# Change in Angular Velocity While Orbiting With No Torque

uSee2
Homework Statement:
A planet orbits a star in an elliptical orbit, is the planet faster when it is farther away from the planet or closer to the planet? Explain in terms of momentum
Relevant Equations:
##\tau = Fr\sin(\theta)##
The planet is faster when it is closer to the planet because when it is closer to the planet it has less rotational inertia, and rotational momentum is conserved in this system, so less rotational inertia means a greater angular velocity. This explains why it is slower when it is farther away from the planet because has more rotational inertia which means it has slower angular velocity to keep momentum conserved.

I believe that my explanation above is correct, however what I am confused about is to how angular velocity can change without a net torque. There is no net torque done on the planet because the force due to gravity is always pointing towards the star, and since ##\tau = Fr\sin(\theta)## combined with the fact that ##\sin(\theta) = 0## always, so net torque done the system equals 0 as well. However, there is an angular acceleration of the system because angular velocity changes as it orbits.

How could there be an angular acceleration in this case if there was no torque being done?

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## Answers and Replies

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how angular velocity can change without a net torque
As you wrote, in the absence of a torque it is the angular momentum that is constant, not the angular velocity. But the question remains, by what process does the angular velocity reduce as it moves away from the star?
Consider first an object passing a star at constant velocity. As it moves away, its angular velocity about the star diminishes purely as a matter of geometry. It is further away, and an increasing fraction of its velocity is in the radial direction.
For the planet, the velocity is also decreasing.

topsquark and uSee2
uSee2
As you wrote, in the absence of a torque it is the angular momentum that is constant, not the angular velocity. But the question remains, by what process does the angular velocity reduce as it moves away from the star?
Consider first an object passing a star at constant velocity. As it moves away, its angular velocity about the star diminishes purely as a matter of geometry. It is further away, and an increasing fraction of its velocity is in the radial direction.
For the planet, the velocity is also decreasing.
Thank you for answering! If the object is not in orbit, doesn't it not have any angular velocity originally? Also, I'm unsure as to what radial direction is. Could you briefly explain the concept to me?

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Thank you for answering! If the object is not in orbit, doesn't it not have any angular velocity originally? Also, I'm unsure as to what radial direction is. Could you briefly explain the concept to me?
For an object mass m moving at velocity ##\vec v## at position ##\vec r## relative to point P, its angular momentum about P is ##\vec L=m\vec r\times \vec v##. It has angular velocity ##\vec\omega## about P, where ##\vec v=\vec r\times\vec \omega## (I may have the orders of the operands wrong) and ##\vec r\cdot\vec\omega=0##.

The radial direction at any instant is ##\hat r##, and the tangential direction is the direction normal to both ##\hat r## and ##\hat L##.

topsquark
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I may have the orders of the operands wrong
happens to me all the time. This one shows the first one is correct: ##\vec L = \vec r\times\vec p## .
And here we see the second one should be ##\vec v_\perp = \vec\omega \times \vec r##.

##\ ##

topsquark
uSee2
For an object mass m moving at velocity ##\vec v## at position ##\vec r## relative to point P, its angular momentum about P is ##\vec L=m\vec r\times \vec v##. It has angular velocity ##\vec\omega## about P, where ##\vec v=\vec r\times\vec \omega## (I may have the orders of the operands wrong) and ##\vec r\cdot\vec\omega=0##.

The radial direction at any instant is ##\hat r##, and the tangential direction is the direction normal to both ##\hat r## and ##\hat L##.
I see, thank you! But, I'm still confused as to why it geometrically has less angular velocity the farther away it is. Does less velocity in the radial direction mean less angular velocity?

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It has less angular velocity because is both farther away and has a smaller transversal velocity. At the perigee (and also at the apogee) the magnitudes of the three quantities are related by ## \omega=\frac{v}{r}## where v is the linear velocity. Why do you think that the angular velocity should be constant, in the first place?
And when they say "in terms of momentum" probably they mean "angular momentum" and not the momentum of the force (which is usually called "torque", at least in North America).

topsquark
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Does less velocity in the radial direction mean less angular velocity?
No, the radial velocity is irrelevant. It's the tangential velocity divided by the radius. So it could be reducing tangential velocity, or increasing radius, or…

topsquark
uSee2
It has less angular velocity because is both farther away and has a smaller transversal velocity. At the perigee (and also at the apogee) the magnitudes of the three quantities are related by ## \omega=\frac{v}{r}## where v is the linear velocity. Why do you think that the angular velocity should be constant, in the first place?
And when they say "in terms of momentum" probably they mean "angular momentum" and not the momentum of the force (which is usually called "torque", at least in North America).
No, the radial velocity is irrelevant. It's the tangential velocity divided by the radius. So it could be reducing tangential velocity, or increasing radius, or…
Sorry for the late responses. I don't believe that the angular velocity is constant, what I'm confused about is that there is no net torque acting upon the planet-star system, which means that there shouldn't be an angular acceleration, which means there shouldn't be a change in angular velocity. I do see that in terms of non-rotational concepts, the linear velocity increases which means that the angular velocity should totally increase. However, since there is no torque on the system, there couldn't be any change in angular velocity which means angular velocity should be constant. In our textbooks at least, it states that no torque on a system means no angular acceleration of a system. So a system with no external forces acting upon it should have a constant angular velocity.

