I want to see the correct formula for this situation. Any clues please?

[Robittybob1]

It seems that still no one knows what I'm talking about. From that site it says "The viscous drag due to the atmosphere actually speeds up the satellite in its orbit,..." do you think that is correct? It does going through some steps, I'm sure drag in the first instance would slow it down, the slowdown allows it to fall. 1/2 of the energy lost from the object through falling is converted to kinetic energy, the other 1/2 was lost by the drag. Now do you all agree with that?

Why do you think it speeds up when drag should be slowing it down?

[swansont]

It's counter-intuitive, because of the way the quoted bit is thinking about it. The object initially slows down, but then it falls closer to the planet, and that (as you said) is what speeds it up. To say that drag speeds it up is skipping a crucial step.

 

I don't know if it has a name, but the orbital energy is easily derived for a circular orbit. The centripetal force is mv^2/r = GMm/r^2, so it's trivial to see that the KE is half of the magnitude of the PE, and we know PE gets larger in magnitude as you get closer to the planet and that total energy is conserved.

[Robittybob1]

It's counter-intuitive, because of the way the quoted bit is thinking about it. The object initially slows down, but then it falls closer to the planet, and that (as you said) is what speeds it up. To say that drag speeds it up is skipping a crucial step.

 

I don't know if it has a name, but the orbital energy is easily derived for a circular orbit. The centripetal force is mv^2/r = GMm/r^2, so it's trivial to see that the KE is half of the magnitude of the PE, and we know PE gets larger in magnitude as you get closer to the planet and that total energy is conserved.

Thanks. It is rather trivial but it certainly makes it easier to think about how orbiting bodies operate. Look at "mv^2/r = GMm/r^2" can you have that equation as a rule of thumb? I know there are some who are much better at math than myself so I wouldn't be surprised if you could but knowing that every bit of change in GPE is converted 50:50 into energy lost (drag or gravitational radiation) and kinetic energy is rather nifty IMO.

[Janus]

It seems that still no one knows what I'm talking about. From that site it says "The viscous drag due to the atmosphere actually speeds up the satellite in its orbit,..." do you think that is correct? It does going through some steps, I'm sure drag in the first instance would slow it down, the slowdown allows it to fall. 1/2 of the energy lost from the object through falling is converted to kinetic energy, the other 1/2 was lost by the drag. Now do you all agree with that?

Why do you think it speeds up when drag should be slowing it down?

http://farside.ph.utexas.edu/teaching/celestial/Celestialhtml/node94.html

 

Here's the issue. If you caused a instantaneous slow down in an orbiting body's velocity then that body would start at that point with the decreased velocity and then speed up as it fell in towards the Earth. So in this case, you would say that it slowed down and then sped up. But this is not the case with drag. The loss of energy due to drag is not instantaneous. It can't be, because it is due to the body moving through the dragging medium. The body has to travel some distance in its orbit to lose any energy. And in losing that energy it moves in closer to the Earth and gains KE in exchange for PE. And it works outs that up to to a certain limit for the drag, the KE gained in this way will be greater than the KE lost to drag and the velocity always increases.

Note that I said "up to a certain limit for the drag". It is possible to imagine a situation where the drag is so strong that it actually slows the orbiting object down over so short a distance that the object can't fall inward enough for the KE from PE conversion to make up for the loss of KE due to the drag(Even an object falling straight down in a medium will eventually reach terminal velocity.)

[Robittybob1]

http://farside.ph.utexas.edu/teaching/celestial/Celestialhtml/node94.html

 

.....

Note that I said "up to a certain limit for the drag". It is possible to imagine a situation where the drag is so strong that it actually slows the orbiting object down over so short a distance that the object can't fall inward enough for the KE from PE conversion to make up for the loss of KE due to the drag(Even an object falling straight down in a medium will eventually reach terminal velocity.)

Thanks Janus. Good point about the terminal velocity.

[Robittybob1]

comes close, similar idea. https://astro.uni-bonn.de/~astolte/StarFormation/Lecture2012_PMS.pdf

 

virial theorem: 1/2 Egrav heats the star, 1/2 can be radiated away

That is close to what was called the "RB Law", so is the virial theorem a better word for it? Gravitational potential energy is split 50:50. I/2 of the GPE is lost (friction drag etc) and is radiated away and 1/2 goes into kinetic energy of motion. I've never thought of that before but the kinetic energy can't just radiate away. Is the virial theorem just related to stars or all orbiting masses in general?

https://en.wikipedia.org/wiki/Virial_theorem

 

For gravitational attraction, n equals −1 and the average kinetic energy equals half of the average negative potential energy

\langle T \rangle_\tau = -\frac{1}{2} \langle V_\text{TOT} \rangle_\tau.
This general result is useful for complex gravitating systems such as solar systems or galaxies.

