How Does Angular Momentum Conservation Affect Asteroid Collision Dynamics?

[Hak]

Thanks!

 

[Hak]

Sorry to reopen this thread, already very full of posts, it will be for a very short time. We have been arguing for a very long time about the reasonableness of the approximation $m \ll M$, but the author of the problem told me:

"I actually never meant $m \ll M$ and in my solution I never used it. I was just saying in the text that the second asteroid is much smaller, in the sense of space size. But it stands to reason that unless they are made of materials of very different densities, which is not particularly realistic, we can assume that the mass is also much smaller...."

At this point, I cannot understand the meaning of the phrase "much smaller," which I think is a confusing statement. Therefore, I would like to ask if you could help me clarify the conditions under which it is possible to make the observation $m \ll M$. What do densities have to do with it? How is it possible to understand when it is possible to advance $m \ll M$? Thanks.

 

[erobz]

At this point, I cannot understand the meaning of the phrase "much smaller," which I think is a confusing statement.

They already stated it was meant to mean "much smaller spatially". They over explained $r \ll R$, and in doing so caused confusion. It happens.

Therefore, I would like to ask if you could help me clarify the conditions under which it is possible to make the observation $m \ll M$. What do densities have to do with it?

If the smaller asteroid object is very dense, it can be small...spatially, while still being significant in its mass, thus it can imbed without significantly distorting the shape of the larger asteroid.

 

[kuruman]

From what I read, I am guessing that the author meant that one should consider the asteroid as a point mass $m$ of negligible spatial extent, e.g. a lump of neutron star density hitting the planet. If the asteroid were a sphere of radius $r$ embedded at some distance from the center of the planet, the ensuing non-uniform mass density of the combined object would make the calculation of the CM and the moment of inertia unnecessarily complicated. It seems that the author didn't want you to go there and I agree.

 

[Hak]

If the smaller asteroid object is very dense, it can be small...spatially, while still being significant in its mass, thus it can imbed without significantly distorting the shape of the larger asteroid.

From what I read, I am guessing that the author meant that one should consider the asteroid as a point mass $m$ of negligible spatial extent, e.g. a lump of neutron star density hitting the planet. If the asteroid were a sphere of radius $r$ embedded at some distance from the center of the planet, the ensuing non-uniform mass density of the combined object would make the calculation of the CM and the moment of inertia unnecessarily complicated. It seems that the author didn't want you to go there and I agree.

Thank you very much. I understand the argument of regarding the asteroid as having a certain mass while considering it spatially smaller than the planet. I, however, would like to know another thing: under what conditions is it possible to consider, instead, the mass $m$ negligible compared to the mass $M$? The author said he didn't mean it, but would such an observation fit. Here, when is it possible to make it, generally speaking, in such a situation? Thank you.

 

[erobz]

Thank you very much. I understand the argument of regarding the asteroid as having a certain mass while considering it spatially smaller than the planet. I, however, would like to know another thing: under what conditions is it possible to consider, instead, the mass $m$ negligible compared to the mass $M$? The author said he didn't mean it, but would such an observation fit. Here, when is it possible to make it, generally speaking, in such a situation? Thank you.

In my opinion, to ignore $m$ ( or imply its negligible w.r.t. $M$ ), just means an impulse was given to the larger asteroid, by the small one.

 

[Hak]

To ignore $m$ ( or imply its negligible w.r.t. $M$ ), just means an impulse was given to the larger asteroid, by the small one.

In what sense "a pulse"? Sorry, I don't understand. Also, the author says that, "unless the two bodies are of very different densities, there can be $m \ll M$." What is that supposed to mean? Thank you.

 

[erobz]

In what sense "a pulse"?

I mean , you don't account for the combined center of mass axis of rotation. The center of mass is the center of the large asteroid.

An angular impulse of magnitude $mvR$ is given to the larger asteroid, nothing else though.

 

[Hak]

I mean , you don't account for the combined center of mass axis of rotation. The center of mass is the center of the large asteroid.

An angular impulse of magnitude $mvR$ is given to the larger asteroid, nothing else though.

Ok, I understand now. But what about the densities? I cannot understand why...

 

[Frabjous]

You have a problem. Your skill at manipulating equations is greater than your understanding of them. You need to start thinking about the meaning of what you are manipulating. You have a question about densities. Plug values in and start developing a physical intuition.

 

[Hak]

You have a problem. Your skill at manipulating equations is greater than your understanding of them. You need to start thinking about the meaning of what you are manipulating. You have a question about densities. Plug values in and start developing a physical intuition.

