Jamie123

If the Earth were hit by a comet...

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11 minutes ago, Mores said:

Yes, you have to have those numbers memorized.  I don't know how many people go around with those numbers memorized.  Most people just look them up when they are needed.  Remember, this is an ENTRANCE exam.  This was not for people who have already completed a degree and use these numbers all the time.

So, without those numbers memorized, how would you calculate that?  You can't.  It all hinges upon whether or not that change in mass will create a condition where escape velocity is achieved.  And you can't know that without knowing what the relative masses, velocities, and distances are.  You cannot do this problem via a symbolic proof. It is situation specific.

What numbers would you need to memorize? GMm/2r=GMm/2r whatever G, M  m and r are equal to. The argument holds whatever the numeric values are.

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21 minutes ago, Jamie123 said:

What numbers would you need to memorize? GMm/2r=GMm/2r whatever G, M  m and r are equal to. The argument holds whatever the numeric values are.

The question that needs answering is if the moon will have escape velocity or not.  And the question pivots upon the mass of the earth being cut in half.  Without knowing the actual values, we cannot know if halving the earth's mass will cause the moon to achieve escape velocity or not.  The fact that we now have the numbers which tell us that it is right on the cusp should tell you that. 

If the current "v" were just a tad bit bigger, then it would go flying off.  If it were a tad smaller, the orbit would certainly be ok.  A symbolic solution will not tell you that.

If, however, you have the "relationships" between m1, m2, v, and r, then you can calculate whether escape velocity is achieved.  But you would have to have all those variables given in terms of a single variable.  You don't have that.  So, you need numbers to provide you that relationship.

If you don't believe me, go ahead and run through the calculations symbolically using only the generalized formulae you posited earlier.  You'll find that you won't be able to determine orbit if "half" of the current earth's mass were removed.

Edited by Mores

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19 minutes ago, Mores said:

Without knowing the actual values, we cannot know if halving the earth's mass will cause the moon to achieve escape velocity or not.  The

Oh yes we can. We do not need to know the actual kinetic energy of the moon - only how it compares with the energy needed to escape. If the former is equal to or greater than the latter then the moon escapes.

23 minutes ago, Mores said:

A symbolic solution will not tell you that.

It will too.

24 minutes ago, Mores said:

You don't have that.  So, you need numbers to provide you that relationship.

Sorry but you are wrong. The expressions for the energy needed and the energy possessed are identical and will be equal whatever numbers you substitute into them.

27 minutes ago, Mores said:

You'll find that you won't be able to determine orbit if "half" of the current earth's mass were removed.

You won't be able to calculate the exact orbit no, but you can ascertain that it is not a closed orbit and that the moon never comes back 

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Perhaps, as it was an entrance exam, that particular question was not to query whether someone had a formula correct or not, or whether the moon actually would escape, but to see HOW someone would actually formulate and answer the question.

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On 11/29/2019 at 10:15 AM, Scott said:

Charon is roughly 1/8 the mass of Pluto and still orbits Pluto.   They are close enough though that Charon tidal locks Pluto's position, rather than vice versa as is the case with all of the rest of the planets and moons in the Solar System.

Charon (the moon of Pluto) is roughly about half the size of Pluto according to information I read.  I believe the mass of Charon is also roughly half of Pluto's mass as well. 

I also read the following about Pluto and Charon's orbit:  Pluto also keeps its same face toward Charon. Like two dancers embracing, these two constantly face each other as they spin across the celestial dance floor.  Astronomers call this a double tidal lock.

Read information at:  https://courses.lumenlearning.com/astronomy/chapter/pluto-and-charon/

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On 11/29/2019 at 10:32 AM, MormonGator said:

@Vort, you might have an answer for this! 

I think Jamie, Mores, and Scott have covered it pretty well.

From a modern perspective, the question is poorly worded, starting with the laughable (from a modern perspective) idea of a comet knocking half of the earth's mass away. Then there is the confusion as to whether the earth's loss in mass is effectively instantaneous, and if not, whether the mass left the earth along the axis of the moon's orbital plane or whether it was more along the plane. And, of course, the whole thing is silly, because any impact with enough energy to literally strip away half of the earth's mass would undoubtedly create a condition where the remaining earth was reduced to a rubble cloud, one that would then eventually recoalesce, almost certainly including the moon in its body. The moon would likely become the new center of accretion; it could hardly avoid being pelted by very large pieces of the earth's rubble cloud.

