Off-center collision of two celestial bodies.

[dougettinger]

How is the rotational velocity of a larger celestial body affected if a smaller body strikes it on its equator but off-center longitudinally by "x" degrees ?
Consider that the collision is almost totally inelastic and most of the smaller body is absorbed.

Doug Ettinger
Pittsburgh, PA

[Chris Peterson]

How is the rotational velocity of a larger celestial body affected if a smaller body strikes it on its equator but off-center longitudinally by "x" degrees ?
Consider that the collision is almost totally inelastic and most of the smaller body is absorbed.

Angular momentum is conserved in an inelastic collision. Since you are assuming that the smaller body is absorbed, the resulting combined body will simply have the total angular momentum of the original system. So both its orbit and rotation will be altered- but there's no way to say just how without actually specifying the details of the two bodies up front. In reality, the energy released in such a collision is likely to be huge, so there is a real possibility that both bodies will be completely disrupted, and much of their material may ultimately be lost.

[dougettinger]

How is the rotational velocity of a larger celestial body affected if a smaller body strikes it on its equator but off-center longitudinally by "x" degrees ?
Consider that the collision is almost totally inelastic and most of the smaller body is absorbed.

Angular momentum is conserved in an inelastic collision. Since you are assuming that the smaller body is absorbed, the resulting combined body will simply have the total angular momentum of the original system. So both its orbit and rotation will be altered- but there's no way to say just how without actually specifying the details of the two bodies up front. In reality, the energy released in such a collision is likely to be huge, so there is a real possibility that both bodies will be completely disrupted, and much of their material may ultimately be lost.

My assumptions are that absorption is met because the larger body's interior is very molten and appropriately viscous. The smaller body is much less massive due to its composition having ligher elements. Their velocity vectors would be more oblique to each other rather than opposing each other. Of course, there will be collisional debris, but insignificant for the basic momentum changes.

I am really looking for the idealistic equation that would approximate the final rotational velocity of the system(?) I realize there would also be an orbital change, too. There initial masses and velocity vectors would be proposed.

Doug Ettinger
Pittsburgh, PA

[alter-ego]

There are more details to be defined before a single-equation solution can had. The densities and post-collision distribution of the masses are needed. A simpler, tractable model could be:
1. The smaller colliding mass hits on the equator (equatorial plane perpendicular to the rotation axis).
2. The trajectory also contained in the equatorial plane, and hits at some incident angle to the surface.
2. The smaller mass is not rotating.
3. The masses have the same density.
4. The post-collision mass is spherical.
5. No debri loss (100% inelastic).

If this is along the lines you were thinking, then an ideal equation is straight forward. You are looking for a case where the Center of Gravity location and resultant mass distribution are described in closed (analytic) form. There are other scenarios that have ideal equation solutions, but this is one of the simpler examples.

[dougettinger]

There are more details to be defined before a single-equation solution can had. The densities and post-collision distribution of the masses are needed. A simpler, tractable model could be:
1. The smaller colliding mass hits on the equator (equatorial plane perpendicular to the rotation axis).
2. The trajectory also contained in the equatorial plane, and hits at some incident angle to the surface.
2. The smaller mass is not rotating.
3. The masses have the same density.
4. The post-collision mass is spherical.
5. No debri loss (100% inelastic).

If this is along the lines you were thinking, then an ideal single-equation is straight forward. You are looking for a case where the Center of Gravity location and resultant mass distribution are described in closed (analytic) form. There are other scenarios that have ideal equation solutions, but this is one of the simpler examples.

Yes, the colliding mass hits on the equator perpendicular to the rotation axis. We can also assume that the colliding mass' trajectory is in the equatorial plane and hits at some incident angle to the surface. Consider that the debris loss and the rotation of the impactor are insignificant and are to be ignored. The post collision is roughly spherical. Consider uniform density for each body, but the densities would be different. The larger body would have a density similar to Earth's and the smaller body would have a density similar to Ganymede's.

If you can provide a generic, ideal single-equation or multiple equation solution for this special case, it would be greatly appreciated.

Doug Ettinger
Pittsburgh, PA

[alter-ego]

There are more details to be defined before a single-equation solution can had. The densities and post-collision distribution of the masses are needed. A simpler, tractable model could be:
1. The smaller colliding mass hits on the equator (equatorial plane perpendicular to the rotation axis).
2. The trajectory also contained in the equatorial plane, and hits at some incident angle to the surface.
2. The smaller mass is not rotating.
3. The masses have the same density.
4. The post-collision mass is spherical.
5. No debri loss (100% inelastic).

