4.5 Billion Orbits? How?

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#41 AFJ

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Posted 30 January 2010 - 04:27 AM

The collision that produced the moon almost certainly wasn't head on and I'm pretty sure the theory doesn't involve the mars sized object sticking to earth.Ã‚Â  While the impact would definitely have affected earth orbit there's no reason to suspect it would make orbit around the sun impossible.Ã‚Â  Without a head on collision only a fraction of the objects energy would be transferred to the earth and a significant amount of the energy that did get transferred would have gone into altering the rotation of the earth rather than it's orbit.

I have read Mars size if my memory serves me correct. Irregardless, another just so story that borrows from the fact that the universe is in order. There's alot of stuff going on at the same time and it all fits. But you would say it's all just laws and forces. What dictates those laws all worked together?

I know enough to know that by Newton's third law, all the mass of the moon would have caused a great reaction even if came out perpendicular to the orbit.

Because velocity is both direction and speed, you can change your velocity if only your speed changes, or if only your direction changes, or if both your speed and direction changes.Ã‚Â  In other words it's possible to change velocity without getting any faster or slower.
Gravity is changing the direction of earth's inertia, it isn't diminishing the amount of inertia it has.Ã‚Â  Remember that velocity is both speed and direction, and you can change velocity without changing speed.

That is exactly what I said about velocity. It could be both--in a decaying orbit both speed and direction are changing.

Inertia wants to go straight. When anything pulls against inertia there is resistance--if not there would be no orbit.

Obviously, numbers, you have studied physics more than I. And I never claimed to have background in it. And I am certainly not going to argue against current theories not having the background.

So the earth here orbiting at 67,000 mph. I realize the theory of spinning gasses, condensing, and then asteroids building the earth. How did all these asteroids come together when the earth was small. Were they magnetic?

The inside of the earth is believed to be iron--so it wasn't iron gas--that could only happen inside a star--not spinning gas in space. So it was hypothetically built by asteroids. Many of them it seems would have defied Newton's Third Law when they stuck together. It seems there would have been alot of banging around, but not alot of building.

#42 Guest_tharock220_*

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Posted 30 January 2010 - 10:12 AM

The speed of the earth's rotation on it's axis is being slowed down due to friction between the Earth and the tides(which are caused by the moons)
Has nothing to do with where the moon is in it's rotation around the Earth directly, only indirectly through the tides.

I'm aware of that. It's energy lost in Earth's spin is being transferred to the moon moving it into a higher orbit. I wasn't talking about that. My point was that if the moon was slowing down the Earth's orbit around the sun when it was behin the Earth, it would surely speed up it when it was in front of the Earth. Point being, the moon isn't going to screw up the Earth's orbit around the sun.

#43 Scanman

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Posted 30 January 2010 - 11:52 AM

I'm aware of that.  It's energy lost in Earth's spin is being transferred to the moon moving it into a higher orbit.  I wasn't talking about that.  My point was that if the moon was slowing down the Earth's orbit around the sun when it was behin the Earth, it would surely speed up it when it was in front of the Earth.  Point being, the moon isn't going to screw up the Earth's orbit around the sun.

This appears to be a 'pulling yourself up by your own bootstraps' argument.

The earth-moon combo should be looked at as a paired masses (linked together) in relationship to their orbits around the sun.

Picture this:

You are sitting on a small block of ice in the middle of a perfectly flat ice-skating rink.
Let's also imagine that there is absolutely no friction between the block of ice and the rink and no friction from air.
If you were to wobble yourself back and forth, you would find that the block of ice would move back and forth also...
...but no matter how much you wobble or attempt to 'scoot', you will always find yourself exactly at the same point that you started at.
(In real life, with friction, you could move a little bit.)

The earth-moon combo (regardless of what they do) will continue to orbit the sun at the same average velocity and trajectory.

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Posted 30 January 2010 - 01:46 PM

This appears to be a 'pulling yourself up by your own bootstraps' argument.

The earth-moon combo should be looked at as a paired masses (linked together) in relationship to their orbits around the sun.

Picture this:

You are sitting on a small block of ice in the middle of a perfectly flat  ice-skating rink.
Let's also imagine that there is absolutely no friction between the block of ice and the rink and no friction from air.
If you were to wobble yourself back and forth, you would find that the block of ice would move back and forth also...
...but no matter how much you wobble or attempt to 'scoot', you will always find yourself exactly at the same point that you started at.
(In real life, with friction, you could move a little bit.)

