# Satellite mechanics: linear and rotational momentum

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vasya
satellite mechanics: linear and rotational momentum
I'm trying to better understand classical mechanics, and came up with a question:
Say we have a squared satellite weighting 100kg, 1 meter on each side. it has a thruster on it's side, shown in picture
thruster quickly ejects 100g of propellant with a speed of 1000m/s giving a satellite 100kg m/s of momentum. the question is how much of it will be given to rotation, and how much - to linear motion

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You have already applied conservation of linear momentum to get the linear speed. Perhaps there is another conservation law that would make sense to use to find a similar expression for angular speed around a particular relevant point in the satellite?

vasya
The question is about how linear momentum of propellant will transform into angular and linear momentum of satellite

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satellite mechanics: linear and rotational momentum
I'm trying to better understand classical mechanics, and came up with a question:
Say we have a squared satellite weighting 100kg, 1 meter on each side. it has a thruster on it's side, shown in picture
thruster quickly ejects 100g of propellant with a speed of 1000m/s giving a satellite 100kg m/s of momentum. the question is how much of it will be given to rotation, and how much - to linear motion
Linear momentum is conserved. It's not "transformed into rotation".

vasya
You are right. But only if we are talking about point body.

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You are right. But only if we are talking about point body.
Conservation of linear momentum applies to extended bodies where the momentum is given by the mass of the body times the velocity of its centre of mass.

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Note that conservation of linear and angular momentum both are valid. That is, you can determine the linear speed as you have already done, and on top of that you can determine the angular speed around the CM by observing that the total angular momentum around the CM for both satellite and the ejected mass must be conserved across the ejection.

Also, when the ejected mass is significant relative to the total mass then you may want to include the effect of CM shift, especially if the satellite was originally rotating and the ejected mass due to a placement far from CM already had significant angular momentum.

vasya
Thank for your replies, but I don't understand you)
I'm new to subject. It would be helphul, if you explained, how to solve problems like this.

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The question is about how linear momentum of propellant will transform into angular and linear momentum of satellite
100% of the linear momentum of the exhausted propellant will manifest as equal and opposite linear momentum of the satellite. That's Newton's third law or conservation of momentum. They are the same thing.

100% of the angular momentum of the exhausted propellant will manifest as equal and opposite angular momentum of the satellite. That's conservation of angular momentum. [Or Newton's third law plus a proviso that the lines of action of third law partner forces always coincide]

The exhaust stream has angular momentum given by ##m\vec{r} \times \vec{v}## where ##\vec{r}## is the displacement from your chosen reference point (or reference axis in 2 dimensions).

In this case one would probably want to use a reference axis at the center of mass of the satellite. This would avoid having to account for any angular momentum associated with the linear motion of the satellite.

• PeroK
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[Edit: Silly mistake struck-through (see Post ##11).] If the OP still wants to know how to calculate the satellite’s angular momentum, I would suggest first reading-up ‘moment of inertia’ (which hasn’t yet been mentioned).

[WAFFLE]
On a different note, can I point out a potential source of confusion in this question? The ‘1000m/s’ can be interpreted as the propellant’s speed in the system’s centre of mass frame, or as the exhaust velocity.

If 1000m/s is the exhaust velocity it’s instructive to consider 2 cases:
Case A: thruster is central (producing no torque about satellite’s CoM).
Case B: thruster is offset (producing torque about satellite’s CoM).

Assume that the thruster converts the same amount of energy (chemical to kinetic) in both cases A and B. This means the linear kinetic energy of the satellite must be less for case B than for case A because some of the energy has been converted to rotational kinetic energy.

This implies the satellite’ s final linear momentum will be different in cases A and B.

Is momentum being conserved? [Of course it is!]
What’s going on? [Rhetorical.]

In the system’s centre of mass frame, the propellant has a slower speed for case B compared to case A. Why? It’s due to the rotation in case B. Propellant laves the thruster at 1000m/s. But, due to the rotation, for a short period the thruster is moving faster than the satellite’s centre of mass. As a result, in the system’s centre of mass frame, the propellant moves slower in case B compared to case A.
[\WAFFLE]

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If 1000m/s is the exhaust velocity it’s instructive to consider 2 cases:
This is indeed an interesting complication and the more complete analysis does address the seeming violation of conservation of energy that could result from an inadequate analysis.

