Jamie123

If the Earth were hit by a comet...

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I've just come across this question, which was apparently first asked in a Cambridge entrance examination:

Quote

"If half the Earth were taken off by the impulse of a comet, what change would be produced in the moon's orbit?"

Unless I've made any stupid mistakes, I think I know the answer they were looking for. (Though I've made the simplifying assumption that half the Earth simply vanished -  that the impulse on the remaining half and the gravitational effects of the departing fragment can be ignored - which I think is reasonable given that we've not been told anything else about what happened.)

Anyone else think they know?

Edited by Jamie123

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I think your assumptions are correct, otherwise we would need to account for the position of the moon and the velocity of the ripped-away half-earth and comet and whether they would collide..,

But assuming that we ignore all that I think you would see the moon drift away since its current velocity, momentum, and centripetal force would exceed that of the newly reduced gravitational force of the earth.  But hey, I wasn't the kid getting an A in physics and it is 2AM so I'm probably wrong.

Edited by clwnuke
grammar

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44 minutes ago, clwnuke said:

I think your assumptions are correct, otherwise we would need to account for the position of the moon and the velocity of the ripped-away half-earth and comet and whether they would collide..,

But assuming that we ignore all that I think you would see the moon drift away since its current velocity, momentum, and centripetal force would exceed that of the newly reduced gravitational force of the earth.  But hey, I wasn't the kid getting an A in physics and it is 2AM so I'm probably wrong.

That's what I thought too - but the big question is would it drift away permanently, or would it stay in a captive orbit around the remaining fragment of Earth?

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I think one would have to be able to do some physics equations for that.  As I'm a historian and not a physicist, I'm not sure.

I think, it highly depends on whether the change is instantaneous or gradual.  If instantaneous, with the momentum I'd say it could drift away permanently, but it would depend on which vector it was traveling at the time.  If it was on a opposing vector where it would go the opposite direction of the Earth I'd say it is possible it would escape the gravity of the remaining Earth, but if it happened to be on a parallel vector of the halved Earth, it would remain affected by the Earth.  In this situation it would be possible that both the Earth and the Moon would now affect each other more deeply and it would become more of a binary planetoid orbit between the two than a strict orbit of the Moon around the Earth.

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39 minutes ago, JohnsonJones said:

I think one would have to be able to do some physics equations for that.  As I'm a historian and not a physicist, I'm not sure.

I think, it highly depends on whether the change is instantaneous or gradual.  If instantaneous, with the momentum I'd say it could drift away permanently, but it would depend on which vector it was traveling at the time.  If it was on a opposing vector where it would go the opposite direction of the Earth I'd say it is possible it would escape the gravity of the remaining Earth, but if it happened to be on a parallel vector of the halved Earth, it would remain affected by the Earth.  In this situation it would be possible that both the Earth and the Moon would now affect each other more deeply and it would become more of a binary planetoid orbit between the two than a strict orbit of the Moon around the Earth.

I think that the change is instantaneous. (We are not told this explicitly, but if the change were gradual we would need more information to do any calculation.) Also I think we can reasonably model the Moon's regular orbit as circular (which it very nearly is) - in which case its velocity is always at a right angle the line between Earth and Moon. 

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1 hour ago, Jamie123 said:

I think that the change is instantaneous. (We are not told this explicitly, but if the change were gradual we would need more information to do any calculation.) Also I think we can reasonably model the Moon's regular orbit as circular (which it very nearly is) - in which case its velocity is always at a right angle the line between Earth and Moon. 

I'm not a physicist, but what I meant in the case of it's vector is in relation to the Earth's orbit around the Sun.  If, at the moment of halving, the Moon is going in the opposing vector than the Earth (for example, let's give orbits the directions of N, S, E, and W, even though the don't really exist in Space, this can give us a frame of reference) by going E while the Earth's orbit is a Westward direction around the Sun I imagine the Moon breaks from the gravity of the Earth.  On the otherhand, if the Moon at that particular moment of orbit is also going Eastward while the Earth is moving in that particular direction, though the Moon moves away, it is still affected by the Earth's gravity as it is not going pull directly against it in the opposite direction.