The earth-moon combo (regardless of what they do) will continue to orbit the sun at the same average velocity and trajectory.

Peace

I understand that. My point was that if someone wants to argue the moon would slow the Earth down I could just as easily say it the moon could speed the Earth up as well.

#45 Adam Nagy

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Posted 30 January 2010 - 02:20 PM

You are sitting on a small block of ice in the middle of a perfectly flatÃ‚Â  ice-skating rink.
Let's also imagine that there is absolutely no friction between the block of ice and the rink and no friction from air.
If you were to wobble yourself back and forth, you would find that the block of ice would move back and forth also...
...but no matter how much you wobble or attempt to 'scoot', you will always find yourself exactly at the same point that you started at.

This is false. If you can wobble around and move the ice by shifting your weight, you can also invoke a pattern of motion that would move the ice away from a specific location and even conceivably increase its momentum in a specific direction. Consider the act of pumping your legs to swing on a swing at a playground. Similar principles of motion can also be incorporated to move a block of ice in a frictionless environment. If any movement allows brief momentary momentum, a pattern of movement can cause a linear motion. Your example is a huge fail.

I guess the main problem with your example is that it includes friction by the nature of what is achieved.

#46 Adam Nagy

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Posted 30 January 2010 - 03:24 PM

The earth-moon combo (regardless of what they do) will continue to orbit the sun at the same average velocity and trajectory.

Actually everything that the earth-moon combo does has an equal and opposite reaction. If the Earth were to store up internal pressure until the side facing the direction of travel were to release pressure like a bottle rocket than it would effect its own orbit by the force exerted from that pressure release. So the 'regardless of what they do' scenario, is also a fail.

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Posted 30 January 2010 - 03:47 PM

If the Earth were to store up internal pressure until the side facing the direction of travel were to release pressure like a bottle rocket than it would effect its own orbit by the force exerted from that pressure release. So the 'regardless of what they do' scenario, is also a fail.

Er...okay, but why on earth [ha, a pun!] would it ever do that? I mean, sure, if Megatron leapt into our solar system and started blowing on the moon, it'd start to move out of pattern. If the Asia and Europe landmass broke off from the earth and started flying off into space, the earth would move out of the proposed orbit. If the Death Star blew up the moon, things'd happen. But why would those things ever happen?

#48 Scanman

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Posted 30 January 2010 - 07:05 PM

This is false. If you can wobble around and move the ice by shifting your weight, you can also invoke a pattern of motion that would move the ice away from a specific location and even conceivably increase its momentum in a specific direction. Consider the act of pumping your legs to swing on a swing at a playground. Similar principles of motion can also be incorporated to move a block of ice in a frictionless environment. If any movement allows brief momentary momentum, a pattern of movement can cause a linear motion. Your example is a huge fail.

I guess the main problem with your example is that it includes friction by the nature of what is achieved.

Actually it is not false. Removing friction removes the ability to work horizontally against a stationary body(such as the surface of the ice-skating rink).
Any apparent movement back and forth is centered around the body's/mass's (including the block of ice) center of gravity.
This 'center' of gravity' will not move side to side or back and forth...hiowever, it could move up and down in relationship to the ice rink.
Force exerted by the body/iceblock in any direction/vector is cancelled out by movement around the center of gravity in the opposite direction/vector.
An astronaut free floating in space does not have the ability to affect his position, speed, vector, no matter how hard he kicks and flails with his feet and arms.

The icerink analogy is probably not the best, I gave it because of the greatly reduced friction implied by using ice.
Should I have used teflon? LOL

I like the astronaut analogy best.

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#49 Adam Nagy

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Posted 30 January 2010 - 09:28 PM

An astronaut free floating in space does not have the ability to affect his position, speed, vector, no matter how hard he kicks and flails with his feet and arms.

I'll have to think about this. At the very least an astronaut would be able to achieve somersaults but the linear motion would need some form of propulsion, without a surface like ice to interact with.

#50 Scanman

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Posted 30 January 2010 - 11:07 PM

I'll have to think about this. At the very least an astronaut would be able to achieve somersaults but the linear motion would need some form of propulsion, without a surface like ice to interact with.

I guess the ice analogy, even though I tried to take friction out of the equation, is hard to envision.

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#51 numbers

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Posted 31 January 2010 - 12:22 AM

I know enough to know that by Newton's third law, all the mass of the moon would have caused a great reaction even if came out perpendicular to the orbit.

