When two planets collide what is their terminal velocity?

[pavelcherepan]

unlikely - no all you showed it was unstable. Look at it this way if Theia was at L3 and Earth opposite the Sun both planets are at the L3 Lagrangian point of the other. Earth formed even though it was at the L3 of the other planet, so either way it is possible, but unstable but even then how long would it take to come around behind the Earth it is a total of 942500 km around one half of the orbit

and if it started off at 1 km/year it will still take a while to impact even if in the final stages it is moving at 7 km/sec.

So it is building up mass and speed on its journey too. No one says the Earth and Theia were fully formed planets at the time of collision.

 

[latex]2 \pi r = 2*3.14*1.5*10^{11} m = 9.42 * 10^{12} m [/latex] roughly for the entire orbit and [latex]4.71*10^{12}m[/latex] for half orbit.

 

At say 4 km/s it will take ~37.4 years. At 1 km/s it will be ~149 years. Laughable in geological scales. You can't build much mass within such a short a timeframe as shown in the article for planetary accretion I linked in post #51.

 

And hence the best option is that Theia had formed in stable L4 or L5 where it could accrete mass for a long period of time and then got thrown out of there due to influence of other celestial bodies.

Edited April 5, 2015 by pavelcherepan

[Robittybob1]

 

[latex]2 \pi r = 2*3.14*1.5*10^{11} m = 9.42 * 10^{12} m [/latex] roughly for the entire orbit and [latex]4.71*10^{12}m[/latex] for half orbit.

 

At say 4 km/s it will take ~37.4 years. At 1 km/s it will be ~149 years. Laughable in geological scales. You can't build much mass within such a short a timeframe as shown in the article for planetary accretion I linked in post #51.

 

And hence the best option is that Theia had formed in stable L4 or L5 where it could accrete mass for a long period of time and then got thrown out of there due to influence of other celestial bodies.

Yes I should not have multiplied it by 2 thanks for that correction. *(Kicking myself)*

It would not have started off at 1 km/sec (it starts from basically zero) so how do the figures work out if you start from 1 km/year?

[pavelcherepan]

It would not have started off at 1 km/sec (it starts from basically zero) so how do the figures work out if you start from 1 km/year?

 

Oh, haven't noticed that it was 1 km/year. In that case it would obviously take 471 million years. The relative velocity will only start increasing very-very slowly when Theia gets as close as 50 million km from Earth and even then it would hardly be noticeable. So it's pretty safe for the rough calculation to use 1 km/yr for the entire trip.

 

Impact would occur at more or less 10-11 km/s relative and at ~4.1 bya which largely contradicts all geological evidence.

Edited April 5, 2015 by pavelcherepan

[Robittybob1]

 

Oh, haven't noticed that it was 1 km/year. In that case it would obviously take 471 million years. The relative velocity will only start increasing very-very slowly when Theia gets as close as 50 million km from Earth and even then it would hardly be noticeable. So it's pretty safe for the rough calculation to use 1 km/yr for the entire trip.

 

Impact would occur at more or less 10-11 km/s relative and at ~4.1 bya which largely contradicts all geological evidence.

How can we work that out mathematically using a formula and increase the initial rate to give a realistic time

period of roughly say 300,000 years?

I see you put the decimal point in the wrong place, did you mean 0.41 billion years???

In fact I don't know what you mean by your last sentence. Could you explain it please?

 

Note you can't use the escape velocity of the current Earth as the terminal speed as the Earth then had not gained its final mass at that stage in it development M could be say 75% or thereabouts.

Edited April 5, 2015 by Robittybob1

[pavelcherepan]

I see you put the decimal point in the wrong place, did you mean 0.41 billion years???

 

It said 4.1 billion years ago (bya).

 

 

How can we work that out mathematically using a formula and increase the initial rate to give a realistic time

period of roughly say 300,000 years?

 

Can't you do ratios? If at 1 km/yr the trip of 471 million km takes 471 million years, then in x km/yr it will take 300,000 years. What is x?

 

 

 

In fact I don't know what you mean by your last sentence. Could you explain it please?