Okay, but what values should I substitute? Density is directly proportional to mass, so the ratio of the density of the smaller asteroid to the density of the larger asteroid should be the smaller the ratio $\frac{m}{M}$ is, right? From this, what should I deduce?

 

[Frabjous]

Okay, but what values should I substitute? Density is directly proportional to mass, so the ratio of the density of the smaller asteroid to the density of the larger asteroid should be the smaller the ratio $\frac{m}{M}$ is, right? From this, what should I deduce?

You need to plug in values and deduce for yourself. What are the minimum and maximum densities that make sense? What is(are) the most probable density(ies)? What do plugging in different values for the small/large asteroid change in the results?

I am not trying to be contrarion, but you are on the path of getting high grades on your homework while getting significantly lower grades on your tests.

 

[Hak]

You need to plug in values and deduce for yourself. What are the minimum and maximum densities that make sense? What is(are) the most probable density(ies)? What do plugging in different values for the small/large asteroid change in the results?

I am not trying to be contrarion, but you are on the path of getting high grades on your homework while getting significantly lower grades on your tests.

Thanks, I will try to enter some values (I haven't figured out which ones yet, though). I wanted to clarify, though, that this is not homework or test preparation, it is just a problem that I am trying to dissect in every way possible and imaginable. The question of masses and densities is just a curiosity of mine, which I am trying to satisfy by learning from you, not a prerequisite of mine (I have not yet dealt with these topics, either in high school or college). The problem has already been solved and I have already understood the physical situation. You are right about this kind of attitude, but this is not my case. I have always gotten high marks in any kind of test, precisely because I try to understand what I am doing (again, I have not yet tackled these topics even in high school). Thank you for your reply, I hope it is understood that I am not a do-nothing...

 

[Hak]

You need to plug in values and deduce for yourself. What are the minimum and maximum densities that make sense? What is(are) the most probable density(ies)? What do plugging in different values for the small/large asteroid change in the results?

First, let us assume that the asteroid has a spherical shape and uniform mass distribution, as you stated. Then, its density $\rho$ is given by $\rho = \frac{M}{\frac{4}{3}\pi R^3}$, where $M$ is its mass and $R$ is its radius. The density of an asteroid depends on its spectral type, which reflects its composition. According to NASA Science, there are three main types of asteroids: C-type (carbonaceous), S-type (silicaceous), and M-type (metallic). The C-type asteroids are the most common, accounting for about 75% of known asteroids. They have a low albedo (reflectivity) and a dark surface. They are composed of clay and silicate rocks, and have an average density of about 1.38 $g/cm^3$. The S-type asteroids are the second most common, accounting for about 17% of known asteroids. They have a moderate albedo and a bright surface. They are composed of silicate minerals and nickel-iron, and have an average density of about 2.71 $g/cm^3$. The M-type asteroids are the rarest, accounting for about 8% of known asteroids. They have a high albedo and a metallic surface. They are composed of pure nickel-iron, and have an average density of about 5.32 $g/cm^3$. These values are based on Wikipedia.

Now, let us consider the collision between the large asteroid and the small asteroid. We can assume that the small asteroid has a negligible radius compared to the large one, as you stated. However, we cannot assume that the small asteroid has a negligible mass compared to the large one, unless we know their densities and relative sizes. For example, if the small asteroid is an M-type asteroid with a high density, it could have a significant mass even if it is much smaller than the large asteroid. On the other hand, if the small asteroid is a C-type asteroid with a low density, it could have a negligible mass even if it is not much smaller than the large asteroid.

To illustrate this point, let us plug in some values for the large and small asteroids. Suppose that the large asteroid has a radius of 10 km and a mass of 1015 kg, which corresponds to a density of about 2.38 $g/cm^3$. This is close to the average density of S-type asteroids. Suppose also that the small asteroid has a radius of 1 km and a mass of 1012 kg, which corresponds to a density of about 5.24 $g/cm^3$. This is close to the average density of M-type asteroids. In this case, the mass ratio between the small and large asteroids is 0.001, which is not negligible. The collision between them could have a significant impact on the large asteroid’s motion and shape.

However, suppose that the small asteroid has a radius of 0.5 km and a mass of 109 kg, which corresponds to a density of about 1.26 $g/cm^3$. This is close to the average density of C-type asteroids. In this case, the mass ratio between the small and large asteroids is 0.000001, which is negligible. The collision between them could have a negligible impact on the large asteroid’s motion and shape.