For another reference to understand better what I am saying...let's say you have a pen and you drop it.  Gravity pulls it to the Earth.  At the moment of the Earth being halved, if you drop it on the East Side it drops through the Earth as the gravity drops and continues on westward away from the Earth.  If this had the mass of the moon after dropping through the earth it keeps on going as it's going the opposite direction of the Earth's orbit around the Sun.

On the otherhand, once again, gravity affects the pen as you drop it, but you drop it on the West Side and so it travels through the earth and starts away from the gravity.  However, as the Earth is also in that general direction, even as the pen with the Moon Mass has gone as effected by gravity and the rotation of the Earth, it's orbit is still on a vector that is in conjunction with the Earth in a closer relationship, thus it and the Earth are both still affected by the gravity.

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It's an interesting question. I'm not sure what assumptions about the scenario we are supposed to use. First we are assuming the existence of a comet large enough to cleave the earth in two? Then that the half the planet plus the object simply disappear? Making the right assumptions, I suppose I could guess an answer, but it feels more like one of those "guess what I am thinking" questions rather than a real thought experiment.

Do you think there are special style points if you frame your answer in the style of "If You Give a Mouse a Cookie"?

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50 minutes ago, JohnsonJones said:

I'm not a physicist, but what I meant in the case of it's vector is in relation to the Earth's orbit around the Sun.  If, at the moment of halving, the Moon is going in the opposing vector than the Earth (for example, let's give orbits the directions of N, S, E, and W, even though the don't really exist in Space, this can give us a frame of reference) by going E while the Earth's orbit is a Westward direction around the Sun I imagine the Moon breaks from the gravity of the Earth.  On the otherhand, if the Moon at that particular moment of orbit is also going Eastward while the Earth is moving in that particular direction, though the Moon moves away, it is still affected by the Earth's gravity as it is not going pull directly against it in the opposite direction.

For another reference to understand better what I am saying...let's say you have a pen and you drop it.  Gravity pulls it to the Earth.  At the moment of the Earth being halved, if you drop it on the East Side it drops through the Earth as the gravity drops and continues on westward away from the Earth.  If this had the mass of the moon after dropping through the earth it keeps on going as it's going the opposite direction of the Earth's orbit around the Sun.

On the otherhand, once again, gravity affects the pen as you drop it, but you drop it on the West Side and so it travels through the earth and starts away from the gravity.  However, as the Earth is also in that general direction, even as the pen with the Moon Mass has gone as effected by gravity and the rotation of the Earth, it's orbit is still on a vector that is in conjunction with the Earth in a closer relationship, thus it and the Earth are both still affected by the gravity.

I think I get part of what you're saying - at some points on the Moon's orbit it would receive a "slingshot" effect of the Moon's velocity plus the Earth's, while at other's it would be the Moon's velocity minus the Earth's. This is quite true and it is why rockets are generally launched from near the equator in the direction of the Earth's rotation - to get an "assist" from that rotation. However, if we are concerned with moon's orbit relative to the Earth (rather than to the Sun) I don't think this is relevant. For practical purposes we can assume the Earth-Moon system to be in an inertial frame of reference

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18 minutes ago, MrShorty said:

It's an interesting question. I'm not sure what assumptions about the scenario we are supposed to use. First we are assuming the existence of a comet large enough to cleave the earth in two? Then that the half the planet plus the object simply disappear? Making the right assumptions, I suppose I could guess an answer, but it feels more like one of those "guess what I am thinking" questions rather than a real thought experiment.

Do you think there are special style points if you frame your answer in the style of "If You Give a Mouse a Cookie"?

I agree - it's a badly worded question. It would make more sense to say "if a wizard waved a magic wand that made half the mass of the Earth suddenly disappear". To make the situation sound a bit more feasible, the examiner has suggested a situation which creates a lot of ambiguity.

It reminds me a bit of this Kenny Everett sketch: https://www.youtube.com/watch?v=DCNJNAJtpCk

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Assuming this isn't a trick question (there are no comets big enough to take out half of the earth), I think they're just looking for a simple answer.  The simple answer is that the Moon's orbit would slow after the collision, at least if there were still enough gravity to hold the moon in orbit.  I would assume they just want the student to calculate the differences in masses in relationship to orbital equations.

Otherwise the question would be too complicated to answer because we don't know all of the factors, i.e. was 1/2 of the material blown away completely or is some still within the earth's gravitational pull?  How close was the collision to the Moon's position?