 

Relative velocity of the impact will be 10-11 km/s which depends on the escape velocity from Earth's surface and hence on mass of the Earth at that time.

Edited April 5, 2015 by pavelcherepan

[Robittybob1]

 

It said 4.1 billion years ago (bya).

 

 

Can't you do ratios? If at 1 km/yr the trip of 471 million km takes 471 million years, then in x km/yr it will take 300,000 years. What is x?

 

 

Relative velocity of the impact will be 10-11 km/s which depends on the escape velocity from Earth's surface and hence on mass of the Earth at that time.

Note: you can't use the escape velocity of the current Earth as the terminal speed as the Earth then had not gained its final mass at that stage in it development M could be say 0.75 current mass or thereabouts.

I feel uneasy using ratios sorry.

you still haven't explained "Impact would occur.... and at ~4.1 bya which largely contradicts all geological evidence."

What is this evidence you are talking about?

Edited April 5, 2015 by Robittybob1

[pavelcherepan]

Note: you can't use the escape velocity of the current Earth as the terminal speed as the Earth then had not gained its final mass at that stage in it development M could be say 0.75 current mass or thereabouts.

I feel uneasy using ratios sorry.

 

That's why I said 10-11. Actually, using [latex]v_e=\sqrt{2GM/r}[/latex] and using mass at 75% of current we get escape velocity at surface of 9.98 km/s. So then impact velocity will be roughly 10 km/s.

 

 

 

"Impact would occur.... and at ~4.1 bya which largely contradicts all geological evidence."

 

What do you not understand from this? Geological evidence (we have rocks dated at 4.04 bya, plenty of zircon crystals etc.) show that conditions on the planet were different from what you'd expect from 60 million years post impact.

 

By the way, what about:

 

 

I'd really love to see one or all of the following from you:

1) Link or quote from any reference that suggests that Theia could've formed at L3 Lagrangian point.

2) Link or quote that suggests that 10⁵ years is enough to form a Mars-sized planet

3) Calculation that proves that object coming from L3 would impact proto-Earth at higher velocity than object coming from L4 or L5 with same initial velocity.

[Robittybob1]

 

That's why I said 10-11. Actually, using [latex]v_e=\sqrt{2GM/r}[/latex] and using mass at 75% of current we get escape velocity at surface of 9.98 km/s. So then impact velocity will be roughly 10 km/s.

 

 

What do you not understand from this? Geological evidence (we have rocks dated at 4.04 bya, plenty of zircon crystals etc.) show that conditions on the planet were different from what you'd expect from 60 million years post impact.

 

 

60 million years???

 

Theia /θiːə/ is a hypothesized ancient planet in the early Solar System that according to the giant impact hypothesis collided with the Early Earth around 4.533 billion years ago (BYa).[1]

is that right?

http://www.amnh.org/education/resources/rfl/web/essaybooks/earth/cs_zircon_chronolgy.html

 

For example, a few grains of zircon found in the early 1990s in a sandstone from western Australia dates back 4.2–4.3 billion years, and we know from meteorites that the Earth is not much older at 4.56 billion years.

difference? That is an interval of 327 - 227 million year period.

Edited April 5, 2015 by Robittybob1

[pavelcherepan]

60 million years???

4.1*10⁹yr - 4.04*10⁹ = 0.06*10⁹ yr = 60*10⁶ yr.

 

is that right?

 

That's what current models say.

 

difference? That is an interval of 327 - 227 million year period.

 

Difference between what and what? You got me confused here.

 

 

* I can't see a planet forming at L4 or L5. I know objects get lodged there but do they form there? There is more chance of a planet forming at L3 even if it is ultimately unstable.

*100,000 years maybe enough time.

*Even so if a planet did form at L4 or L5 and became unstable and drawn to the Earth it would not impact at a speed similar to an object coming in from infinity.

 

Any evidence for those coming my way?

[Robittybob1]

4.1*10⁹yr - 4.04*10⁹ = 0.06*10⁹ yr = 60*10⁶ yr.

 

 

That's what current models say.

 

 

Difference between what and what? You got me confused here.

