When two planets collide what is their terminal velocity?

[Robittybob1]

This post results from the problem posed in another thread:

http://www.scienceforums.net/topic/88301-earth-what-is-the-real-age/page-6#entry861043

 

If two planets each 1/2 the mass of the Earth collide what is their terminal velocity?

 

This would depend if one planet was falling toward the Sun so it could have a velocity over and above that caused by the mutual attraction. (So we will have to ignore this additional velocity in the meantime.)

 

I saw one statement that said: https://books.google.co.nz/books?id=0iggAwAAQBAJ&pg=PA427&lpg=PA427&dq=potential+energy+planets+colliding&source=bl&ots=MQYOmb0bj2&sig=cLZQMP-V8pwzhPi220UakLldSVA&hl=en&sa=X&ei=TTYaVZ-iBYXxmAWc8oD4BQ&ved=0CEQQ6AEwBzgK#v=onepage&q=potential%20energy%20planets%20colliding&f=false

 

"the (relative) impact speed is at least as large as the escape velocity"

 

Ve = sqrt(2*G*M)/r)

 

(I can see where Enthalpy converts this to V^2 = 2*G*M/r below)

 

r being the sum radii of the two planets for they would be touching at that point. There would be additional potential energy being released as they merge together.

 

Please, can anyone help me estimate the speed that the 2 half Earth-sized planets would impact each other and what the kinetic energy would do to the mass of the two planets (assume they were on average 1000 degrees Kelvin to start with)?

Edited March 31, 2015 by Robittybob1

[Strange]

Are you assuming they fall towards each other starting from an initial speed of zero?

 

If not, you would need to know their initial speeds (and I suspect for planetary bodies that would be the dominant factor; off the top of my head, I don't think the acceleration due to gravity would add much).

[Robittybob1]

Are you assuming they fall towards each other starting from an initial speed of zero?

 

If not, you would need to know their initial speeds (and I suspect for planetary bodies that would be the dominant factor; off the top of my head, I don't think the acceleration due to gravity would add much).

"the (relative) impact speed is at least as large as the escape velocity" It could be a lot higher than that too.

Let them start at opposite sides of a similar sized orbit around the Sun if you like. Just to see what the acceleration due to gravity would do to start with please. One planet could just catch up with the other so even though the whole process is taking place at something like 30 km/sec it is the relative motion that will determine how much kinetic energy will heat the planets.

 

 

30 km/s

Since the orbital velocity of the Moon about the Earth (1 km/s) is small compared to the orbital velocity of the Earth about the Sun (30 km/s), this never occurs. There are no rearward loops in the Moon's solar orbit.

Edited March 31, 2015 by Robittybob1

[Enthalpy]

About the initial speed difference of the planets:

 

Only objects near enough to the central mass orbit in the same plane, and only if the system is old enough. Around Jupiter and Saturn, the outer moons have random orbit inclinations, half of them being retrograde. Around our Sun, objects of Oort's cloud make a spherical shell. And in our Galaxy, the globular clusters are outside the plane.

 

Objects out of the plane, or with very elliptic orbits, can have speeds very different from Earth's one, in magnitude or direction. We see it when a meteorite comes in our atmosphere or when a near-Earth object passes by.

 

Two half-Earths falling against an other:

 

The Sun makes it complicated. A trailing accelerated planet will pass over the other, a leading braked one will pass below. But let's imagine planets far from any star, since such planets have been observed.

 

Tidal forces will break both planets before they collide, since these forces suffice to make a single planet spherical. But let's imagine a superglue.

 

The attraction force of spheres near to an other is stronger than the distance of the centers tell, because R^-2 is... concave or convex, I don't care - anyway, it increases more steeply than it decreases, so the sum over both volumes exceeds what concentrated masses tell. But let's neglect this as well.