Therefore, we can see that the condition for considering the mass m negligible compared to the mass M depends on both their densities and relative sizes. A general rule of thumb is that if $\frac{m}{M} \ll \frac{\rho_M}{\rho_m}$, where $\rho_M$ and $\rho_m$ are the densities of the large and small asteroids respectively, then we can consider $m$ negligible compared to $M$.

As for the last question, plugging in different values for the small/large asteroid can change the results in terms of how much momentum and energy are transferred during the collision, how much angular momentum is conserved or changed, how much deformation or fragmentation occurs on both asteroids, and how their orbits are affected by their new velocities and directions.

What can I really gain from this analysis? Thanks.

 

[Frabjous]

First, let us assume that the asteroid has a spherical shape and uniform mass distribution, as you stated. Then, its density $\rho$ is given by $\rho = \frac{M}{\frac{4}{3}\pi R^3}$, where $M$ is its mass and $R$ is its radius. The density of an asteroid depends on its spectral type, which reflects its composition. According to NASA Science, there are three main types of asteroids: C-type (carbonaceous), S-type (silicaceous), and M-type (metallic). The C-type asteroids are the most common, accounting for about 75% of known asteroids. They have a low albedo (reflectivity) and a dark surface. They are composed of clay and silicate rocks, and have an average density of about 1.38 $g/cm^3$. The S-type asteroids are the second most common, accounting for about 17% of known asteroids. They have a moderate albedo and a bright surface. They are composed of silicate minerals and nickel-iron, and have an average density of about 2.71 $g/cm^3$. The M-type asteroids are the rarest, accounting for about 8% of known asteroids. They have a high albedo and a metallic surface. They are composed of pure nickel-iron, and have an average density of about 5.32 $g/cm^3$. These values are based on Wikipedia.

Now, let us consider the collision between the large asteroid and the small asteroid. We can assume that the small asteroid has a negligible radius compared to the large one, as you stated. However, we cannot assume that the small asteroid has a negligible mass compared to the large one, unless we know their densities and relative sizes. For example, if the small asteroid is an M-type asteroid with a high density, it could have a significant mass even if it is much smaller than the large asteroid. On the other hand, if the small asteroid is a C-type asteroid with a low density, it could have a negligible mass even if it is not much smaller than the large asteroid.

To illustrate this point, let us plug in some values for the large and small asteroids. Suppose that the large asteroid has a radius of 10 km and a mass of 1015 kg, which corresponds to a density of about 2.38 $g/cm^3$. This is close to the average density of S-type asteroids. Suppose also that the small asteroid has a radius of 1 km and a mass of 1012 kg, which corresponds to a density of about 5.24 $g/cm^3$. This is close to the average density of M-type asteroids. In this case, the mass ratio between the small and large asteroids is 0.001, which is not negligible. The collision between them could have a significant impact on the large asteroid’s motion and shape.

However, suppose that the small asteroid has a radius of 0.5 km and a mass of 109 kg, which corresponds to a density of about 1.26 $g/cm^3$. This is close to the average density of C-type asteroids. In this case, the mass ratio between the small and large asteroids is 0.000001, which is negligible. The collision between them could have a negligible impact on the large asteroid’s motion and shape.

Therefore, we can see that the condition for considering the mass m negligible compared to the mass M depends on both their densities and relative sizes. A general rule of thumb is that if $\frac{m}{M} \ll \frac{\rho_M}{\rho_m}$, where $\rho_M$ and $\rho_m$ are the densities of the large and small asteroids respectively, then we can consider $m$ negligible compared to $M$.

As for the last question, plugging in different values for the small/large asteroid can change the results in terms of how much momentum and energy are transferred during the collision, how much angular momentum is conserved or changed, how much deformation or fragmentation occurs on both asteroids, and how their orbits are affected by their new velocities and directions.

What can I really gain from this analysis? Thanks.

Nothing, given that you did not actually analyze the problem you stated.

You answer also reads like it was generated by an AI.

 

[Hak]

Nothing, given that you did not actually analyze the problem you stated.

You answer also reads like it was generated by an AI.

Could you give me some advice on how to proceed?

No, the answer is not generated by an artificial intelligence. I only used an Italian-English translator, since I do not master English very well, and researched the numerical values from the sources I cited. I would just appreciate some advice.

 

[Frabjous]

You are given R,M and m.
You are told that r is negligible compared to R.
Maximizing the density ratio $\frac {\rho_{small}} {\rho_{large}}=5.32/1.38= 3.86$

So $MAX(m/M) = \frac {\rho_{small}r^3} {\rho_{large}R^3} = 3.86 \frac {r^3} {R^3}$

What does this say about m/M?