The mass of the moon (1/80th of earth) only moved about 240,000 miles away, that means the ejection of the moon only moved earth about 3000 miles from it's pre-moon position.

That is exactly what I said about velocity.  It could be both--in a decaying orbit both speed and direction are changing.
Inertia wants to go straight.  When anything pulls against inertia there is resistance--if not there would be no orbit.
Obviously, numbers, you have studied physics more than I. And I never claimed to have background in it.  And I am certainly not going to argue against current theories not having the background.

You say you aren't going to argue against current theories immediately after arguing against current theories. I'm going to try summarizing one last time before letting this go.
1.) Changing the energy of a planet changes the path it takes as it orbits.
2.) We understand and can describe the change in energy associated with any process you'd care to name. Some of the processes you have named don't work the way you think they do.
3.) Small changes in orbit (a few hundred, or a few thousand, or even a few million miles) don't make the earth uninhabitable.
4.) The processes that are affecting the planets (including earth) are so insignificant that the sun will reach it's red giant phase before enough energy has been lost to move the earth's orbit into a uninhabitable area.

So the earth here orbiting at 67,000 mph.  I realize the theory of spinning gasses, condensing, and then asteroids building the earth.  How did all these asteroids come together when the earth was small.  Were they magnetic?

The inside of the earth is believed to be iron--so it wasn't iron gas--that could only happen inside a star--not spinning gas in space.  So it was hypothetically built by asteroids. Many of them it seems would have defied Newton's Third Law when they stuck together.  It seems there would have been alot of banging around, but not alot of building.

http://en.wikipedia....i/Planetesimals
http://en.wikipedia....etary_formation

I'm not going to try and explain planetary formation if you are still having trouble understanding that there's no reason for the earth to not be able to orbit the sun for billions of years. The information you are asking for is available on wikipedia and other sources. Google is your friend.

#52 Adam Nagy

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Posted 31 January 2010 - 06:54 AM

I guess the ice analogy, even though I tried to take friction out of the equation, is hard to envision.

Peace

Yeah, but the astronaut example offered some redemption for you.

#53 AFJ

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Posted 31 January 2010 - 07:55 AM

The mass of the moon (1/80th of earth) only moved about 240,000 miles away, that means the ejection of the moon only moved earth about 3000 miles from it's pre-moon position.

Aren't you forgetting to multiply the velocity at which the moon left the earth? Or are you going to let the present state of things, of which the cause is an assumption, be your guide for calculation? In other words, the impact that was strong enough to knock a moon size chunk out of out of earth had aot of momentum, not just mass.

It's like a bullet--I can throw it at wall and it will bounce off, but if I shoot it, it will go through the wall.

Also the moon leaving us had a reactive impact on the earth--but you don't know what either one these variables are. So how can you do accurate math?

I'm not going to try and explain planetary formation if you are still having trouble understanding that there's no reason for the earth to not be able to orbit the sun for billions of years. .

I have seen the models. Just because I don't know the math, doesn't mean I don't understand the concept of physics and/or formation of earth. So could you please stop with your derogatory insinuations?

You say that, with your knowledge of physics, but yet how can you account for asteroids sticking together, which would defy Newton's Third Law. I don't need math to understand that concept.

#54 numbers

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Posted 31 January 2010 - 11:06 AM

Aren't you forgetting to multiply the velocity at which the moon left the earth?Ã‚Â  Or are you going to let the present state of things, of which the cause is an assumption, be your guide for calculation?Ã‚Â  In other words, the impact that was strong enough to knock a moon size chunk out of out of earth had aot of momentum, not just mass.Ã‚Â

The velocity of the moon leaving the earth doesn't matter since this is a simple center of mass calculation. Jump up and down at any velocity you want. The height you reach and your mass determines how far you are moving the earth at the peak of your jump.

http://en.wikipedia..../World_Jump_Day
It is impossible to permanently change the Earth's orbit using the planet's own mass (which includes that of the world's population) unless such mass is ejected from the Earth at escape velocity (see Newton's third law of motion). The center of gravity of the system containing the earth and its population of humans will remain in the exact same orbit it was always in throughout the jump

We can state quite confidently that the moon didn't reach escape velocity, as it is still here. The center of mass of the earth/moon system is at the same place it was before the lunar ejection (the actual impact of the Mars sized body would have moved the center of mass, but the ejection of the moon did not)

It's like a bullet--I can throw it at wall and it will bounce off, but if I shoot it, it will go through the wall.