Jamie's proposed amendment is much better from a modern perspective: What if Harry Potter magically vanished away half of the earth's mass? As has already been discussed. it looks like the result would be indeterminate, with a good probability that the moon would escape the remnant earth's gravitation. Except, again, that's not a good answer if you take it further. Because the moon will still remain in the same basic orbit as it was with the earth, and the remnant earth will likewise remain in that same orbit. Both orbits will probably become somewhat elliptical, but they'll stay in the same general area, with their orbits intersecting at one or two points. So it's only a matter of time until the moon collides with the remnant earth or rubble cloud, and eventually you'll get a new, much smaller planet where the earth once orbited.

The first rule of solving any physics problem is to know what is being asked for. A probem like this is hard exactly because it's not clear what's being asked for. In this case, the difficulty is more precisely that you're not sure what the assumptions are supposed to be, and until those assumptions are clarified, you're not likely to arrive at a good solution.

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10 hours ago, JohnsonJones said:

Perhaps, as it was an entrance exam, that particular question was not to query whether someone had a formula correct or not, or whether the moon actually would escape, but to see HOW someone would actually formulate and answer the question.

I think you're right - it's a sloppy formulated question, but that can have its advantages. It helps examiners separate the duller candidates - the sort of people who give lazy answers like "how long is a piece of string?" - from those with a spark of inquisitiveness and speculation. A better candidate will think "OK I don't have all the information I would like, but maybe if I make some assumptions I can still go somewhere with this!"

Edited by Jamie123

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OK I'm finally back in my office and can use the scanner again. This was my complete solution...

Untitled.png.a4f00a5035027eeaa4ae8100f287cf03.png

Expressions (1) and (2) are identical, so we know the Moon has just enough energy to escape the Earth regardless of what the actual numbers are. However, Mores was quite right about one thing: the solution is on a knife edge, and the slightest change could affect it. This got me thinking while I was on the bus to work: by assuming M>>m, we are effectively assuming some gigantic Godlike (though massless) "hand" holding the Earth still so only the moon moves. In reality we know the Earth and Moon both revolve around a common centre of gravity. Though this wouldn't cause more than a slight "wobble" in the Earth's position, given the "knife edge" nature of the solution, could this affect the answer?

After a bit of back-of-envelope scribbling I think I know: from moments we find that the radius of the Earth's "wobble" R=(m/M)r, so the gravitational force between the Earth and Moon is now GMm/(r+R)^2=GMm/[r^2(1+m/M)^2] . Equating this (as before) to the centripetal force mv^2/r we find that v^2=GM/[r(1+m/M)^2], so the kinetic energy is now GMm/[2r(1+m/M)^2]. Now the energy needed to escape (given that the Earth's mass is now M/2) is the integral of G(M/2)m/[x^2(1+m/{M/2})^2] between x=r and x=infinity, which is GMm/[2r(1+2m/M)^2], which is slightly less than the kinetic energy - so now the Moon is more likely to escape.  I keep seeing more errors in this - I'll post again when I've had time to sort it out properly.

OK let's try again: I think the previous expression for the Moon's initial kinetic energy is correct: GMm/[2r(1+m/M)^2]. Let's write this as (GMm/2r)/(1+k)^2 where k=m/M. Now when half the Earth/s mass is removed, the initial distance of the moon from the new common centre of gravity becomes r(1+k)/(1+2k) - so for the moon to escape the kinetic energy must be greater than the integral between this and infinity of GMm/[2r^2(1+k)^2] which is (GMm/2r)(1+2k)/(1+k)^3. Dividing this by the kinetic energy we get (1+2k)/(1+k) (approximately 1+k for small k), which is >1 for all k>0. Therefore kinetic energy loses and (all other things being equal) the moon does not escape.

I think that's right now. Our original solution was the special case where k=0, and the ratio of two energies becomes 1.

This is still without numbers, so although we know the moon fails to escape, we don't yet know by what margin. Just for fun let's put some numbers in: the mass of the moon is around 7E22 kg and the earth 6E24 kg, so k is approximately 0.012, so the Moon's kinetic energy is about 1.2% too small for it to escape.

