A few sample physics questions for Reiku

Reiku:

Your eigenvectors for Question 1 are correct.

Although there were a couple of gaps in your working, your answers to Question 1 were correct.

Now, let's look at Question 2.

Let's work through your answer...

K is defined in the question. It's just a constant.

There's no mention of or need for Gauss's law here. The question itself gives a hint about what needs to be done right at the start, by citing Newton's shell theorem. You know about that, right?

All you have done here is put \(\mu = GM\), which introduces a new constant for no useful reason, as far as I can see.

Also, your expression does not give the gravitational field stength. More about that in a moment. What you have actually given is something like the gravitational potential, but your expression for that is incorrect too.

No. That's wrong. There's no \(r\) in the given value of \(K\), and you can't just put one in for no reason.

Your first expression for \(g\) is somewhat correct, though not entirely correct for reasons I'll give in a moment. But in the next line you just make an assumption based on nothing. Your second line is also incorrect, because it has no dependence on \(r\).

Why did you make the assumption? You don't need to assume anything in this problem, other that Newton's law of gravity and a few other basic principles of physics.

That's just a repeat of what the question is asking you to show.

Yes. That seems very obvious. But this part of the question should be doable by somebody in the second-last or last year of school. So why do you need to guess? You're writing a book on advanced quantum physics and consciousness etc., aren't you? And you're the guy who said he could skip a lot of the elementary stuff because he already knows it.

That might have got you some marks in an exam if you could justify it, but here you have given no justification at all. It almost looks like you've searched the web for similar expressions to the one for K and then pasted that formula here.

So, overall for this part of the question, this is a fail. You simply haven't done what the question asked you to do.

Here is the solution:

Start with Newton's law of gravity (and the shell theorem mentioned at the start of the question). By the shell theorem, all mass of the Earth outside radius \(r\) doesn't contribute to the gravitational field strength at radius \(r\). Therefore, the Earth's field at radius \(r\) inside the Earth is:

\(g=-\frac{GM(<r)}{r^2}\)

where \(M(<r)\) is the amount of the Earth's mass located within a sphere of radius \(r\) and the minus sign indicates that the field is directed towards the Earth's centre.

Here's where you need to connect to the density. The amount of mass within radius \(r\) is \(M(<r)=\rho V(<r)\), where \(V(<r)\) is the volume of the sphere inside radius \(r\).

And so you need the expression for the volume of a sphere:

\(V=\frac{4\pi r^3}{3}\)

Putting the pieces together we find:

\(g=-\frac{GM(<r)}{r^2} = -G\rho \frac{4 \pi r^3}{3}\frac{1}{r^2} = -G\rho \frac{4 \pi r}{3} = -Kr\)

This is the complete solution.

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Let's look at your solution.

Your answer is wrong. For a start, you have \(M^2\), but what is \(M\)? If it is the mass of the Earth, then it is wrong for two reasons. First, the entire mass of the Earth doesn't come into play due to the shell theorem and the result in part (a). Second, the interaction is not between the Earth and itself but between the Earth and the small mass mentioned in the question.

The correct answer goes as follows:

First, we make use of part (a) of the question. Since the gravitational force on a mass \(m\) is always \(F=mg\), we have, quite simply:

\(F=-mKr\)

If you want to substitute the value of K, we have:

\(F = -m\frac{4 \pi G \rho}{3}r\)

Clearly, this force is linear in \(r\), whereas your answer goes as \(r^{-2}\).

You would score no marks for this part of the question.

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Your answer:

This answer would get you approximately one third of the available marks. Your general description of the force and the motion is correct.

However, you failed to identify the motion as simple harmonic motion, and you did not give a mathematical reason for why the motion would be simple harmonic motion. The reason is that the force is a linear, restoring force - i.e. linear in \(r\).

Your answer:

This is a fundamental error, because the equation you have cited only works if the acceleration \(g\) is constant. In this case, the value of \(g\) varies with position \(r\).

Since you didn't identify the motion as simple harmonic motion in the previous part, really you had no hope of getting this part of the question correct.