 

So I take M and R for each half-Earth, then the gravitational energy at contact is GM²/2R (put signs if you like) and spreads over M+M as kinetic energies, so each speed is V²=GM/2R. To compare, a small object m falling on one half-Earth M would convert GMm/R into V²=2GM/R. So two identical planets (whether half-Earths or not) get relative to an other (2*V) a bit more speed (due to mass spread over a volume) than an object falling on one planet.

[Robittybob1]

About the initial speed difference of the planets:

 

Only objects near enough to the central mass orbit in the same plane, and only if the system is old enough. Around Jupiter and Saturn, the outer moons have random orbit inclinations, half of them being retrograde. Around our Sun, objects of Oort's cloud make a spherical shell. And in our Galaxy, the globular clusters are outside the plane.

 

Objects out of the plane, or with very elliptic orbits, can have speeds very different from Earth's one, in magnitude or direction. We see it when a meteorite comes in our atmosphere or when a near-Earth object passes by.

 

Two half-Earths falling against an other:

 

The Sun makes it complicated. A trailing accelerated planet will pass over the other, a leading braked one will pass below. But let's imagine planets far from any star, since such planets have been observed.

 

Tidal forces will break both planets before they collide, since these forces suffice to make a single planet spherical. But let's imagine a superglue.

 

The attraction force of spheres near to an other is stronger than the distance of the centers tell, because R^-2 is... concave or convex, I don't care - anyway, it increases more steeply than it decreases, so the sum over both volumes exceeds what concentrated masses tell. But let's neglect this as well.

 

So I take M and R for each half-Earth, then the gravitational energy at contact is GM²/2R (put signs if you like) and spreads over M+M as kinetic energies, so each speed is V²=GM/2R. To compare, a small object m falling on one half-Earth M would convert GMm/R into V²=2GM/R. So two identical planets (whether half-Earths or not) get relative to an other (2*V) a bit more speed (due to mass spread over a volume) than an object falling on one planet.

 

So two identical planets (whether half-Earths or not) get relative to an other (2*V) a bit more speed (due to mass spread over a volume) than an object falling on one planet.

 

Can you prove your last sentence please?

 

If Ve = sqrt(2*G*M)/r)

(I can see where Enthalpy converts this to V^2 = 2*G*M/r )

 

Get rid of the square root by squaring both sides

V^2 = 2*G*M/r

 

Divide BS by 2

 

1/2 * V^2 = G*M/r

Multiply BS by m (mass of incoming body)

1/2 * m * V^2 = G * M * m / r

 

In the case where the two masses are equal and they will form a combined mass the size of the Earth so r = radius of the Earth.

1/2 * m * V^2 = Kinetic Energy = G * M ^ 2 / r

 

Do you agree with that?

Edited March 31, 2015 by Robittybob1

[Robittybob1]

Are you assuming they fall towards each other starting from an initial speed of zero?

 

If not, you would need to know their initial speeds (and I suspect for planetary bodies that would be the dominant factor; off the top of my head, I don't think the acceleration due to gravity would add much).

Using the above equations the relative velocity due to gravitational attraction would be 7.9 km/sec. That is quite fast, but that is if the other planet came in from infinity. If it just starts falling from the other side of the Sun it might not get to be going so fast.

An interesting figure was the Kinetic Energy /unit Kg = 1.56E+07 Joules/kg.

 

 

4. Similarly to calculate how much heat is needed to heat then melt a quantity of rock I assumed that:

- We need to raise the temperature of the rock by 500 C. Virtually all rocks are molten by 1000 C. And the surface temperature of the Earth is 15 C. So rocks at the surface need to be warmed by around 1000 C while those at the greatest depths need hardly be warmed at all. As we move down through the crust the temperature increases approximately linearly with depth. So the warming needed, averaged over the full depth of the crust would be around 500 C.

- The specific heat values vary for different rock types (see page 87) but they are mostly in a range from 0.6 to 1.1 KJoules/(Kg C). So I have used a value of 0.85 K Joules/(Kg C)

- For rock we are melting it not vaporizing it so we need to use the Latent Heat of Fusion. This varies for different types of rock (see Table 4) but is typically around 420 K Joules/Kg.