If this is along the lines you were thinking, then an ideal single-equation is straight forward. You are looking for a case where the Center of Gravity location and resultant mass distribution are described in closed (analytic) form. There are other scenarios that have ideal equation solutions, but this is one of the simpler examples.

Yes, the colliding mass hits on the equator perpendicular to the rotation axis. We can also assume that the colliding mass' trajectory is in the equatorial plane and hits at some incident angle to the surface. Consider that the debris loss and the rotation of the impactor are insignificant and are to be ignored. The post collision is roughly spherical. Consider uniform density for each body, but the densities would be different. The larger body would have a density similar to Earth's and the smaller body would have a density similar to Ganymede's.

If you can provide a generic, ideal single-equation or multiple equation solution for this special case, it would be greatly appreciated.

FYI, per your request.
The result at the bottom is a simple ideal equation; 100% inelastic and no heat energy loss(!) - all tangential kinetic energy is converted to rotational energy. The assumptions are defined (above), and the primary and impactor mass densities are allowed to be different. The absolute parameters you need to input are: impactor velocity, primary body radius, primary angular velocity before impact and incident impact angle (note when theta <0, angular momentum decreases, but negative angular velocity is not possible with this equation). The rest are unitless ratios: mass and density.

[dougettinger]

Thank you very much, alter-ego. If the impact angle, theta, were less than zero, I could simply substract E(k) from the energy conservation equation to obtain the resulting angular velocity. Of course, the solution for w(1) would necessarily be different. Is this correct ? Also, if the primary body has an initial orbital velocity then the final translational kinetic energy would be E(f) = 1/2 M x (v(primary))(2) + 1/2 m x (v(impactor))(2) x cos (theta) assuming the two velocity vectors do not oppose each other. Is this correct ?

I will experiment with real values and see how the final angular velocity is affected.

Doug Ettinger
Pittsburgh, PA

[alter-ego]

Thank you very much, alter-ego. If the impact angle, theta, were less than zero, I could simply substract E(k) from the energy conservation equation to obtain the resulting angular velocity. Of course, the solution for w(1) would necessarily be different. Is this correct ? Also, if the primary body has an initial orbital velocity then the final translational kinetic energy would be E(f) = 1/2 M x (v(primary))(2) + 1/2 m x (v(impactor))(2) x cos (theta) assuming the two velocity vectors do not oppose each other. Is this correct ?

Yes, basically you are right. W1 will be less when the impactor velocity component opposes the primary surface velocity (velocity vectors subtract, theta <0), or will increase when the component velocity vectors add (theta >0). Energy is conserved, but the impact angle, theta, is not defined wrt the primary orbital velocity vector. E(k)xcos(theta) is always the primary radial component of energy, so it's affect on orbital change is unknown with the current definitions. At any point in the orbit, the impactor can hit the equator at any longitude (0 to 360deg) while theta = constant, i.e. we don't know whether the engery vectors add or subtract. Therefore the orbital change is indeterminant. You can surmise that the orbit will be perturbed most likely in its eccentricity and the orbital plane angle. The details of how the orbit changes are not considered here.

[dougettinger]

By adding more initial conditions I should be able to determine orbital changes with conservation of linear momentum and conservation of kinetic energy equations.

A more difficult equation for me that is still required to complete my collision scenario is the case when the impactor strikes the primary body at a certain latitudinal angle either above or below the equator. Assume the angle of impact is normal to the surface and directed toward the center of the primary body. All other initial conditions are similar to the first case except the angle of the impactor and its location of impact. How do you determine the resulting change of tilt in the primary body's spin axis.

This problem is similar to changing the angular momentum vector of a spinning bicycle wheel on an extended axle. But, instead of a human trying to change the axis angle, the force of an impactor changes the angle. The elusive equation need only address the resulting amount of axis tilt and not necessarily address the final angular momentum vector angle using a vector cross product. Can you come my aid once more ? I may be asking too much. But you have to admit that is fun to play God and derive genesis. I hope you will want to play the game.

Doug Ettinger
Pittsburgh, PA

[The Code]

Hi

I understand that the earth is spinning because of the above .

Questions

Why is mars Spinning?

When did the Moon stop spinning And why?