AlsoÃ‚Â  the moon leaving us had a reactive impact on the earth--but you don't know what either one these variables are.Ã‚Â  So how can you do accurate math?

It's not like your bullet example at all. By shooting a bullet the gun moves at a velocity opposite the bullet. If the bullet is eventually stopped by the gravity of the gun, then the gun will stop at a distance determined by the bullets final distance and mass.

Em*Ed=Mm*Md

Em = Earth/gun mass
Ed = Earth/gun distance from center of mass
Mm = Moon/bullet mass
Md = Moon/bullet distance from center of mass

You say that, with your knowledge of physics, but yet how can you account for asteroids sticking together, which would defy Newton's Third Law. I don't need math to understand that concept.

Asteroids have gravity. If dust/asteroids aren't moving at escape velocity relative to another asteroid they will stick together. As more and more dust/asteroids stick together escape velocity becomes larger and more and more dust/asteroids stick together.

#55 AFJ

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Posted 31 January 2010 - 05:10 PM

The velocity of the moon leaving the earth doesn't matter since this is a simple center of mass calculation.  Jump up and down at any velocity you want.  The height you reach and your mass determines how far you are moving the earth at the peak of your jump.

http://en.wikipedia..../World_Jump_Day
It is impossible to permanently change the Earth's orbit using the planet's own mass (which includes that of the world's population) unless such mass is ejected from the Earth at escape velocity (see Newton's third law of motion). The center of gravity of the system containing the earth and its population of humans will remain in the exact same orbit it was always in throughout the jump

We can state quite confidently that the moon didn't reach escape velocity, as it is still here.  The center of mass of the earth/moon system is at the same place it was before the lunar ejection (the actual impact of the Mars sized body would have moved the center of mass, but the ejection of the moon did not)
It's not like your bullet example at all.  By shooting a bullet the gun moves at a velocity opposite the bullet.  If the bullet is eventually stopped by the gravity of the gun, then the gun will stop at a distance determined by the bullets final distance and mass.

As for 1/80th of the mass of the earth, this is apparently be assumption based on the hypothesis we are now discussing:

The Moon has approximately 1/4 Earth's diameter, 1/50 Earth's volume, and 1/80 Earth's mass. Earth is very dense overall (it is the densest planet in the Solar System), but the Moon is light for its size. The difference is partly because Earth has a large core of iron and other heavy metallic elements, while the Moon has only a small core, if it has a core at all. The Moon's surface gravity is 1/6 of Earth's, and escape velocity from the surface is about 1/5 of Earth's. Source 1

Since no one has drilled through the earth or moon, here are the facts we can see. The moon is 1/4th the diameter of earth. It's surface gravity is 1.62 m/s^2 while the earth's is 9.78 m/s^2. So it's diameter is 25% and it's surface gravity is 16.5% of the Earth's.
Sourced from source 1.

If the moon is 1/80th of the mass of the Earth, why does it have 1/6th of it's gravity?

Asteroids have gravity.  If dust/asteroids aren't moving at escape velocity relative to another asteroid they will stick together.  As more and more dust/asteroids stick together escape velocity becomes larger and more and more dust/asteroids stick together.

Here's a source I found, but perhaps they forgot to tell us about the asteroids sticking together in the asteroid belt between Mars and Jupiter. But then if this isn't happening you only have an unproven hypothesis.

"Asteroid belt
Thousands of asteroids swarm across the 20 million miles of space between the planets Mars and Jupiter. This 'asteroid belt' marks the junction between the inner and outer Solar System and houses 90 - 95% of all asteroids.

Others orbit close to the Sun and some have been captured by the gravity of planets like Jupiter, Mars and the Earth.

Asteroids with moons
Some asteroids have others revolving around them, just as the Moon revolves around the Earth. Ida, an asteroid about 56km in diameter has its own moon - a tiny body only 1km in size. Other asteroids may well have moons of their own waiting to be discovered.