So the virial theorem holds for large gravitationally bound objects as well as gases in the protoplanetary dust disks.

I realize now the RB Law is a simple form of the Virial Theorem.

Edited February 22, 2016 by Robittybob1

[J.C.MacSwell]

It's counter-intuitive, because of the way the quoted bit is thinking about it. The object initially slows down, but then it falls closer to the planet, and that (as you said) is what speeds it up. To say that drag speeds it up is skipping a crucial step.

 

I don't know if it has a name, but the orbital energy is easily derived for a circular orbit. The centripetal force is mv^2/r = GMm/r^2, so it's trivial to see that the KE is half of the magnitude of the PE, and we know PE gets larger in magnitude as you get closer to the planet and that total energy is conserved.

 

I knew we had discussed this before...just hadn't realized it was over 10 years ago!

 

http://www.scienceforums.net/topic/13482-is-the-earth-losing-energy/

 

From Swansont vintage 2005 (answer might seem opposite but it is opposite of a decaying orbit as it is with regard to the moon getting further away)

 

 

Slowing down. Of course this means it's increasing in total energy - receding at about 4 cm/year, currently. (This may seem counterintuitive to some - the kinetic energy is getting smaller, but the potential energy, which is negative, is getting smaller in magnitude at ~twice the rate. If you solve for KE+PE assuming a circular orbit it is easy to see what's going on)

http://farside.ph.utexas.edu/teaching/celestial/Celestialhtml/node94.html

 

Here's the issue. If you caused a instantaneous slow down in an orbiting body's velocity then that body would start at that point with the decreased velocity and then speed up as it fell in towards the Earth. So in this case, you would say that it slowed down and then sped up. But this is not the case with drag. The loss of energy due to drag is not instantaneous. It can't be, because it is due to the body moving through the dragging medium. The body has to travel some distance in its orbit to lose any energy. And in losing that energy it moves in closer to the Earth and gains KE in exchange for PE. And it works outs that up to to a certain limit for the drag, the KE gained in this way will be greater than the KE lost to drag and the velocity always increases.

Note that I said "up to a certain limit for the drag". It is possible to imagine a situation where the drag is so strong that it actually slows the orbiting object down over so short a distance that the object can't fall inward enough for the KE from PE conversion to make up for the loss of KE due to the drag(Even an object falling straight down in a medium will eventually reach terminal velocity.)

Is this is exactly right? A slowly decaying orbit speeds up. If you do a free body diagram on the satellite the net force will be ever so slightly forward, not of it's would be no drag path, but of it's new spiral path. Until that change in direction occurs (from,say perfectly circular orbit) the net force is still slightly aft, and a one time slowing would occur as described by Swansont.

Edited February 22, 2016 by J.C.MacSwell

[Robittybob1]

 

I knew we had discussed this before...just hadn't realized it was over 10 years ago!

 

http://www.scienceforums.net/topic/13482-is-the-earth-losing-energy/

 

From Swansont vintage 2005 (answer might seem opposite but it is opposite of a decaying orbit as it is with regard to the moon getting further away)

 

 

This is exactly right. A slowly decaying orbit speeds up. If you do a free body diagram on the satellite the net force will be ever so slightly forward, not of it's would be no drag path, but of it's new spiral path.

That spiral path only continues while there is the transfer of momentum, it is not on some set spiral path. Stop the tidal acceleration and the Moon will stop drifting away.

 

Slowing down. Of course this means it's increasing in total energy - receding at about 4 cm/year, currently. (This may seem counterintuitive to some - the kinetic energy is getting smaller, but the potential energy, which is negative, is getting smaller in magnitude at ~twice the rate.

There are a lot of commas in that sentence but to me it reads "the potential energy is getting smaller in magnitude at approximately twice the rate".

RB law says the "the potential energy is getting smaller in magnitude at exactly twice the rate". Is there a fault in it?

Edited February 22, 2016 by Robittybob1

[J.C.MacSwell]

That spiral path only continues while there is the transfer of momentum, it is not on some set spiral path. Stop the tidal acceleration and the Moon will stop drifting away.

There are a lot of commas in that sentence but to me it reads "the potential energy is getting smaller in magnitude at approximately twice the rate".

RB law says the "the potential energy is getting smaller in magnitude at exactly twice the rate". Is there a fault in it?

That's correct. If on that path the drag stops the orbit would revert to an ellipse, speeding up initially.