- So to heat then melt 1 kg of rock requires (0.850 * 500) + 420 = 845 KJoules

Seems too much heat That is enough heat to bring the whole Earth to an average temp of nearly 18,000 degrees!

Rocks would be evaporating at those sort of temperatures.

Edited March 31, 2015 by Robittybob1

[Mordred]

Most of that would be at the surface, in the form of plasma at high enough temperatures. Can't offer much on this particular topic. Never looked at the numbers involved. Though from a glance looks reasonable, (in flight from work to home,so will looker closer on.

 

 

You might try learning how to latex your formulas. Judging from your prior posts, your an avid poster, with a desire to learn. It would be a valuable aspect to learn how to post in latex. It's also easier to read. (Though the formulas above aren't.)

This site has a guide on how,

 

http://www.scienceforums.net/topic/3751-quick-latex-tutorial/#entry53693

 

I find this list handy

 

http://web.ift.uib.no/Teori/KURS/WRK/TeX/symALL.html.

PS (Some latex I still struggle with,lol or as they say "eat the Apple one bite at a time")

Edited April 1, 2015 by Mordred

[Robittybob1]

Most of that would be at the surface, in the form of plasma at high enough temperatures. Can't offer much on this particular topic. Never looked at the numbers involved. Though from a glance looks reasonable, (in flight from work to home,so will looker closer on.

 

 

You might try learning how to latex your formulas. Judging from your prior posts, your an avid poster, with a desire to learn. It would be a valuable aspect to learn how to post in latex. It's also easier to read. (Though the formulas above aren't.)

This site has a guide on how,

 

http://www.scienceforums.net/topic/3751-quick-latex-tutorial/#entry53693

 

I find this list handy

 

http://web.ift.uib.no/Teori/KURS/WRK/TeX/symALL.html.

PS (Some latex I still struggle with,lol or as they say "eat the Apple one bite at a time")

Yes I think you have hit the nail on the head. There was nothing stopping the surfaces, where all the forces would be initially applied, reaching phenomenal temperatures.

But once I start allowing small amounts of material turning to plasma and radiating away I'm never going to estimate the overall average effect.

Looking at sources that mention "vaporization of rock" http://en.wikipedia.org/wiki/Giant_impact_hypothesis

 

There remain several questions concerning the best current models of the giant impact hypothesis, however.[6] The energy of such a giant impact is predicted to have heated Earth to produce a global 'ocean' of magma; yet there is no evidence of the resultant planetary differentiation of the heavier material sinking into Earth's mantle. At present, there is no self-consistent model that starts with the giant impact event and follows the evolution of the debris into a single moon. Other remaining questions include when the Moon lost its share of volatile elements and why Venus, which also experienced giant impacts during its formation, does not host a similar moon.

 

This plasma and radiation is not going to radiate unless it able to escape instantaneously as might happen in an atomic explosion, turning matter into energy.

If all that kinetic energy was releasable as light, that seems impossible to imagine, yet an artistic impression is given http://en.wikipedia.org/wiki/Giant_impact_hypothesis#/media/File:Artist%27s_concept_of_collision_at_HD_172555.jpg

So it might be possible that there is an extreme jet of material escaping sideways. Radiating out in a 360 degree field.

 

Only now am I realizing that this picture shows a small planet impacting a larger one, but if they were both of similar mass sharing the kinetic energy there would be an extreme flattening and material would be flung out into space orthogonal to the line of impact. That would be useless material in the ultimate Moon formation process, for ultimately we are going to need the material to be in the ecliptic and truly averaged on the Moon's final inclination.

So for this type of collision the other planet would have to be sourced out of the Solar System to get the ejected material in the right plane to form the Moon. (That seems to be ruled out just because it is so improbable.)