Is it true, another 30 thousand miles out, and the Moon is lost forever?

Mark

[bystander]

Why is mars Spinning?

Same reason all planets spin, conservation of angular momentum. See:
http://asterisk.apod.com/vie ... =8&t=19153
http://asterisk.apod.com/vie ... =8&t=19209

When did the Moon stop spinning And why?

It didn't. Its rotational and orbital periods are such that one side is always facing earth. It's called tidal locking.

[Chris Peterson]

I understand that the earth is spinning because of the above

The Earth is not spinning because of the hypothesized collision that produced the Moon.

Why is mars Spinning?

All objects in the Solar System are spinning because they conserve the angular momentum of the nebula from which the system formed.

When did the Moon stop spinning And why?

It didn't. The Moon rotates on its axis once every 27 days. Because of tidal forces, that period is just the same as its orbital period around the Earth, so from our viewpoint (and only our viewpoint) we don't really notice the spin. But it's there.

Is it true, another 30 thousand miles out, and the Moon is lost forever?

No.

[dougettinger]

I understand that the earth is spinning because of the above

The Earth is not spinning because of the hypothesized collision that produced the Moon.

Why is mars Spinning?

All objects in the Solar System are spinning because they conserve the angular momentum of the nebula from which the system formed.

When did the Moon stop spinning And why?

It didn't. The Moon rotates on its axis once every 27 days. Because of tidal forces, that period is just the same as its orbital period around the Earth, so from our viewpoint (and only our viewpoint) we don't really notice the spin. But it's there.

Is it true, another 30 thousand miles out, and the Moon is lost forever?

No.

These answers are well constructed and quick for all those "spin" questions. Why is Venus hardly spinning and in retrograde ? The rotation period is 243 days and the orbital period is 224 days. What went wrong with Venus in your estimation ? Did some collision almost stop its rotation or did tidal locking by the Sun play some role ?

Doug Ettinger
Pittsburgh, PA

[Chris Peterson]

Why is Venus hardly spinning and in retrograde ? The rotation period is 243 days and the orbital period is 224 days. What went wrong with Venus in your estimation ? Did some collision almost stop its rotation or did tidal locking by the Sun play some role?

That isn't known. Most people think Venus started out with a prograde rotation, and was perturbed or actually impacted, resulting in the axis flipping and producing a retrograde rotation. The planet isn't tidal locked yet, but will be eventually. There hasn't been enough time for Venus to reach its current slow rotation rate by tidal forces alone, assuming that it has always been at its current location- another unknown.

My own view is that questions like this may never be answered. The Solar System is chaotic, and that makes it impossible to trace its history backwards beyond a certain point- all the more so since bodies that were involved in its dynamics are no longer present. At best, we will have a well enough developed theory of planetary system formation to look at a planet like Venus and describe the possible influences on it, but not necessarily the specific ones that were actually present.

[dougettinger]

Chris, you are your own Wikipedia. I am curious as to why theorists think there may have been an axis flipping ? Could not an off-center collision against its rotational direction have stopped its rotation and slightly reversed it. Perhaps that kind of scenario would have destroyed the planet in the mind of physicists ?

How is axis flipping theorized ? Does the older, stronger magnetic couple between Venus and that of another large celestical body passing closely make this happen ? I am also thinking the Earth's crust and mantle was jerked about the liquid core by the amount of the distance between its spin axis and the magnetic poles by a similar magnetic coupling of a close encounter with another celestial body.

Chris, I am never looking for definite answers. I know well that there are none. But any good guess should be plausible within the framework of the known laws of nature. And I am certainly not the expert on the laws of nature. But, I love to study and think about them.

Doug Ettinger
Pittsburgh, PA

[Chris Peterson]

Chris, you are your own Wikipedia.

Thanks, but I should point out that in this discussion, we're kind of in my home territory. My research is heavily focused on meteoritics, which encompasses both orbital dynamics as well as the structure and evolution of the Solar System.

I am curious as to why theorists think there may have been an axis flipping ? Could not an off-center collision against its rotational direction have stopped its rotation and slightly reversed it. Perhaps that kind of scenario would have destroyed the planet in the mind of physicists ?