 

Sorry if I pulled things slightly off topic when quoting Swansont from 2005. If you compare circular orbits at different heights, the PE difference is twice that the KE difference. While slowly decaying it would be approximately that.

Edited February 22, 2016 by J.C.MacSwell

[Robittybob1]

That's correct. If on that path the drag stops the orbit would revert to an ellipse, speeding up initially.

 

Sorry if I pulled things slightly off topic when quoting Swansont from 2005. If you compare circular orbits at different heights, the PE difference is twice that the KE difference. While slowly decaying it would be approximately that.

No that's fine the thread had nearly run its course.

[swansont]

That spiral path only continues while there is the transfer of momentum, it is not on some set spiral path. Stop the tidal acceleration and the Moon will stop drifting away.

There are a lot of commas in that sentence but to me it reads "the potential energy is getting smaller in magnitude at approximately twice the rate".

RB law says the "the potential energy is getting smaller in magnitude at exactly twice the rate". Is there a fault in it?

 

 

The factor of two is exact for circular, and for the average of other elliptical orbits. But instantaneously it's not. The moon's orbit is not circular.

 

It's also the steady-state condition, with the only force being gravity. Other forces will cause deviation — an obvious example is that a rocket with its engines firing will not follow the rule.

[Robittybob1]

 

 

The factor of two is exact for circular, and for the average of other elliptical orbits. But instantaneously it's not. The moon's orbit is not circular.

 

It's also the steady-state condition, with the only force being gravity. Other forces will cause deviation — an obvious example is that a rocket with its engines firing will not follow the rule.

How could we tell if it is not instantaneous? I used to think in those stages (slows > falls >speeds up) but now I'm tending to the thought (at least) it is instantaneous, but how that could be is baffling.

Like when they refuel the ISS do they notice those stages? Forward thrust > increases speed > it increases in orbit due to the increased speed > as it goes up and gains GPE it slows. Do they see the sequence like this? Does anyone know?

In elliptical orbits there definitely seems to be a sequence as there is this exchange of energy going on between velocity and gravitational effects. Is this represented by a tilted path on those rubber sheet models of spacetime? Nothing on YT to give me the picture of how it would work in elliptical orbits.

Edited February 23, 2016 by Robittybob1

[swansont]

How could we tell if it is not instantaneous? I used to think in those stages (slows > falls >speeds up) but now I'm tending to the thought (at least) it is instantaneous, but how that could be is baffling.

Like when they refuel the ISS do they notice those stages? Forward thrust > increases speed > it increases in orbit due to the increased speed > as it goes up and gains GPE it slows. Do they see the sequence like this? Does anyone know?

With thrust (or drag) you have to look at the force acting on the object. That will tell you its motion. There is no general answer, because the direction and amount of the additional force matters.

 

In elliptical orbits there definitely seems to be a sequence as there is this exchange of energy going on between velocity and gravitational effects. Is this represented by a tilted path on those rubber sheet models of spacetime? Nothing on YT to give me the picture of how it would work in elliptical orbits.

I find it by looking at the equations.

[Robittybob1]

With thrust (or drag) you have to look at the force acting on the object. That will tell you its motion. There is no general answer, because the direction and amount of the additional force matters.

 

 

I find it by looking at the equations.

One day I might be so lucky as to look at an equation and see the physical image in my mind's eye. I can do this for the inverse square law and not much else.

[swansont]

One day I might be so lucky as to look at an equation and see the physical image in my mind's eye. I can do this for the inverse square law and not much else.

 

 

It's not luck. It the result of effort and practice.

[Robittybob1]

 

 

It's not luck. It the result of effort and practice.

That was why I was trying to understand how the three interchange. The three being gravitational potential energy, deceleration via heat loss and kinetic energy gain and whether it was stepwise or instantaneous (in which case you wouldn't see some steps).

[swansont]

That was why I was trying to understand how the three interchange. The three being gravitational potential energy, deceleration via heat loss and kinetic energy gain and whether it was stepwise or instantaneous (in which case you wouldn't see some steps).

 

 

It won't be stepwise, because Work + Energy will be continually adding up properly, but as long as work is being done, and/or you aren't in a circular orbit, you can't say that the KE = 1/2|PE| relationship holds at any instant.

[J.C.MacSwell]

That was why I was trying to understand how the three interchange. The three being gravitational potential energy, deceleration via heat loss and kinetic energy gain and whether it was stepwise or instantaneous (in which case you wouldn't see some steps).

Do a free body diagram of a body in free circular orbit entering a drag zone. It slows down (temporarily) as their is a aftward component of the net force from the drag, with the gravitational force being normal to the direction of travel.

