Particle spiraling into the Sun due to "drag" force?

• petreza
In summary, the particle loses angular momentum when it orbits the Sun because the force of gravity is always opposite the velocity.
petreza
Hello,

Orbital mechanics is a bit of mystery to me.

Let us have a particle with mass m orbiting the Sun (with mass M)in a circular orbit with radius R. The velocity vector V is always tangential to the orbit. Let us also have a weak, constant ("drag") force Fd that is always pointing in the direction opposite of V.

At a first glance it would seem that Fd would reduce the magnitude of V (less K - kinetic energy) and as a result push the particle to a higher orbit (more U - potential energy). What actually happens is the opposite: the particle speeds up and moves to a lower orbit. And since the force is constant, it keeps loosing R, spiraling into the Sun.

Is there a simple geometric explanation why this happens? After all, there are only two forces here Fd and FG - Sun's force of gravity.

I know that what happens is that the particle looses angular momentum - L = mVR (in this case). Why the increase of V is proportionally less than the decrease of R?

I know that for circular orbits K = -(1/2)U and that total mechanical energy E is:
E = K + U = (1/2)(GMm)/R - (GMm)/R

Still, conceptually I don't see why the particle would react in that particular way to those two forces.

petreza said:
At a first glance it would seem that Fd would reduce the magnitude of V (less K - kinetic energy) and as a result push the particle to a higher orbit (more U - potential energy).
Why do you say so? The total mechanical energy is only conserved if there are no non-conservative forces and the drag certainly is non-conservative. Thus, you cannot make this argument.

Orodruin said:
Why do you say so? The total mechanical energy is only conserved if there are no non-conservative forces and the drag certainly is non-conservative. Thus, you cannot make this argument.

Of course, I only meant it as a (pre-) intuitive view of the problem, hence the "At first glance"
The initial (wrong) view could be like this: higher orbits have shower velocities; the force acts opposite the velocity and so must reduce it; consequently the particle must be moving to a higher orbit.
But as you say energy is not conserved.

Still, I don't see geometrically why when the force is opposite V energy is lost.
(and why when the force has same direction as V energy is gained)

----------------------------------

Now that I thought of it a little bit more at least in terms of angular momentum it makes sense:
I can see from the equations for kinetic energy = (1/2)mV^2 and = (1/2)(GMm)/R that the relation between R and V is inverse and V is squared:

1/R ~ V^2
so if we put an X coefficient in front of R
for K we get a 1/SQRT(X) coefficient in front of V
and for L we get LX = L0( X/SQRT(X) )

If R is reduced (that is 0 < X < 1 ) then X/SQRT(X) < 1 so I can see how LX is less than L0. In other words loss of angular momentum necessarily leads to reduction in R and vice versa (for orbits).

Yet again, why do the two forces combine in such a way to reduce both angular momentum and R?

Thanks!

petreza said:
Still, I don't see geometrically why when the force is opposite V energy is lost.
(and why when the force has same direction as V energy is gained)
This is a basic and fundamental concept in mechanics, the definition of work and the work-energy theorem.

Orodruin said:
This is a basic and fundamental concept in mechanics, the definition of work and the work-energy theorem.

I understand it in simple one dimensional motion. Force acts against velocity, velocity is reduced, energy (kinetic) is reduced.

Here, force acts against velocity, velocity is increased, kinetic energy is increased yet total energy is decreased because reasons.

I suspect it has to do with torque, its direction and the direction of the angular momentum.

petreza said:
Here, force acts against velocity, velocity is increased
No it isn't, at least not due to the drag force. The force that increases velocity is gravity.

petreza said:
Here, force acts against velocity,
YEs, it does but there is another force at work, too. The centripetal force accounts for the increase in velocity, not the drag force. If the object were not in orbit, the drag force would just slow it down.

Orodruin said:
No it isn't, at least not due to the drag force. The force that increases velocity is gravity.

sophiecentaur said:
YEs, it does but there is another force at work, too. The centripetal force accounts for the increase in velocity, not the drag force. If the object were not in orbit, the drag force would just slow it down.

Let me be more specific in my terms here - I am concerned about the SPEED of the particle - velocity always changes because its direction always changes - with or without the drag force.