P.S. If Scott is right about this question being 200+ years old, this couldn't have been the original solution since the modern formula for kinetic energy was not developed until 1829. It would be interesting to know how the original candidates attempted it.

Edited by Jamie123
Errors in last part

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On 11/30/2019 at 8:30 PM, Still_Small_Voice said:

Charon (the moon of Pluto) is roughly about half the size of Pluto according to information I read.  I believe the mass of Charon is also roughly half of Pluto's mass as well. 

Where did you read that Charon's mass is roughly half of Pluto's?  It is 1/8 of Pluto's and Charon is only 1.71 times denser than water.  

https://space-facts.com/moons/charon/

Charon isn't even remotely close to half of Pluto's mass and is much less dense.

Most of Charon's mass is ice.  Most of Pluto's mass is rock.
 

Edited by Scott

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21 hours ago, Jamie123 said:

OK I'm finally back in my office and can use the scanner again. This was my complete solution..OK let's try again: I think the previous expression for the Moon's initial kinetic energy is correct: GMm/[2r(1+m/M)^2]. Let's write this as (GMm/2r)/(1+k)^2 where k=m/M. Now when half the Earth/s mass is removed, the initial distance of the moon from the new common centre of gravity becomes r(1+k)/(1+2k) - so for the moon to escape the kinetic energy must be greater than the integral between this and infinity of GMm/[2r^2(1+k)^2] which is (GMm/2r)(1+2k)/(1+k)^3. Dividing this by the kinetic energy we get (1+2k)/(1+k) (approximately 1+k for small k), which is >1 for all k>0. Therefore kinetic energy loses and (all other things being equal) the moon does not escape.

This was similar to what I was coming up with.

Quote

P.S. If Scott is right about this question being 200+ years old, this couldn't have been the original solution since the modern formula for kinetic energy was not developed until 1829. It would be interesting to know how the original candidates attempted it.

Unfortunately the book in Google Books only gives the questions rather than the answers.

Apparently the question came from William Dealtry.   Most of the information online about William Dealtry has to do with his clergy work, but he was the Proxime Accessit at Cambridge in 1796.

What that means is that he was the #2 mathematics undergrad at Cambridge (the top postition of Senior Wrangler is considered as the "the greatest intellectual achievement attainable in Britain)."   See here for 1796 and for Proxime Accessit:

https://en.wikipedia.org/wiki/Senior_Wrangler_(University_of_Cambridge)

If this was just an exam question for a lower level class, I would assume a simple answer such as the orbit would slow or if in a physics/dynamics class something similar to what you have above, but since came from William Dealtry, I assume it would be some sort of proposal to work on and answer?

 

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1 hour ago, Scott said:

What that means is that he was the #2 mathematics undergrad at Cambridge (the top postition of Senior Wrangler is considered as the "the greatest intellectual achievement attainable in Britain)."

That's interesting - I read a book a few years ago called "Einstein's Heroes" and was all about Isaac Newton, Michael Faraday and James Clark Maxwell. Maxwell was the Proxime Accessit (the term the book used was "Second Wrangler") of his year, but what I didn't realize until 2 minutes ago was that the guy who beat him - Edward Routh - was same Routh who discovered the Routh Stability Criterion and the Routh-Hurwitz theorem.

In those days, if you scored first class honours in the mathematics examination you were called a "wrangler" and if you got second or third class honours you were called an "optime". (If you passed without honours you were called a "poll man".) The lowest-scoring successful honours candidate was also called the "wooden spoon"; back in the day they would dangle an ornately-painted wooden spoon over his head as he came out from receiving his degree.

Edited by Jamie123

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On 12/2/2019 at 3:28 AM, Jamie123 said:

OK I'm finally back in my office and can use the scanner again. This was my complete solution...

Untitled.png.a4f00a5035027eeaa4ae8100f287cf03.png

Yes, I see what I was missing. 

We said that we assumed that the moon's orbit was circular.  I didn't realize that the circular orbit had a gravitational attraction that was exactly double of the farthest elliptical orbit before reaching escape velocity.  And, yes, the KE equation is crucial to that understanding.  It was that relationship that made "half the earth's mass" a very important statement.

Well done!

However, if it is any elliptical orbit other than circular, then this symbolic solution wouldn't be possible.  And since the moon's orbit is actually ellpitical (slightly) it would depend where in the orbit it is at the moment of impact.

Edited by Mores

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