Do a second of a body in almost the same orbit but in a decaying orbit from being in a drag zone for some time. It speeds up as their is a persistent forward component of the net force, since their will be a gravitational forward component that is approximately twice the drag force.

 

Put the two together with a smooth transition in between and you have essentially what Swanaont described earlier:

 

The object initially slows down, but then it falls closer to the planet, and that (as you said) is what speeds it up.

[Robittybob1]

 

 

It won't be stepwise, because Work + Energy will be continually adding up properly, but as long as work is being done, and/or you aren't in a circular orbit, you can't say that the KE = 1/2|PE| relationship holds at any instant.

... because you will have changed the eccentricity of the orbit. Is that the rest of your sentence?

"It won't be stepwise, because Work + Energy will be continually adding up properly, but as long as work is being done, and/or you aren't in a circular orbit, you can't say that the KE = 1/2|PE| relationship holds at any instant because you will have changed the eccentricity of the orbit."

Do a free body diagram of a body in free circular orbit entering a drag zone. It slows down (temporarily) as their is a aftward component of the net force from the drag, with the gravitational force being normal to the direction of travel.

Do a second of a body in almost the same orbit but in a decaying orbit from being in a drag zone for some time. It speeds up as their is a persistent forward component of the net force, since their will be a gravitational forward component that is approximately twice the drag force.

 

Put the two together with a smooth transition in between and you have essentially what Swanaont described earlier:

 

I seem to get you are saying.

But rather having a zone of drag how would your description go if the object just hit only one piece of interstellar dust and that bounced off the craft but the impact created a net backward force?

I have tried to make the modifications to your description:

"It slows down (temporarily instantly or does it?) as there is a aftward component of the net force from the drag, with the gravitational force being normal to the direction of travel.

Do a second of a body in almost the same orbit but in a decaying orbit from being hit by a single particle causing a "drag force". It speeds up as there is a persistent forward component of the net force [this is the bit I have trouble filling in], since there will be a gravitational forward component that is twice the drag force [that bit as well]."

Edited February 23, 2016 by Robittybob1

[J.C.MacSwell]

... because you will have changed the eccentricity of the orbit. Is that the rest of your sentence?

"It won't be stepwise, because Work + Energy will be continually adding up properly, but as long as work is being done, and/or you aren't in a circular orbit, you can't say that the KE = 1/2|PE| relationship holds at any instant because you will have changed the eccentricity of the orbit."

I seem to get you are saying.

But rather having a zone of drag how would your description go if the object just hit only one piece of interstellar dust and that bounced off the craft but the impact created a net backward force?

It slows down (temporarily) as their is a aftward component of the net force from the drag, with the gravitational force being normal to the direction of travel. I have tried to make the modifications:

"a body in almost the same orbit but in a decaying orbit from being hit by a single particle causing a "drag force". It speeds up as there is a persistent forward component of the net force [this is the bit I have trouble filling in], since there will be a gravitational forward component that is twice the drag force."

You will have impulsed it to slow it down, then left it in an eccentric orbit with no further drag. It should come around to that exact same spot and same velocity you left it. (assuming Newton not Relativity)

Edited February 23, 2016 by J.C.MacSwell

[Robittybob1]

You will have impulsed it to slow it down, then left it in an eccentric orbit with no further drag. It should come around to that exact same spot and same velocity you left it. (assuming Newton not Relativity)

So I'm tending to the view it is stepwise for the impact slows it down instantly, it then falls enough to regain additional velocity to maintain a lower orbit.

Are you sure about the same spot and velocity? Surely it must be a different spot for if it continued having impacts it will fall.

[J.C.MacSwell]

So I'm tending to the view it is stepwise for the impact slows it down instantly, it then falls enough to regain additional velocity to maintain a lower orbit.

Are you sure about the same spot and velocity? Surely it must be a different spot for if it continued having impacts it will fall.

You suggested only one impact in the post I replied to. My reply was with respect to that.

[Robittybob1]

You suggested only one impact in the post I replied to. My reply was with respect to that.

I was trying to understand your phrase "exact same spot" is that at the time before or after the impact?

Are you looking at it as no instantaneous radius change but just a change of direction? If it was previously in a circular orbit the change of direction occurring (at what becomes the aphelion) making it take an elliptical orbit from then on.

Edited February 23, 2016 by Robittybob1

[J.C.MacSwell]

I was trying to understand your phrase "exact same spot" is that at the time before or after the impact?

Are you looking at it as no instantaneous radius change but just a change of direction? If it was previously in a circular orbit the change of direction occurring (at what becomes the aphelion) making it take an elliptical orbit from then on.

After impact and forever in an ideal Newtonian case, once per elliptical orbit, until something changes the system. GR is different but that would be off topic.