I don't think you can say that the drag force has no effect on the SPEED of the particle because without the drag force you would not have the change of the speed.
You might say that it does not have direct effect on the speed but this leads back to my question: What is the mechanism by which a force opposite in direction to the direction of the velocity of a particle in circular orbit, the particle speeds up and looses mechanical energy?

Could you please comment on my conjecture:
I suspect it has to do with torque, its direction and the direction of the angular momentum.
Please take a look at this wikipedia animation:
https://en.wikipedia.org/wiki/Angular_momentum#/media/File:Torque_animation.gif
My drag force is the dark blue vector (during the part of the animation where it causes the object to slow down).
Of course, this is a rotating rigid body so it does not spiral inwards but I think this accurately portrays the relationship between the torque produced (the pale blue vector) and how it reduces the angular momentum (the pale green vector).

There is no torque involved with a point mass. The gravitational force, along with the reduced orbit radius, constitutes a Force Times Distance. Work is done on the object which increases its KE as it falls to a lower orbit.
But why would you let your intuitive view take over from the perfectly correct and valid maths of the situation. If you see a conflict in the explanations then that’s a personal matter and it’s only you who can sort it out in a way that’s acceptable to you. There is no point in saying your intuition must be right because it cannot be. You will never find a valid argument that makes you correct.
It may help to consider an elliptical unit where the speed varies according to the radial distance.

sophiecentaur said:
There is no torque involved with a point mass.
There is a torque (relative to the sun), angular momentum is decreasing due to the drag force (and the torque is the time derivative of the angular momentum). Otherwise I agree. If the orbit is circular, the particle will first loose some kinetic energy due to the drag force, but this leads to a smaller radius of curvature and therefore the particle will move closer to the sun, meaning that the gravitational force will do positive work on the particle that leads to an increase in the kinetic energy.

I was hoping for a simple explanation in terms of basic "properties" - forces, distances and mass.
For example we can say that without the drag force, R (as a position vector) is always counter-parallel to FG so the particle stays in the same circular orbit with the same V.
On the other hand, with the drag force Fd, the resultant force FG+d is not only slightly bigger in magnitude, but also points slightly "back" from the center of the Sun. It is as if the Sun moved diagonally - first closer to the particle (FG+d > FG) and then in the direction of Fd. The particle's new orbit is a circular orbit around this imaginary Sun' but that brings it closer to the real Sun. As soon as it moves along this new path the imaginary Sun' also moves accordingly which again alters the orbit and (and gives a boost to FG+d) and so on. The end result is the spiraling orbit into the real Sun.
See, no need for higher-level, derived, abstraction "properties", such as linear momentum, torque, energy. But this only explains the geometry not the reason why the particle accelerates. Maybe it is because the imaginary Sun' itself keeps getting closer to the particle.

petreza said:
The particle's new orbit is a circular orbit around this imaginary Sun' but that brings it closer to the real Sun
What makes you think that the orbit is necessarily circular?

Orodruin said:
What makes you think that the orbit is necessarily circular?

At each instant FG+d points toward the center of Sun' which is counter-parallel to R' (position vector relative to Sun' ) - it is the same situation as with the regular Sun and the particle but without the drag force - circular orbit (which I chose as a starting orbit for simplicity)

BTW, I am talking about Poynting-Robertson Effect here - the drag force will actually increase as the particle gets closer to the Sun which will accelerate the process.

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Obviously, the path is not circular since the particle is spiralling into the Sun and ##r## is changing ...

I think this thread is a great example of the problems you can get into if you insist on approaching a problem in one particular way. The Energy Argument is just as valid and, in just one step, explains what will happen and gives a perfectly valid reason for the increase in speed. You lose PE and gain KE, unless you are traveling through treacle, in which case you will soon reach a terminal velocity and your path will rapidly approach the radial. Why not just approach the question in the way that gives the 'easiest' answer? More convoluted arguments may be fun to follow but they really aren't necessary.

Orodruin said:
Obviously, the path is not circular since the particle is spiralling into the Sun and ##r## is changing ...