In think they're just looking for the simple answer and that you are supposed to calculate the change in orbit to the Moon if the Earth had half of its mass, or if they wanted an even simple answer, that would be that the Moon's orbit would become slower.

 

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6 hours ago, clwnuke said:

But assuming that we ignore all that I think you would see the moon drift away

If the earth lost half of its mass, it would still have about 40 times the mass of the Moon.  This would still be enough to keep the moon in orbit.

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.

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23 minutes ago, Scott said:

Assuming this isn't a trick question (there are no comets big enough to take out half of the earth), I think they're just looking for a simple answer.  The simple answer is that the Moon's orbit would slow after the collision, at least if there were still enough gravity to hold the moon in orbit.  I would assume they just want the student to calculate the differences in masses in relationship to orbital equations.

Otherwise the question would be too complicated to answer because we don't know all of the factors, i.e. was 1/2 of the material blown away completely or is some still within the earth's gravitational pull?  How close was the collision to the Moon's position?

In think they're just looking for the simple answer and that you are supposed to calculate the change in orbit to the Moon if the Earth had half of its mass, or if they wanted an even simple answer, that would be that the Moon's orbit would become slower.

If you make the simplifying assumptions that (i) the Earth does not accelerate around the Sun, (ii) the Moon's original orbit is exactly circular, (iii) the mass of the Moon is negligible relative to that of the Earth so the Earth does not "wobble" due to the Moon's gravity - then the answer I got is so elegant I think it must be what the examiner was looking for. (Though who knows - maybe there were extra marks for pulling the assumptions apart!)

P.S. Having thought about it a bit assumption (ii) may not be necessary to get the same result - but the maths for an elliptical orbit would be a lot more complicated I suspect.

Edited by Jamie123

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10 hours ago, clwnuke said:

I think your assumptions are correct, otherwise we would need to account for the position of the moon and the velocity of the ripped-away half-earth and comet and whether they would collide..,

But assuming that we ignore all that I think you would see the moon drift away since its current velocity, momentum, and centripetal force would exceed that of the newly reduced gravitational force of the earth.  But hey, I wasn't the kid getting an A in physics and it is 2AM so I'm probably wrong.

9 hours ago, Jamie123 said:

That's what I thought too - but the big question is would it drift away permanently, or would it stay in a captive orbit around the remaining fragment of Earth?

CLNuke is correct.  It would fly off.  Consider what causes an object to maintain an orbit.

  • Its tangential velocity is such that its rate of "fall" is equal to the projected change in radius.
  • The fall is dictated by the graviational attraction beetween the two objects.
  • The gravitational attraction is based on the masses of the two objects and the distance between their centers of mass.

Thus when the earth's mass is dramatically and "instantaneously" removed from the system, the gravitational attraction no longer pulls the moon that same distance required to keep it in orbit.  Your follow up question is answered by the following.

  • Gravitational attraction decreases with the square of distance
  • "Required fall" to maintain orbit decreases linearly with the distance.

Gravity loses faster than the required fall.  Thus, the moon shoots off into either the ball of fire or the wall of fire depending on the trajectory at the moment of impact.

BTW, I'm going to have to correct a word that is VERY important in a question like this.  "Impact", not "Impulse".  Impulse is a very important concept in physics and doesn't make sense in this situation.

Edited by Mores

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19 minutes ago, Mores said:

BTW, I'm going to have to correct a word that is VERY important in a question like this.  "Impact", not "Impulse".  Impulse is a very important concept in physics and doesn't make sense in this situation.

I quoted "impulse" directly from the original and (if you'll excuse me contradicting you) I think it makes sense. Impulse means force acting over time and is equal to change of momentum. The force of a comet's impact acting over the duration of that impact would equal the change of momentum of the fragment knocked off the Earth. It's already been pointed out there's no comet big enough to do that but I agree it's not really a very realistic scenario being proposed anyway.

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1 hour ago, Mores said:
  • Gravitational attraction decreases with the square of distance
  • "Required fall" to maintain orbit decreases linearly with the distance.

Gravity loses faster than the required fall. 