However, I suspect that the poster does not want to contemplate a continuous burn with an exhaust stream that has changed velocity during the burn due to the thruster's acceleration over that time frame.

As I read the original post, the 1000 m/s it is not the exhaust velocity exactly. We do not have a conventional rocket ejecting a continuous stream. Instead we have some sort of idealized impulsive event which involves the exhaust moving off at a fixed 1000 m/s. Presumably relative to the craft's original center-of-mass frame. One also imagines that the craft mass of 100 kg was net, not gross. All of the unburned fuel is positioned right at the thruster and does not count toward the center of mass calculation. [And yes, I realize that it is physically unrealistic to expect the same thruster with the same fuel load to deliver exactly the same 1000 m/s result regardless of where it is positioned on the craft or regardless of the craft's moment of inertia].

So we started with 100 kg of craft and 100 g of fuel at rest. We end with 100 g of exhaust moving off at 1000 m/s and 100 kg of craft moving off at (OP needs to do the calculation) in the opposite direction.

The calculation of the craft's angular momentum about its center of mass now requires no knowledge of the craft's moment of inertia. All we need is the offset of the thruster from the craft's center of mass.

If we want the satellite's rotation rate, we need to divide the calculated angular momentum by the craft's moment of inertia.

• vasya
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The calculation of the craft's angular momentum about its center of mass now requires no knowledge of the craft's moment of inertia. All we need is the offset of the thruster from the craft's center of mass.
Yes indeed. My silly mistake.

vasya
This is indeed an interesting complication and the more complete analysis does address the seeming violation of conservation of energy that could result from an inadequate analysis.

However, I suspect that the poster does not want to contemplate a continuous burn with an exhaust stream that has changed velocity during the burn due to the thruster's acceleration over that time frame.

As I read the original post, the 1000 m/s it is not the exhaust velocity exactly. We do not have a conventional rocket ejecting a continuous stream. Instead we have some sort of idealized impulsive event which involves the exhaust moving off at a fixed 1000 m/s. Presumably relative to the craft's original center-of-mass frame. One also imagines that the craft mass of 100 kg was net, not gross. All of the unburned fuel is positioned right at the thruster and does not count toward the center of mass calculation. [And yes, I realize that it is physically unrealistic to expect the same thruster with the same fuel load to deliver exactly the same 1000 m/s result regardless of where it is positioned on the craft or regardless of the craft's moment of inertia].

So we started with 100 kg of craft and 100 g of fuel at rest. We end with 100 g of exhaust moving off at 1000 m/s and 100 kg of craft moving off at (OP needs to do the calculation) in the opposite direction.

The calculation of the craft's angular momentum about its center of mass now requires no knowledge of the craft's moment of inertia. All we need is the offset of the thruster from the craft's center of mass.

If we want the satellite's rotation rate, we need to divide the calculated angular momentum by the craft's moment of inertia.
Offset is shown in picture attached. It's 25cm from side, it is pointed down. And yes. I'm assuming very quick ejection of propellant)

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Offset is shown in picture attached. It's 25cm from side, it is pointed down. And yes. I'm assuming very quick ejection of propellant)
What exactly are you asking? Are you asking for a worked example? What source are you using to study mechanics? Does that not have examples to show you how to handle this sort of problem?

vasya
What exactly are you asking? Are you asking for a worked example? What source are you using to study mechanics? Does that not have examples to show you how to handle this sort of problem?
I'm asking what principles do I need to understand to solve that type of problems, I know some basics like what is momrnt of inertia is. But my knowledge is limited.
Btw moment of inertia here is 16.666.
https://en.m.wikipedia.org/wiki/List_of_moments_of_inertia

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I'm asking what principles do I need to understand to solve that type of problems,
Conservation of linear momentum; conservation of angular momentum.

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Offset is shown in picture attached. It's 25cm from side, it is pointed down. And yes. I'm assuming very quick ejection of propellant)
Quick does not matter as it turns out for the feature that @Steve4Physics mentions. No matter how quickly the propellant burns the craft will be changing velocity by the same amount during the burn. But yes, this will be a small discrepancy that we can ignore and instead concentrate on conservation of angular momentum.

So we have 100 grams of exhaust moving at 1000 m/s downrange at an offset of 0.25 meters from the center of mass. How much angular momentum is carried away in the exhaust stream? I gave the formula in #9 above.

vasya
still, i don't understand how to solve it

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