It is fairly easy to flip a planet's axis. Even without being perturbed by a passing body, most planets have somewhat unstable axes (the Moon keeps the Earth's very stable... luckily for us). But a passing planet can easily produce an angular momentum change that can flip the axis of rotation more than 90°, which means that it switches from prograde to retrograde rotation (or vice versa). However, as you surmise, a collision that transfers enough energy to change the direction of rotation would release enough energy to melt and break apart the planet. It's not impossible that something like that happened; we can't know for sure without closely examining Venusian materials. But most people think the first scenario is much more likely.

How is axis flipping theorized ? Does the older, stronger magnetic couple between Venus and that of another large celestical body passing closely make this happen ? I am also thinking the Earth's crust and mantle was jerked about the liquid core by the amount of the distance between its spin axis and the magnetic poles by a similar magnetic coupling of a close encounter with another celestial body.

It is unrelated to magnetic fields. Even the strongest planetary magnetic fields are insignificant in transferring angular momentum. Gravity is the force that is involved here- it is orders of magnitude stronger.

[The Code]

Hi Chris

mark swain wrote:I understand that the earth is spinning because of the above


The Earth is not spinning because of the hypothesized collision that produced the Moon.

Check this out .
Quote:

A large rock, about the size of Mars, is doomed. It's heading toward Earth, destined for a slightly off-center impact that will set everything that isn't already rotating into a frenzy of spin.

Upon impact, material from the incoming object and from the new Earth is cast into space. A ring of debris orbits the planet, and in an amazingly short amount of time -- about one day -- it begins to coalesce into a satellite. It takes somewhere between 1 and 100 years for the Moon to gather most of the stuff into a ball.

There are other theories for how the Moon was born, but this one is widely accepted as the most plausible.


http://www.space.com/scienceastronomy/m ... 03018.html

Mark

[Chris Peterson]

Quote:

A large rock, about the size of Mars, is doomed. It's heading toward Earth, destined for a slightly off-center impact that will set everything that isn't already rotating into a frenzy of spin.

Upon impact, material from the incoming object and from the new Earth is cast into space. A ring of debris orbits the planet, and in an amazingly short amount of time -- about one day -- it begins to coalesce into a satellite. It takes somewhere between 1 and 100 years for the Moon to gather most of the stuff into a ball.

There are other theories for how the Moon was born, but this one is widely accepted as the most plausible.

Yes, that's the most widely accepted theory regarding the formation of the Moon. It doesn't, however, suggest that the Earth is spinning because of the collision, nor that the Moon's slow spin is a product of that collision. I'm not sure what point you are trying to make.

[The Code]

The point i was trying to make, is the fact the Earth and Mars both have a similar rotation time, yet Mars has no big Moon to slow it down. So why are they the same?

Mark

[Chris Peterson]

The point i was trying to make, is the fact the Earth and Mars both have a similar rotation time, yet Mars has no big Moon to slow it down. So why are they the same?

Just a coincidence. Every planet has its own rotation rate, determined by the details of its formation and history. Tidal drag has slowed the Earth from perhaps as little as 6 hours per day to 24 hours per day. Nobody really knows what the initial rotation rates were for any of the planets. In any case, Mars is quite different from the Earth in terms of size and mass, so different initial rates wouldn't be odd.

[alter-ego]

.... 100% inelastic and no heat energy loss(!)- all tangential kinetic energy is converted to rotational energy.

This statement is plum wrong, and I apologize for that stumble, and a big one at that! The correct statement is: For a 100% inelastic collision, ALL residual energy AFTER Momentum Conservation (Angular + Linear) is converted to Heat. Clearly this will lead to a different analytical result.

Once again, and the last time, I present the same kind of summary analysis, this time conserving momentum instead of energy. There are three bottom-line equations at the end of the attachment below.
- Equation #1 shows the change in angular velocity, different mass densities allowed.
- Equation #2 shows fraction of impactor energy converted linear momentum, mass densities constrained to be equal.
- Equation #3 shows fraction of impactor energy converted to angular momentum, mass densities constrained to be equal.

Eq. #2 & #3 are insightful in that they show just how little impactor energy is converted to momentum. It is undoubtedly clear why high/hyper-velocity impacts can be so devasting, from bullets and flesh to asteroids and planets..

[dougettinger]

Thanks for these corrections. The fraction of types of energy transfer should be somewhat dependent upon the velocity vectors of the two bodies. If the vectors are largely in the same direction then less heat energy is produced as opposed to the two vectors opposing each other.

I will be testing your equations very soon.

Doug Ettinger
Pittsburgh, PA