This is assuming that the path of the moon remains circular, with a larger radius. The sudden reduction in centripetal force would cause the moon to recede, but its path would no longer be circular. It's exactly the same situation as when a satellite breaks out of low earth orbit when it fires its second stage booster - and enters an elliptical transfer orbit to its higher (permanent) orbit - but it still remains gravitational bound to the Earth. The only difference here is that instead of gaining kinetic energy, the moon suddenly needs less kinetic energy than it already has to maintain its circular path - so it continues in a non - circular path just as the satellite does. The moon's trajectory still bends towards the earth  slowing as it converts its excess kinetic energy to potential energy. But given that exactly half the gravitational force has gone, will it eventually curve around and come back?

Edited by Jamie123
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I wouldn't directly answer the question - instead, I would point out all the faults in the question and explain what information the question lacked and still needed in order to give a precise answer. Being able to give a comprehensive answer to what information is missing and why it is needed to answer the question shows that you have the knowledge required to give an answer and also might motivate the examiners to construct their questions more carefully in future.

Edited by askandanswer

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4 hours ago, Jamie123 said:

I quoted "impulse" directly from the original and (if you'll excuse me contradicting you) I think it makes sense. Impulse means force acting over time and is equal to change of momentum. The force of a comet's impact acting over the duration of that impact would equal the change of momentum of the fragment knocked off the Earth. It's already been pointed out there's no comet big enough to do that but I agree it's not really a very realistic scenario being proposed anyway.

I suppose it is a matter of semantics.  But the problem with "impulse" is that your assumption is incorrect.  You assumed that the missing half "vanished".  Impulse means that it "moved" the other half away.  And if it moved it, then there would be many more variables required to give a reasonable analysis.  Thus, I said that it wouldn't make sense.

4 hours ago, Jamie123 said:

This is assuming that the path of the moon remains circular, with a larger radius. The sudden reduction in centripetal force would cause the moon to recede, but its path would no longer be circular.

Ok, the elliptical orbit is a possibility.  For it to be valid, the change in mass cannot be sufficient enough to cause the moon to then have escape velocity.  So, let's look at that.

  • Currently, the moon's average velocity is 1.02 Km/s.
  • With the earth's current mass, and the moon's current distance, escape velocity is 1.44 Km/s.
  • With half the earth's mass gone, escape velocity would be factored by 0.707.  = 1.02 Km/s. (escape velocity is proportional to the square root of the mass).

Only a few significant figures.  So, it is right on the cusp.  Slight variables come into play.

  • The moon's current orbit is actually an ellipse as it is.  It is very close to a circle.  But it is slightly elliptical.
  • If the earth were hit at its apogee, it "could" be enough to keep it in orbit.
  • If at its perigee, it will go flying off.

But given the vague nature of the question (i.e. not a lot of variables mentioned; the only number we're given is "half"; No idea how many significant figures we're talking about) chances are that the answer is "it would be too close to call".

EDIT:  The more I think about it, the question is more of your knowledge of astronomical trivia.  If you had gone over this in the past, you would already know that the answer would be that it was too close to call.  If you didn't know, then you'd make some explanations of some possibility.

Maybe??

Edited by Mores

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2 hours ago, Mores said:

So, it is right on the cusp.  

You got it! Theoretically - and with the assumptions made - the moon would only just escape. To prove this you only need the inverse square law and the formulas for kinetic energy and centripetal force and the fact that gravity is proportional to mass. Oh - and you need basic integration too. I'm not sure that counts as "astronomical trivia".

Next time I have access to a scanner (probably Monday) I'll scan and post my own solution.

2 hours ago, Mores said:

You assumed that the missing half "vanished".  Impulse means that it "moved" the other half away.  And if it moved it, then there would be many more variables required to give a reasonable analysis. 

If it were "moved away" very rapidly - in a time too short for the moon to significantly respond before the missing portion were too far away to have any gravitational effect, and if none of the momentum were transferred to the remaining portion then it would in effect have "vanished". Its not very likely - for one thing i suspect the "comet" would need to be terrifyingly large and travelling close to the speed of light - but it's the only way (I think) you can get a precise answer out of this - but like you rightly point out that answer has to be qualified.

I think it's quite a good question - though not as far as eliciting a cut-and dried answer - but it does force you to make the best of incomplete information - something a scientist very often needs to do.