 

[As an after thought about the play on words of hitting a nail on the head, you sometimes get this effect when hitting nails on their heads they will fracture into small extremely hot pieces of iron flying off from under the hammer head. Sparks are an example of this.]

"Re: What is the heat of vaporization of Earth's crust?"

http://www.madsci.org/posts/archives/mar2002/1015040902.Es.r.html

 

 

What is the amount of heat required to actually vaporize a cubic meter of

the Earth's crust? Around 10 GJ

All that energy would be enough to vaporise 1/8 the mass of the Moon. So vaporisation would be a way to deal will a sizable chunk of the heat.

 

With the incoming relative speed of 7.9 km/sec it would not be hard to imagine liquified rock being blasted away a higher than escape velocity. http://en.wikipedia.org/wiki/Escape_velocity

 

11.2 km/s (approx. 40,320 km/h, or 25,000 mph) is required; (escape velocity from the Earth)

Edited April 1, 2015 by Robittybob1

[Robittybob1]

What would happen if it was a Mars sized planet whacking into a nearly Earth sized planet?

[pavelcherepan]

What would happen if it was a Mars sized planet whacking into a nearly Earth sized planet?

 

From Impact Earth using parameters of Mars and Earth and impact at 40^o angle and 20 km/sec impact velocity:

 

 

Energy before atmospheric entry: 1.19 x 10^32 Joules = 2.84 x 10^16 MegaTons TNT

 

 

The Earth is strongly disturbed by the impact, but loses little mass.

57.93 percent of the Earth is melted

Depending on the direction and location the collision, the impact may make a noticeable change in the tilt of Earth's axis (< 5 degrees).

Depending on the direction and location of impact, the collision may cause a change in the length of the day of up to 229 hours.

The impact does not shift the Earth's orbit noticeably.

[Robittybob1]

 

From Impact Earth using parameters of Mars and Earth and impact at 40^o angle and 20 km/sec impact velocity:

 

 

Does that mean it could just about stop the Earth spinning a day being longer than a week long after impact.

[pavelcherepan]

Does that mean it could just about stop the Earth spinning a day being longer than a week long after impact.

 

But as it said there it really depends on the direction and location of the impact. I would say that 229 hour day scenario was for the impact in the direction exactly against the direction of rotation of the Earth.

[Robittybob1]

 

But as it said there it really depends on the direction and location of the impact. I would say that 229 hour day scenario was for the impact in the direction exactly against the direction of rotation of the Earth.

Ultimately we want the situation that forms the Moon. Now with the Moon orbiting the Earth in the same direction as the Earth spins, that impact scenario did not happen. It would need to hit in a direction that would tend to speed up the rotation of the Earth wouldn't it?

[pavelcherepan]

Ultimately we want the situation that forms the Moon. Now with the Moon orbiting the Earth in the same direction as the Earth spins, that impact scenario did not happen. It would need to hit in a direction that would tend to speed up the rotation of the Earth wouldn't it?

 

Yeah, and that is also one of the reasons why Giant Impact hypothesis is preferred nowadays over many others - it does better job in explaining angular momentum of Earth-Moon system.

[Robittybob1]

 

Yeah, and that is also one of the reasons why Giant Impact hypothesis is preferred nowadays over many others - it does better job in explaining angular momentum of Earth-Moon system.

So what can we put into Impact Earth site to mimic the Giant Impact? We want the impactor to glance off the surface. It has to exchange material and leave 7/8 mass behind and have enough energy to climb out to at least 50,000 km to reform the Moon.

Can the trajectory be made to curve around the Earth so the remnants have enough angular momentum to stay up there?

It is not just a matter of lifting it to 50,000 km but it has to be able to orbit as well.

Each piece of molten ejecta will be cooling in space before regrouping. Will the regrouping material still have enough energy to stratify the Moon into a mantle outer core and inner core?