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In our textbooks at least, it states that no torque on a system means no angular acceleration of a system.
Such a statement could be made for rigid bodies attached with a fixed axis.

However, the correct general statement is that no net torque on a system means no change in the system's angular momentum, not its angular velocity.

In the case of a rigid body, angular momentum can be computed as angular velocity multiplied by the body's fixed moment of inertia. So constancy of angular momentum in the absence of net torque means constancy of angular velocity as well.

A satellite in orbit is not a rigid body attached with a fixed axis.

uSee2, topsquark, nasu and 1 other person
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As an example of @jbriggs444's point re angular momentum v. angular velocity, consider a skater spinning on the spot while drawing her arms in closer to her body. Her angular momentum is constant but her angular velocity increases because her moment of inertia reduces.

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In our textbooks at least, it states that no torque on a system means no angular acceleration of a system. So a system with no external forces acting upon it should have a constant angular velocity.
This is what happens when you take a statement out of context. It is very important to understand for what conditions a specific statement holds. You see that in the chapter about the motion of a rigid body. Which means that the distance between any point and the axis is fixed.

uSee2, jbriggs444 and topsquark
uSee2
This is what happens when you take a statement out of context. It is very important to understand for what conditions a specific statement holds. You see that in the chapter about the motion of a rigid body. Which means that the distance between any point and the axis is fixed.
Oh I see, thank you! So there can be an angular acceleration without a torque? (in terms of non-rigid bodies) Does this mean that the equation for Newton's Second Law in rotational form, ##\tau = I\alpha## doesn't apply to non-rigid body rotational motion?

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Oh I see, thank you! So there can be an angular acceleration without a torque? (in terms of non-rigid bodies) Does this mean that the equation for Newton's Second Law in rotational form, ##\tau = I\alpha## doesn't apply to non-rigid body rotational motion?
Correct. Using the product rule for derivatives, the law can be written as ##\tau = \frac{dL}{dt} = \frac{d (I \omega)}{dt} = I \alpha + \omega \frac{dI}{dt}##

As long as the moment of inertia (##I##) is held constant, the second term vanishes and you get the formula from the textbook.

Note that for a non-rigid object such as a whirlpool or a falling cat, it may not even be possible to define a "rotation rate". The correctness of ##L=I\omega## is questionable in the case of a non-rigid object, even if that object has a definable rotation rate. One may have to fall back on the definition of angular momentum as ##\int_\text{volume} \rho \vec{r} \times \vec{v}##.

uSee2
Correct. Using the product rule for derivatives, the law can be written as ##\tau = \frac{dL}{dt} = \frac{d (I \omega)}{dt} = I \alpha + \omega \frac{dI}{dt}##

As long as the moment of inertia (##I##) is held constant, the second term vanishes and you get the formula from the textbook.

Note that for a non-rigid object such as a whirlpool or a falling cat, it may not even be possible to define a "rotation rate" and make the above formulation meaningful.
Thank you! So basically, a changing rotational inertia makes things a lot more complicated haha. I have no knowledge on derivatives (I'm not yet in precalculus either), would it be possible if you could explain what the d means in the equation you showed above in simple terms? Or should I just leave this question alone until I get to calculus?

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Thank you! So basically, a changing rotational inertia makes things a lot more complicated haha. I have no knowledge on derivatives (I'm not yet in precalculus either), would it be possible if you could explain what the d means in the equation you showed above in simple terms? Or should I just leave this question alone until I get to calculus?
The "##d##" is a notation that can be understood roughly as "tiny change in". Yes, that is calculus. Strictly speaking, it is pure notation. But often it can be treated algebraicly as just another term for purposes of cancellation when a like factor is present in both numerator and denominator.

If I write ##\frac{dI}{dt}##, that means the rate of change in ##I## compared to the rate of change of ##t##. Or "how fast is the moment of inertia changing over time".

If I write ##\alpha = \frac{d \omega}{dt}##, I am saying that angular acceleration is the rate of change of the rotation rate over time.

If I write ##a = \frac{d v}{dt}## and ##v = \frac{d x}{dt}## I can then write ##a = \frac{d^2 x}{{(dt)}^2}## which is to say that "acceleration is the second derivative of position with respect to time".