I am sorry I didn't make my point more clear. I was talking about my imaginary friend here - Sun' - note the prime symbol at the end. It is a point like sun with the same mass M as the Sun and that always stays in the direction of FG+d but just a little bit closer to the particle to account for the difference between FG+d and FG - or alternatively you can say the same distance from the particle but just a bit more massive to account for the difference between the two forces ( I imagine only one of these two scenarios is 100% true - the other is approximation) ( probably the second one is the True one - yes, definitely the second one)

So there are two ways to explain what is going on here:

1 - the formulaic way - trust the equations - calculate E loss and L loss and R loss and V increase - all fine and dandy

2 - the geometric way - draw the superposition of the two forces - how can you interpret the new vector FG+d - it is as if the Sun moved a tinny bit and became a tinny bit more massive ( Sun' ) - what happens at the next moment? - the particle changes direction just a little bit more (starts to spiral in) and accelerate (more gravity) and the direction of FG+d changes again - so you move the Sun' again (keeping same distance R from the particle as the distance to regular Sun) and repeat the cycle for each instance - the end result? - particle spirals inward and Sun' spirals outward - of course the particle impacts the real Sun, not Sun'
The spiraling out of Sun' explains the trajectory (spiraling in) of the particle. The increase of mass of Sun' accounts for the increase of V. Hey, I know, self-moving Sun' with self-adjusting mass, but can anyone give me a simpler, geometric, imaginary explanation?

petreza said:
2 - the geometric way - draw the superposition of the two forces - how can you interpret the new vector FG+d
The "geometric way" is more qualitative than quantitative. If the Maths is "all fine and dandy" then why go another way? I get the impression that you somehow find the Mathematical Approach lacking in some way. But geometry is only another form of Maths and, in this case, it couldn't be achieved with great enough accuracy to yield any useful answer.
For a bit of self entertainment, then try any approach you want but standing up in a hammock is not the best way for doing most jobs.

sophiecentaur said:
The "geometric way" is more qualitative than quantitative. If the Maths is "all fine and dandy" then why go another way? I get the impression that you somehow find the Mathematical Approach lacking in some way. But geometry is only another form of Maths and, in this case, it couldn't be achieved with great enough accuracy to yield any useful answer.
For a bit of self entertainment, then try any approach you want but standing up in a hammock is not the best way for doing most jobs.

Ha-ha, so true about the hammock. Haven't even tried it - I know the answer subconsciously.
In my defense, geometric representations often give additional insight than just mechanical algebraic calculation but unfortunately in this case the geometric representation might be considered convoluted.

@petreza If the orbit is circular, the gravitational force, which would be completely centripetal and perpendicular to the path, would not cause an increase in speed. If the orbit is circular before any applied drag force, and then the drag force is suddenly introduced, its very initial effect would be to slow the velocity/speed momentarily, before the non-circular spiral path with a gravitational pull somewhat in the forward direction causes it to speed up. (In the extreme case, the drag force could bring the orbiting object to a stop, and then it would get subsequently get pulled straight into the larger body).## \\ ## There is a related problem that you might find of interest that also produces an inward spiral, but in this related problem, there is no increase in speed. A centripetal force is applied that increases linearly with time, but always remains normal to the path. I wrote it up several months ago in an Insights Article: https://www.physicsforums.com/insights/frenet-equations-2-d-result-cornu-spiral/

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What is the "drag" force?

The "drag" force is a type of frictional force that acts on a particle as it moves through a fluid, such as air or water. It is caused by the collision of the particle with the molecules of the fluid.

How does the "drag" force affect a particle spiraling into the Sun?

The "drag" force acts opposite to the direction of motion of the particle and slows it down. This causes the particle's orbit to gradually decrease in size and spiral closer to the Sun.

Why does a particle spiral into the Sun due to the "drag" force?

The "drag" force is dependent on the density and velocity of the fluid, as well as the size and shape of the particle. As the particle gets closer to the Sun, the density of the solar atmosphere increases, leading to a stronger "drag" force that pulls the particle further into the Sun.

What factors affect the strength of the "drag" force on a particle?

The strength of the "drag" force depends on the properties of the fluid, such as its density and viscosity, as well as the particle's size, shape, and speed. Other factors that can affect the "drag" force include the angle of attack and the roughness of the particle's surface.

Can the "drag" force be counteracted to prevent a particle from spiraling into the Sun?

Yes, the "drag" force can be counteracted by a balancing force, such as the particle's own propulsion or the gravitational pull of a larger body. However, in the case of a particle spiraling into the Sun, the "drag" force is typically much stronger and cannot be counteracted by other forces.

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