[Acme]

Computer simulations by A.G.W. Cameron, Harvard College Observatory found here: >> Origin of the Earth and Moon @NASA

 

Other numeric simulations here: >> Simulations of a Late Lunar-Forming Impact

 

Edit: Abstract from Simulations of a Late Lunar-Forming Impact

ABSTRACT

We present results of about 100 hydrodynamic simulations of potential Moon-forming impacts, focusing on the "late impact" scenario in which the lunar forming impact occurs near the very end of Earths accretion (Canup & Asphaug 2001). A new equation of state is utilized that includes a treatment of molecular vapor (M-ANEOS; Melosh 2000). We assess the sensitivity of impact outcome to collision conditions, in particular how the mass, angular momentum, composition and origin (target vs. impactor) of the material placed into circumterrestrial orbit vary with impact angle, speed, impactor-to-target mass ratio, and initial thermal state of the colliding objects. The most favorable conditions for producing a sufficiently massive and iron-depleted protolunar disk involve collisions with an impact angle near 45 degrees and an impactor velocity at infinity < 4 km/sec. For a total mass and angular momentum near to that of the current Earth-Moon system, such impacts typically place about a lunar mass of material into orbits exterior to the Roche limit, with the orbiting material composed of 10 to 30% vapor by mass. In all cases, the vast majority of the orbiting material originates from the impactor, consistent with previous findings. By mapping the end fate (escaping, orbiting, or in the planet) of each particle and the peak temperature it experiences during the impact onto the figure of the initial objects, it is shown that in the most successful collisions, the impactor material that ends up in orbit is that portion of the object that was in general heated the least, having avoided direct collision with the Earth. Using these and previous results as a guide, a continuous suite of impact conditions intermediate to the "late impact" (Canup & Asphaug 2001) and "early Earth" (Cameron 2000, 2001) scenarios is identified that should also produce iron-poor, ~lunar-sized satellites and a system angular momentum similar to that of the Earth-Moon system. Among these, we favor those that leave the Earth > 95% accreted after the Moon-forming impact, implying a giant impactor mass between 0.11 and 0.14 Earth masses.

Edited April 3, 2015 by Acme

[Robittybob1]

Computer simulations by A.G.W. Cameron, Harvard College Observatory found here: >> Origin of the Earth and Moon @NASA

 

Other numeric simulations here: >> Simulations of a Late Lunar-Forming Impact

 

Edit: Abstract from Simulations of a Late Lunar-Forming Impact

I found the NASA article very interesting. Thanks for that.

[Acme]

I found the NASA article very interesting. Thanks for that.

Your welcome. I'm reading the other paper and Cameron -who did the computer simulations at the NASA article- is cited frequently. The paper can be downloaded as a PDF and goes into considerable detail.

[Robittybob1]

Your welcome. I'm reading the other paper and Cameron -who did the computer simulations at the NASA article- is cited frequently. The paper can be downloaded as a PDF and goes into considerable detail.

.... an impactor velocity at infinity < 4 km/sec. ....

 

You might get to tell me what they mean by that?

[pavelcherepan]

I think that refers to the velocity of the impactor relative to the Earth at a distance where effects of Earth's gravity are negligible. The actual velocity of the impact would be higher.

[Robittybob1]

I think that refers to the velocity of the impactor relative to the Earth at a distance where effects of Earth's gravity are negligible. The actual velocity of the impact would be higher.

I had a way of calculating impact velocity on another computer so I'll check it out whether a smaller planet will fall faster. But on the surface all things should fall at the same rate. But when they say from infinity that means it is not falling just toward the Earth but toward the Sun as well (Generally) as could happen in chaotic interaction of 3 or more planetary bodies. I've seen simulations of three body problems and not all of them have a stable pattern.

But if a planet came around the back of Jupiter and was slingshot down toward the Sun in a very eccentric orbit it could get very high velocities as it approached the Earth and the Sun combined.