Binary asteroids
Some asteroids travel in pairs, spinning around a common centre of gravity. These are called 'binaries'. Astronomers were surprised by this, because they thought their gravity would be too weak to bind them together." Source 2

#56 AFJ

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Posted 31 January 2010 - 05:12 PM

The velocity of the moon leaving the earth doesn't matter since this is a simple center of mass calculation.Ã‚Â  Jump up and down at any velocity you want.Ã‚Â  The height you reach and your mass determines how far you are moving the earth at the peak of your jump.

http://en.wikipedia..../World_Jump_Day
It is impossible to permanently change the Earth's orbit using the planet's own mass (which includes that of the world's population) unless such mass is ejected from the Earth at escape velocity (see Newton's third law of motion). The center of gravity of the system containing the earth and its population of humans will remain in the exact same orbit it was always in throughout the jump

We can state quite confidently that the moon didn't reach escape velocity, as it is still here.Ã‚Â  The center of mass of the earth/moon system is at the same place it was before the lunar ejection (the actual impact of the Mars sized body would have moved the center of mass, but the ejection of the moon did not)
It's not like your bullet example at all.Ã‚Â  By shooting a bullet the gun moves at a velocity opposite the bullet.Ã‚Â  If the bullet is eventually stopped by the gravity of the gun, then the gun will stop at a distance determined by the bullets final distance and mass.

The bullet illustration is analogous to the asteroid collision which hypothetically caused the moon, not the reaction of the moon sized chunk. You do not have it's (the asteroid's) mass, velocity, nor angle of collision by which to work the math--and have therefore supposed an answer.

As for 1/80th of the mass of the earth (the moon), this is apparently an assumption based on the hypothesis we are now discussing:

The Moon has approximately 1/4 Earth's diameter, 1/50 Earth's volume, and 1/80 Earth's mass.....The difference is partly because Earth has a large core of iron...while the Moon has only a small core, if it has a core at all. [In other words, they don't know] The Moon's surface gravity is 1/6 of Earth's.... Source 1

Since no one has drilled through the earth or moon, here are the facts we can see. The moon is 1/4th the diameter of earth. It's surface gravity is 1.62 m/s^2 while the earth's is 9.78 m/s^2. So it's diameter is 25% and it's surface gravity is 16.5% of the Earth's.
Sourced from source 1.

If the moon is 1/80th of the mass of the Earth, why does it have 1/6th of it's gravity????

Asteroids have gravity.Ã‚Â  If dust/asteroids aren't moving at escape velocity relative to another asteroid they will stick together.Ã‚Â  As more and more dust/asteroids stick together escape velocity becomes larger and more and more dust/asteroids stick together.

Here's a source I found, but perhaps they forgot to tell us about the asteroids sticking together in the asteroid belt between Mars and Jupiter. But then, if this isn't happening, you only have an unproven hypothesis.

"Asteroid belt
Thousands of asteroids swarm across the 20 million miles of space between the planets Mars and Jupiter. This 'asteroid belt' marks the junction between the inner and outer Solar System and houses 90 - 95% of all asteroids.

Others orbit close to the Sun and some have been captured by the gravity of planets like Jupiter, Mars and the Earth.

Asteroids with moons
Some asteroids have others revolving around them, just as the Moon revolves around the Earth. Ida, an asteroid about 56km in diameter has its own moon - a tiny body only 1km in size. Other asteroids may well have moons of their own waiting to be discovered.

Binary asteroids
Some asteroids travel in pairs, spinning around a common centre of gravity. These are called 'binaries'. Astronomers were surprised by this, because they thought their gravity would be too weak to bind them together." Source 2

#57 amandarandom

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Posted 31 January 2010 - 07:42 PM

Since no one has drilled through the earth or moon, here are the facts we can see.  The moon is 1/4th the diameter of earth.  It's surface gravity is 1.62 m/s^2 while the earth's is 9.78 m/s^2.  So it's diameter is 25% and it's surface gravity is 16.5% of the Earth's.
Sourced from source 1.

If the moon is 1/80th of the mass of the Earth, why does it have 1/6th of it's gravity?

Ahh, a (highschool) physics questions, I can do this.
Lets do the basic maths, shall we?

We have basic Newtonian gravitation: F = G*((m1*m2)/r^2)
G is a constant, so we don't need to worry about that. m1 is out testmass, which will be 1, so we can ignore that too.

We are left with F = m2/r^2
If we set m2earth = 1 and rearth = 1 we get
Fearth = 1/1^2 = 1

m2moon = 1/80 and rmoon = 1/4
Fmoon = .0125 / .25^2
Fmoon = .0125 / 0.0625
Fmoon = 0.20

So, on the moon, our 1 unit mass, will weigh 1/5th of what is does on earth.
Now, I assume the difference between 0.20 and .17 is because of sloppy averages and ballpark figures, but really, there's no huge cover-up going on here. Just do the maths, you'll see it adds up.