The product rule for derivatives is pretty much the result of simple polynomial evaluation. One can ask about the product of ##(a + \Delta a)(b + \Delta b)##. That is ##ab + a \Delta b + b \Delta a + \Delta a \Delta b##.

If we want to know how much the product changes when the deltas are added, we subtract the before from the after. That cancels ##ab## term. If we let the intervals ##\Delta a## and ##\Delta b## get very small then their product becomes negligible. It becomes zero in the appropriate limit and we can cancel the trailing term.

Rewrite the ##\Delta## as ##d## in what is left to indicate that we are talking about derivatives and the result becomes: ##d(ab) = a\ db + b\ da##. That is the rule for the first derivative of a product.

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Does this mean that the equation for Newton's Second Law in rotational form, ##\tau = I\alpha## doesn't apply to non-rigid body rotational motion?
Correct
Um…., that equation still applies, no? But if it is not a rigid body then ##I## can change, whereas m cannot.

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Um…., that equation still applies, no? But if it is not a rigid body then ##I## can change, whereas m cannot.
The equation that I am thinking of: ##\tau = I \alpha## equates the rate of change of angular momentum with angular momentum multiplied by a fixed moment of inertia.

The obvious reason for ##\tau = I \alpha## to fail is because the moment of inertia might change over time. However, that is not the only failure mode for that equation. Even with a fixed moment of inertia, and even with an identifiable notion of angular acceleration, the angular momentum of a non-rigid object with moving parts might not satisfy ##L = I \omega##. The torque-free rotation of a falling cat is a one example. With the failure of that equation, the failure of its first derivative is then a natural outcome.

If I understand your point, it is that ##F=ma## is reliable since, as you point out, ##m## is conserved. With that, I certainly agree.

uSee2
The "##d##" is a notation that can be understood roughly as "tiny change in". Yes, that is calculus. Strictly speaking, it is pure notation. But often it can be treated algebraicly as just another term for purposes of cancellation when a like factor is present in both numerator and denominator.

If I write ##\frac{dI}{dt}##, that means the rate of change in ##I## compared to the rate of change of ##t##. Or "how fast is the moment of inertia changing over time".

If I write ##\alpha = \frac{d \omega}{dt}##, I am saying that angular acceleration is the rate of change of the rotation rate over time.

If I write ##a = \frac{d v}{dt}## and ##v = \frac{d x}{dt}## I can then write ##a = \frac{d^2 x}{{(dt)}^2}## which is to say that "acceleration is the second derivative of position with respect to time".

The product rule for derivatives is pretty much the result of simple polynomial evaluation. One can ask about the product of ##(a + \Delta a)(b + \Delta b)##. That is ##ab + a \Delta b + b \Delta a + \Delta a \Delta b##.

If we want to know how much the product changes when the deltas are added, we subtract the before from the after. That cancels ##ab## term. If we let the intervals ##\Delta a## and ##\Delta b## get very small then their product becomes negligible. It becomes zero in the appropriate limit and we can cancel the trailing term.

Rewrite the ##\Delta## as ##d## in what is left to indicate that we are talking about derivatives and the result becomes: ##d(ab) = a\ db + b\ da##. That is the rule for the first derivative of a product.
Correct. Using the product rule for derivatives, the law can be written as ##\tau = \frac{dL}{dt} = \frac{d (I \omega)}{dt} = I \alpha + \omega \frac{dI}{dt}##

As long as the moment of inertia (##I##) is held constant, the second term vanishes and you get the formula from the textbook.

Note that for a non-rigid object such as a whirlpool or a falling cat, it may not even be possible to define a "rotation rate". The correctness of ##L=I\omega## is questionable in the case of a non-rigid object, even if that object has a definable rotation rate. One may have to fall back on the definition of angular momentum as ##\int_\text{volume} \rho \vec{r} \times \vec{v}##.

Thank you! So basically, there is a "secret" term in the ##\tau = I\alpha## equation, that becomes present when ##I## changes. The reason that angular velocity changes is because ##\tau = I \alpha + \omega \frac{dI}{dt}## means that, if ##\tau = 0## then ##-I\alpha = \omega \frac{dI}{dt}##, which means that if ##I## increases, ##\omega## increases. Is this correct?