 

[Hak]

You are given R,M and m.
You are told that r is negligible compared to R.
Maximizing the density ratio $\frac {\rho_{small}} {\rho_{large}}=5.32/1.38= 3.86$

So $MAX(m/M) = \frac {\rho_{small}r^3} {\rho_{large}R^3} = 3.86 \frac {r^3} {R^3}$

What does this say about m/M?

It says that $m$ can be negligible with respect to $M$ (i.e., $\frac{m}{M} \ll 1$) iff $\frac{r}{R} \ll 1$?

 

[Frabjous]

Yes. Notice that if r/R = .1 (which is probably too big), m/M=.004 which is close to what you need according to post 130.

You need to become more comfortable with back of the envelope calculations.

At the fun level you might try
Guesstimation by Weinstein and Adam
Back-of-the-envelope Physics by Schwartz

A little more formal
The Art of Insight in Science and Engineering by Mahajan

 

[Hak]

Yes. Notice that if r/R = .1 (which is probably too big), m/M=.004 which is close to what you need according to post 130.

You need to become more comfortable with back of the envelope calculations.

At the fun level you might try
Guesstimation by Weinstein and Adam
Back-of-the-envelope Physics by Schwartz

A little more formal
The Art of Insight in Science and Engineering by Mahajan

Thank you very much, sorry if I was too impetuous. I agree with your critique, it is really the skill in a posteriori calculations that I am trying to acquire. These books are recommended for a posteriori calculations, right? Thank you, anyway.

Anyway, I get that in order for it to be $\frac{m}{M} \le 0.001$, one must have that $\frac{r}{R} \le 0.06$, approximately. Do you confirm?

 

[Frabjous]

Thank you very much, sorry if I was too impetuous. I agree with your critique, it is really the skill in a posteriori calculations that I am trying to acquire. These books are recommended for a posteriori calculations, right? Thank you, anyway.

Anyway, I get that in order for it to be $\frac{m}{M} \le 0.001$, one must have that $\frac{r}{R} \le 0.06$, approximately. Do you confirm?

Numerically that is correct, but there is not a definitive error percentage. The percentage that we desire depends on other things.

According to post 130, the error goes as (1-8μ) so
μ error
.01 8%
.004 3.2%
.001 0.8%

For example, what if we only knew the velocity to 10%?

 

[Hak]

Numerically that is correct, but there is not a definitive error percentage. The percentage that we desire depends on other things.

According to post 130, the error goes as (1-8μ) so
μ error
.01 8%
.004 3.2%
.001 0.8%

For example, what if we only knew the velocity to 10%?

Thank you. What do you mean "the velocity to 10%"?

 

[Frabjous]

Thank you. What do you mean "the velocity to 10%"?

±10%

 

[Hak]

±10%

OK, but I did not understand the point of this question. Your question is about what speed? The angular velocity or the translational velocity (of which one)? "The velocity at 10%" means "suppose there is a 10% error on the velocity just found?". If so, how do I calculate this possibility if I have no numerical value available? Thank you very much for your patience.

 

[Frabjous]

You have a solution that depends on velocity squared. There is then an associated error due to this of around 20%. Aiming for 1% accuracy in your calculated solution is a bit of overkill in this case. I am ignoring the size of r for the purposes of this discussion.

 

[Hak]

You have a solution that depends on velocity squared. There is then an associated error due to this of around 20%. Aiming for 1% accuracy in your calculated solution is a bit of overkill in this case. I am ignoring the size of r for the purposes of this discussion.

So your previous question asks to consider an associated error of 10% due to speed $v$ instead of one of 20%?

 

[Frabjous]

So your previous question asks to consider an associated error of 10% due to speed $v$ instead of one of 20%?

The error in v is 10%. This gives an error in force of around 20%.

Before someone jumps on me, I know there are more accurate approximations.

 

[Hak]

The error in v is 10%. This gives an error in force of around 20%.

I have to try to figure out why the error in $v$ is 10%. I will try to think about it...

 

[Frabjous]

I have to try to figure out why the error in $v$ is 10%. I will try to think about it...

You are measuring the relative velocity of a small body in the depths of space. There will be some arror associated with it. This was not in the original problem, but I did not want you to think that the error percentage was written in stone.

 

[pbuk]

Continuing this discussion is pointless.

If a question says that radius is negligible then it means precisely this and nothing more: you do not need to include the radius anywhere in your calculations.