To have these low velocities it must have started in an orbit close to the Earth (the word was Trojan planet in one video) so it never really starts from infinity and is relatively close to begin with.

But if the Trojan planet is formed at the same distance to the Earth it could have a similar composition in any case, that is my argument.

http://en.wikipedia.org/wiki/Trojan_%28astronomy%29

 

In astronomy, a trojan is a minor planet or natural satellite (moon) that shares an orbit with a planet or larger moon, but does not collide with it because it orbits around one of the two Lagrangian points of stability (trojan points), L4 and L5, which lie approximately 60° ahead of and behind the larger body, respectively. Trojan objects are also sometimes called Lagrangian objects. They are one type of co-orbital object. In this arrangement, the massive star and the smaller planet orbit about their common barycenter. A much smaller mass located at one of the Lagrangian points is subject to a combined gravitational force that acts through this barycenter. Hence the object can orbit around the barycenter with the same orbital period as the planet, and the arrangement can remain stable over time.[1]

This gels with me particularly at the L3 point.

 

http://en.wikipedia.org/wiki/Near-Earth_object

 

On 13 October 1990 an Earth-grazing meteoroid EN131090 was observed above Czechoslovakia and Poland. It was moving with a speed of 41.74 km/s

As seen from the speed quoted if objects start from further out they can get to be much faster.

Edited April 3, 2015 by Robittybob1

[pavelcherepan]

To have these low velocities it must have started in an orbit close to the Earth (the word was Trojan planet in one video) so it never really starts from infinity and is relatively close to begin with.

 

The term "velocity at infinity" is used in hyperbolic trajectory calculations and simply refers to the fact that the object would coast to infinity:

 

[latex]v^2 = v_{esc}^2 + v_\infty^2[/latex]

 

Where [latex]v_{esc}[/latex] is the local escape velocity, [latex]v_\infty[/latex] is the hyperbolic excess velocity or excess velocity at infinity and [latex]v[/latex] is the orbital velocity. At Lagrangian point L4 or L5, where Theia had supposedly formed the escape velocity relative to the Earth would be:

 

[latex]v_{esc} = \sqrt \frac {2GM}{r} = 225 \, m/s[/latex]

 

So using the number in simulation (4 km/s velocity at infinity), when Theia started moving towards the Earth from either of those points its orbital velocity relative to the Earth would be:

 

[latex]v = \sqrt {0.225^2 + 4^2} = 4.006 \, km/s[/latex],

 

and just before the impact with the Earth:

 

[latex]v = \sqrt{11.2^2 + 4^2} = 11.89 \, km/s[/latex]

 

P.S. Latex trial was a success

Edited April 4, 2015 by pavelcherepan

[Robittybob1]

 

The term "velocity at infinity" is used in hyperbolic trajectory calculations and simply refers to the fact that the object would coast to infinity:

 

[latex]v^2 = v_{esc}^2 + v_\infty^2[/latex]

 

Where [latex]v_{esc}[/latex] is the local escape velocity and [latex]v_\infty[/latex] is the hyperbolic excess velocity or excess velocity at infinity.

 

So using the number in simulation (4 km/s velocity at infinity), in the vicinity of the Earth the orbital velocity of Theia would be:

 

[latex]v = \sqrt{11.2^2 + 4^2} = 11.89 \, km/s[/latex]

 

P.S. Latex trial was a success

If is going faster than escape velocity it is most likely falling toward the Sun as well.

[pavelcherepan]

If is going faster than escape velocity it is most likely falling toward the Sun as well.

 

Yes, it is in fact orbiting the Sun, not the Earth, I just used Earth as a reference to calculate relative velocity. The Earth just happened to be in the wrong place at the wrong time

[Robittybob1]

 

Yes, it is in fact orbiting the Sun, not the Earth, I just used Earth as a reference to calculate relative velocity. The Earth just happened to be in the wrong place at the wrong time

So was Theia a trojan planet before hand or not IYO?