#58 AFJ

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Posted 31 January 2010 - 08:28 PM

Ahh, a (highschool) physics questions, I can do this.
Lets do the basic maths, shall we?

We have basic Newtonian gravitation:  F = G*((m1*m2)/r^2)
G is a constant, so we don't need to worry about that. m1 is out testmass, which will be 1, so we can ignore that too.

We are left with F = m2/r^2
If we set m2earth = 1  and rearth = 1 we get
Fearth = 1/1^2 = 1

m2moon = 1/80 and  rmoon = 1/4
Fmoon = .0125 / .25^2
Fmoon = .0125 / 0.0625
Fmoon = 0.20

So, on the moon, our 1 unit mass, will weigh 1/5th of what is does on earth.
Now, I assume the difference between 0.20 and .17 is because of sloppy averages and ballpark figures, but really, there's no huge cover-up going on here. Just do the maths, you'll see it adds up.

Thanks amanda, it's not that hard, but you have to know the formulas--and I don't (lol). So let me ask you, would you have a calculation on the momentum it would take to knock a chunk out of earth with the mass of the moon?

#59 numbers

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Posted 01 February 2010 - 05:07 PM

The bullet illustration is analogous to the asteroid collision which hypothetically caused the moon, not the reaction of the moon sized chunk.Ã‚Â  You do not have it's (the asteroid's) mass, velocity, nor angle of collision by which to work the math--and have therefore supposed an answer.Ã‚Â
...
different post
Thanks amanda, it's not that hard, but you have to know the formulas--and I don't (lol). So let me ask you, would you have a calculation on the momentum it would take to knock a chunk out of earth with the mass of the moon?

http://www.lpi.usra....08/pdf/2429.pdf
A mars sized object (~10% earth mass) traveling between 100-110% Earth escape velocity (~12km/sec) would posses a momentum of roughly 4% Earth's current linear momentum. All models of the impact involve a significant amount of the energy going toward rotating the earth (angular momentum), as opposed to changing earth orbit (linear momentum).

Here's a source I found, but perhaps they forgot to tell us about the asteroids sticking together in the asteroid belt between Mars and Jupiter.Ã‚Â  But then, if this isn't happening, you only have an unproven hypothesis.

Examples of asteroids orbiting asteroids

Let's see if I have this correct. You are claiming the idea that asteroids can attract and hold other asteroids via gravity is unproven, and to demonstrate this you are providing examples of asteroids attracting and holding other asteroids via gravity?

If two asteroids are orbiting each other that's the same as saying that they are "stuck together" from the perspective of any object outside their orbit. For actual physical contact 1 of 2 things needs to happen. Either one of the asteroids accumulates enough mass to bring the other into contact, or something hits one of the asteroids to bring it into low enough orbit for contact. Both are unlikely in the present day asteroid belt due to the extreme sparsity of asteroids. There are typically hundreds of thousands of kilometers between asteroids in the belt.

#60 amandarandom

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Posted 01 February 2010 - 05:21 PM

Thanks amanda, it's not that hard, but you have to know the formulas--and I don't (lol).  So let me ask you, would you have a calculation on the momentum it would take to knock a chunk out of earth with the mass of the moon?

that's a bit harder...
I can do the maths on how much energy it takes to separate the earth and the moon to the distance they are now, but that doesn't include energy required to loosen the rock, or the energy lost. It'll give a minimum required energy, but I'm sure there's a lot more maths involved than just separating two masses in a vacuum.

So, here goes...
U = -int (inf->r) (F) dr
F = -(G*m1*m2)/r^2

U = -int (inf->r) (-(G*m1*m2)/r^2) dr

U = G*m1*m2)/r

filling in

G = 6.67300 * 10^-11 m^3 kg^-1 s^-2
m1 = 5.9736 * 10^24 kg   (earth)
m2 = 7.347 7 * 10^22 kg (moon)
r = 384 399 000 m

U = ((6.67300 * 10^-11) * (5.9736 * 10^24) * (7.347 7 * 10^22)) / (384 399 000)
U = 7.61949924 * 10^28 joules

7.62 * 10^28 joules
that's about a three hundred and fifty billion of biggest nuke ever, which was 50 megaton worth of TNT.

Minimum.

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