Your equating \(g=K\) is wrong for reasons given previously, and in fact should have been obviously wrong to you just from reading part (a) of the question.

Also, you have given no justification for replacing one symbol with another. You look like you're just shuffling symbols at random to try to work towards the expected answer.

The period of simple harmonic motion is independent of the amplitude of the motion, so the period cannot depend on \(r\). But you were just guessing here anyway.

That's right, because the force law here is akin to Hooke's law for a spring, which is how we identify the motion as simple harmonic motion in the first place.

That is, students doing this question were expected to recognise the mathematical signature of simple harmonic motion and apply that to the problem. You missed it completely.

The correct answer here goes like this:

The force law on the small mass, derived above is:

\(F=-mKr\)

We compare this to the force law for a mass on a spring:

\(F=-kx\)

Now, for a mass on a spring, we know from our knowledge of simple harmonic motion that the period of the motion is:

\(T=2\pi\sqrt{\frac{m}{k}}\)

By analogy between the force laws here, we identify:

\(k=mK\)

in the period expression, so the period for our question is:

\(T=2\pi\sqrt{\frac{m}{mK}}=2\pi\sqrt{\frac{1}{K}}\)

But to go from one side of the Earth to the other is only half a period, so the required time is the one given in the question.

For this part of the question, you may have picked up a mark or so for mentioning Hookes' law near the end. But mostly your working is just wrong.
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So, for Question 2 overall, you would have easily failed in terms of the marks allocated to this exam question.

The story so far:

Question 1: correct - essentially full marks for Reiku.
Question 2: fail - Perhaps 3 marks out of 20.

 

How are questions 3, 4 and 5 coming along, Reiku?

 

Reiku:

You probably wouldn't have found it on the web, since it was invented for this question. There are lots of similar questions to this one on the web, though.

And no, I don't think you would have found anything that helped much on the web, unless you stumbled across a complete solution to a similar problem. See, the thing is, exam questions like this are supposed to test your understanding of physics, and not your ability to regurgitate formulas from memory.

The students who took the class and were given this question as an exam question had never seen \(K\) used that way before. Still, a lot of them got the question right.

Um... no you weren't.

That's no problem. Take your time.

The question asked for a proof of the expression for the time taken to fall through the earth. It did not ask for a number. Your derivation of the expression was invalid. Your number was approximately correct, but that's the sort of information that can be googled quite easily. You didn't show how you calculated the number - if you calculated it.


funkstar:

Oh?

It wouldn't be hard to google "eigenvalue" and work from the wikipedia page, I know.

And you're right about the working.

 

IMO the use of k makes the problem easier to conceptualize, because the problem boils down to F=-kr which paralells nicely with Hooks Law F=-kx; which makes sense when you think about it - they're both describing undamped simple harmonic motion, except instead of the elasticity of a spring providing the restoring force, we've got earths gravity providing the restoring force.

 

The "kid" demanded to be tested, after his level of skill and knowledge was questioned and James told him that if he wanted to test himself, he could, so he said yes... to prove his skills, since you know, he is apparently good enough to write a book about all of this and all.

So James is testing him with beginner level basic questions, at his request, that you'd find on a first year physics exam at university (ie. year twelve physics students should be able to understand this).

 

No.

 

Any physics student worth his or her salt will state any assumptions under which his or her answers will or won't apply. There is certainly enough information in the question to give a reasonable answer. And why add complications that aren't mentioned in the question?

Exam questions aren't usually set to "trick" students. They aim to give a fair test of knowledge.

You don't have to participate in this thread. Start your own crazy threads.

Does it matter? It's just a crazy thread. Don't tear your hair out.

Elementary to you, perhaps. But not elementary to Reiku, who failed Question 2.

Also not elementary to the 1st-year university students who actually sat these questions as an exam. They had a reasonably bell-shaped distribution of marks on the exam, ranging from fails to excellent results, with most students falling somewhere in between.

So you're on a personal vendetta, are you?