Collision between asteroids

[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.