What it does NOT mean is that you can bring in a load of other information from outside the problem statement to decide whether you can omit any other variables from your calculations and make assumptions about what numerical range of error is permissible in an analytic solution (hint: none). If you do this then you will get the wrong answer.

Of course in this question it is nonsense to assume that the mass of the smaller asteroid is negligible: if it were then its angular momentum would also be negligible and the collision would have no effect on the rotation of the larger asteroid.

You got it wrong because you omitted a term you shouldn't have omitted. Don't waste your and everybody else's time trying to invent a justification for this, learn from your mistake and don't do it next time.

 

[Hak]

Continuing this discussion is pointless.

If a question says that radius is negligible then it means precisely this and nothing more: you do not need to include the radius anywhere in your calculations.

What it does NOT mean is that you can bring in a load of other information from outside the problem statement to decide whether you can omit any other variables from your calculations and make assumptions about what numerical range of error is permissible in an analytic solution (hint: none). If you do this then you will get the wrong answer.

Of course in this question it is nonsense to assume that the mass of the smaller asteroid is negligible: if it were then its angular momentum would also be negligible and the collision would have no effect on the rotation of the larger asteroid.

You got it wrong because you omitted a term you shouldn't have omitted. Don't waste your and everybody else's time trying to invent a justification for this, learn from your mistake and don't do it next time.

I am sorry to contradict you, but I guess you have not followed the whole discussion. I did not omit any terms that should not have been omitted, so I do not see where the error lies. I did not assume mass $m$ to be negligible in an absolute sense, but negligible with respect to mass $M$, which is a very different thing. The result itself is different, since it implies no zero angular momentum. If you follow the discussion from the beginning, you will see that far more authoritative people than myself, such as @haruspex, @Chestermiller, @PeroK and others have, before and better than me, advanced reasoning on the condition $m \ll M$. Only today I was made aware that this condition was not binding, because it was not part of the problem, but nevertheless it is in some respects reasonable, and not meaningless as you say. I do not understand why you made this criticism. I advise you to re-read all the previous posts, in no post is a result of a physical quantity of 0 implied.

 

[Frabjous]

I am sorry to contradict you, but I guess you have not followed the whole discussion. I did not omit any terms that should not have been omitted, so I do not see where the error lies. I did not assume mass $m$ to be negligible in an absolute sense, but negligible with respect to mass $M$, which is a very different thing. The result itself is different, since it implies no zero angular momentum. If you follow the discussion from the beginning, you will see that far more authoritative people than myself, such as @haruspex, @Chestermiller, @PeroK and others have, before and better than me, advanced reasoning on the condition $m \ll M$. Only today I was made aware that this condition was not binding, because it was not part of the problem, but nevertheless it is in some respects reasonable, and not meaningless as you say. I do not understand why you made this criticism.

@Hak Don’t argue with mentors or science advisors. They tend to stick together, and even when you win, you lose. Just move on to new physics questions.

 

[Hak]

@Hak Don’t argue with mentors or science advisors. They tend to stick together, and even when you win, you lose. Just move on to new physics questions.

All right, I apologise for any misbehaviour. It was not my intention to offend anyone, nor to create discord. Sorry again.

 

[erobz]

I wanted to clarify, though, that this is not homework or test preparation, it is just a problem that I am trying to dissect in every way possible and imaginable.

It's just unnecessary. You are going to have plenty of practice and chances to evolve your skillset in the peripheral physics surrounding the problem. It's not like educators give you a single problem each term...you get hundreds! If you are only 99% sure on some part you will get a hundred other chances to dive into it as you progress.

 

[Hak]

It's just unnecessary. You are going to have plenty of practice and chances to evolve your skillset in the peripheral physics surrounding the problem. It's not like educators give you a single problem each term...you get hundreds! If you are only 99% sure on some part you will get a hundred other chances to dive into it as you progress.

Thanks for your advice.

 

[haruspex]

You got it wrong

Why do you say that? In post #206, @Hak found $\omega$ correctly without making any approximation, beyond the given one that the size of the smaller object can be ignored.
The remaining question was how to find the subsequent centripetal force. For that, it is just a matter of finding the common mass centre. Because it had originally been stated that m<<M, it sufficed to take this as the centre of the larger body. It turned out that it had not been given that m<<M, yet the original author had made precisely that assumption.
So, in what sense has @Hak got it wrong?

Don't waste your and everybody else's time trying to invent a justification for this

I see no evidence of @Hak trying to justify any approximations or other doubtful steps. Quite the contrary; he has been trying to understand approximations that have been suggested.