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Oh I see, thank you! So there can be an angular acceleration without a torque? (in terms of non-rigid bodies) Does this mean that the equation for Newton's Second Law in rotational form, ##\tau = I\alpha## doesn't apply to non-rigid body rotational motion?
Yes, you can see this even without calculus. If you look ahead and something crosses your line of sight moving with constant velocity, its angular velocity relative to you changes. And there is no force at all so there is no torque to think about. The angle changes the fastest when the moving object is at the minimum distance and then it looks like it slows down as it goes away on the side. Its linear velocity does not change but the angular one decreases. After all, the maximum angle cannot be more than 90 degrees no matter how much time passes. (I mean the angle between your line of sight straight ahead of you and the line to the moving object).

uSee2
Yes, you can see this even without calculus. If you look ahead and something crosses your line of sight moving with constant velocity, its angular velocity relative to you changes. And there is no force at all so there is no torque to think about. The angle changes the fastest when the moving object is at the minimum distance and then it looks like it slows down as it goes away on the side. Its linear velocity does not change but the angular one decreases. After all, the maximum angle cannot be more than 90 degrees no matter how much time passes. (I mean the angle between your line of sight straight ahead of you and the line to the moving object).
In an ellipse, would the angular velocity still be equal to tangential velocity divided by the radius? What would the radius be in an ellipse?

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Even with a fixed moment of inertia, and even with an identifiable notion of angular acceleration, the angular momentum of a non-rigid object with moving parts might not satisfy L=Iω. The torque-free rotation of a falling cat is a one example.
I disagree.
An astronaut can reorient in space by rotating trunk and legs while the arms are extended, drawing in the arms and spreading the legs simultaneously, then rotating back the trunk and arms to realign with the legs. In principle this can all be with constant moment of inertia and the angular momentum never changing, yet the orientation changes.
Unlike buttered toast, an inverted falling cat does something similar.

I see no way to violate ##\tau=I\alpha ##.

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I disagree.
An astronaut can reorient in space by rotating trunk and legs while the arms are extended, drawing in the arms and spreading the legs simultaneously, then rotating back the trunk and arms to realign with the legs. In principle this can all be with constant moment of inertia and the angular momentum never changing, yet the orientation changes.
Unlike buttered toast, an inverted falling cat does something similar.

I see no way to violate ##\tau=I\alpha ##.
Take your own example. We have the astronaut with no torque. ##\tau## = 0.
The astronaut has a non-zero moment of inertia ##I##.
He starts with a zero rate of change of orientation. He changes orientation. He ends with a zero rate of change of orientation. ##\alpha## was non-zero somewhere in there.

That violates ##\tau = I \alpha##.

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That's the trick.. since she is not a rigid body, "##\alpha##" is undefined. She had opposite signs for it in different parts of her body. ##\Sigma m_ir_i^2\alpha_i## was 0 throughout.
https://space.stackexchange.com/questions/2954/how-do-astronauts-turn-in-space
If ##\alpha## is undefined then the ##\tau = I \alpha## automatically fails to hold. I thought that you were claiming that it holds always.

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If ##\alpha## is undefined then the ##\tau = I \alpha## automatically fails to hold. I thought that you were claiming that it holds always.
Good point, but there is a difference between being violated and ceasing to be meaningful. The difference from F=ma is that for a flexible body we can apply that to the motion of the mass centre. There is no rotational equivalent (that I can think of) to a mass centre.
To make ##\tau = I \alpha## meaningful in the non-rigid context we have to replace ## I \alpha## with ##\Sigma m_ir_i^2\alpha_i## or define ##\alpha## as ##\frac{\Sigma m_ir_i^2\alpha_i}{\Sigma m_ir_i^2}##.

jbriggs444
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In an ellipse, would the angular velocity still be equal to tangential velocity divided by the radius? What would the radius be in an ellipse?
"Tangential" means tangent to some curve (or surface). The velocity is always tangential to the trajectory, by definition. So "tangential velocity" is just the velocity. If you consider the line that connects the origin with the position of the object on the trajectory, in general, the velocity has one componentalong this line (##v_r##) and another perpendicular to it (##v_\perp ##). You may see this last one referred to as "tangential" but this is not tangent to the trajectory. This component is related to the instantaneous angular velocity by the relationship ##v_\perp =\omega r ##. At the two endpoints of the major axis, the radial component is zero so the relationship is ##v=\omega r## with v being the magnitude of the full velocity vector.

uSee2
Good point, but there is a difference between being violated and ceasing to be meaningful. The difference from F=ma is that for a flexible body we can apply that to the motion of the mass centre. There is no rotational equivalent (that I can think of) to a mass centre.
To make ##\tau = I \alpha## meaningful in the non-rigid context we have to replace ## I \alpha## with ##\Sigma m_ir_i^2\alpha_i## or define ##\alpha## as ##\frac{\Sigma m_ir_i^2\alpha_i}{\Sigma m_ir_i^2}##.
If ##\alpha## is undefined then the ##\tau = I \alpha## automatically fails to hold. I thought that you were claiming that it holds always.
So essentially, we can't use ##\tau = I\alpha## for non-rigid body motion?

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You may see this last one referred to as "tangential"
Yes, that's how I was using it in post #8, but I should have clarified that.
As you note, the velocity is always tangential to the trajectory, so there is nothing gained in using "tangential velocity" to mean that.