Edited by Jamie123

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2 hours ago, Mores said:

I suppose it is a matter of semantics.  But the problem with "impulse" is that your assumption is incorrect.  You assumed that the missing half "vanished". 

Also, the comet wouldn't vanish either.  The mass of the comet would fuse with the mass of the earth (minus the debris orbiting the earth).   

Another interesting question:  The fastest comet known traveled at ~300 miles per second.  What would the mass of the comet required have to be to destroy half of the earth?  

Quote

Thus when the earth's mass is dramatically and "instantaneously" removed from the system, the gravitational attraction no longer pulls the moon that same distance required to keep it in orbit. 

How did you calculate this?  I'm genuinely curious rather than debating.

Using Newtonian physics at least, the Earth should still be able to keep the moon in orbit if reduced by 1/2 the size.  What am I missing?

 

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10 hours ago, Jamie123 said:

You got it! Theoretically - and with the assumptions made - the moon would only just escape.

Just like friction, you technically need to "exceed" the resisting force.  So, "only just escape" is a 50/50 probability.

Quote

To prove this you only need the inverse square law and the formulas for kinetic energy and centripetal force and the fact that gravity is proportional to mass. Oh - and you need basic integration too. I'm not sure that counts as "astronomical trivia".

If you're sitting in an examination room, you'd have to have a whole bunch of variables memorized... OR you happen to know that the variables are such that if you removed half of the earth's mass, then it would be right on the cusp.  The fact that it is *right at* that point is quite interesting, and worthy of a trivia question.

Quote

If it were "moved away" very rapidly - in a time too short for the moon to significantly respond before the missing portion were too far away to have any gravitational effect, and if none of the momentum were transferred to the remaining portion then it would in effect have "vanished". Its not very likely - for one thing i suspect the "comet" would need to be terrifyingly large and travelling close to the speed of light - but it's the only way (I think) you can get a precise answer out of this - but like you rightly point out that answer has to be qualified.

I think it's quite a good question - though not as far as eliciting a cut-and dried answer - but it does force you to make the best of incomplete information - something a scientist very often needs to do.

We can put all the conditions we want until we end up having a completely different question than we started off with.

Bottom line is that these arguments are about the usage of a word which, according to Scott's post, was not scientifically specific by today's standards because of the date of the original question.

Given that non-specificity, the question is really asking: Considering all other things being equal, "What would happen to the moon's orbit if the earth's mass were half what it is today?"

Edited by Mores

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9 hours ago, Scott said:

How did you calculate this?  I'm genuinely curious rather than debating.

Using Newtonian physics at least, the Earth should still be able to keep the moon in orbit if reduced by 1/2 the size.  What am I missing?

I'm guessing you missed my response to our Limey friend about the elliptical orbit.

If you did read it, then what are you talking about that's different?

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1 hour ago, Mores said:

you're sitting in an examination room, you'd have to have a whole bunch of variables memorized..

Not at all. You can prove it from first principles without memorizing any data whatsoever. I'll attempt the proof textually: Before impact the moon is in a circular orbit so the centripetal force mv^2/r=GMm/r^2 where v is the moon's velocity, r is orbital radius, m is the Moon's mass and M the Earth's mass and G the gravitational constant. This gives us v^2=GM/r so the Moon's kinetic energy mv^2/2=GMm/2r. Now the energy needed to escape the Earth is the integral between r and infinity of GMm/r^2 which is GMm/r, so we halve this (since only half the Earth is left) to get GMm/2r - which by a startling coincidence is exactly how much kinetic energy the moon has.

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2 hours ago, Jamie123 said:

Not at all. You can prove it from first principles without memorizing any data whatsoever. I'll attempt the proof textually: Before impact the moon is in a circular orbit so the centripetal force mv^2/r=GMm/r^2 where v is the moon's velocity, r is orbital radius, m is the Moon's mass and M the Earth's mass and G the gravitational constant. This gives us v^2=GM/r so the Moon's kinetic energy mv^2/2=GMm/2r. Now the energy needed to escape the Earth is the integral between r and infinity of GMm/r^2 which is GMm/r, so we halve this (since only half the Earth is left) to get GMm/2r - which by a startling coincidence is exactly how much kinetic energy the moon has.

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.

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