 

 

 

Any evidence for those coming my way?

I got my figures from websites 4.533 Gya for Theia impact and 4.2 - 4.3 Gya for the Zircon crystals.

Where did you get your figures?

 

Those other questions 1. Insolvable

2. Will depend on how long it takes Theia to go the half billion kms.

3. Will be in few days time.

[Robittybob1]

.....

Even so if a planet did form at L4 or L5 and became unstable and drawn to the Earth it would not impact at a speed similar to an object coming in from infinity. I read about that yesterday that it was easier to send a spacecraft to a trojan at the L4 or L5 position, so presumably to reverse applies if an object comes in from the same location. Earth has a trojan at the L4 Lagrangian Point.

http://en.wikipedia.org/wiki/2010_TK7

.....

 

 

You can show this yourself simply by taking a mass and dropping it.

Step 1. Drop it from 1 meter. measure its terminal speed.

Step 2. Drop it from 2 meters. measure its terminal speed.

Step 3. Drop it from 3 meters. measure its terminal speed.

 

......

Carry that on to infinity. Always the higher the gravitational potential the higher the terminal speed.

Can you deny that?

OK the change in speed gets less and less as the height increases for the gravitational strength declines in the ratio of 1/(r^2) but it is never zero until you reach what is called infinity.

Edited April 5, 2015 by Robittybob1

[pavelcherepan]

You can show this yourself simply by taking a mass and dropping it.

Step 1. Drop it from 1 meter. measure its terminal speed.

Step 2. Drop it from 2 meters. measure its terminal speed.

Step 3. Drop it from 3 meters. measure its terminal speed.

 

......

Carry that on to infinity. Always the higher the gravitational potential the higher the terminal speed.

Can you deny that?

OK the change in speed gets less and less as the height increases for the gravitational strength declines in the ratio of 1/(r^2) but it is never zero until you reach what is called infinity.

 

Sorry, mate, but you clearly have no idea what you're talking about. Both the planet coming from L3 and the planet coming from L4,5 will start at "the infinity" where gravitational influence of the Earth is negligible and will pass through entire Earth's gravitational well and get the same amount of acceleration. Your example is another logical fallacy where an analogy is used but such analogy is not applicable to the situation. Let's do the maths because you clearly can't.

 

Let's define 5 points - point 1 is the L3 Lagrangian point, point 2 is L4, point 3 is 50 million km away, point 4 is 10 million km away and point 5 is the impact with the Earth. And let's use some reasonable starting velocity ~ 1km/s.

 

Point 1.

r = 3*10¹¹m

escape velocity with regards to the Earth v_e = 44.62 m/s

orbital velocity [latex]v_{orbital}=\sqrt{v_e^2 + v_{\infty}^2} = \sqrt{44.62^2 + 1000^2} = 1000.99 m/s[/latex]

 

Point 2.

r~1.5*10¹¹m

v_e=63.10 m/s

[latex]v_{orbital} = 1001.99 m/s[/latex]

 

Point 3

r = 5*10¹⁰ m

v_e = 109.3 m/s

[latex]v_{orbital}=1005.96 m/s[/latex]

 

Point 4

r = 1*10¹⁰ m

v_e= 244.4 m/s

[latex]v_{orbital}=1029.43 m/s[/latex]

 

Point 5

r~6*10⁶m (from the center of the Earth)

v_e ~ 9977.4 m/s

[latex]v_{orbital}=10027.45 m/s[/latex]

 

 

First of all, can you see that even within 10 million km from the Earth it's gravitational influence is tiny - just adds 30 m/s to orbital speed and also the fact is that co-orbital object starting from L3 will have to go through L4 or L5 on the way to Earth and it doesn't matter if the object just starts from there or it's passing by with the same initial velocity the velocity at L4/5 will be the same.

 

And as always:

 

 

I'd really love to see one or all of the following from you:

1) Link or quote from any reference that suggests that Theia could've formed at L3 Lagrangian point.

2) Link or quote that suggests that 10⁵ years is enough to form a Mars-sized planet

3) Calculation that proves that object coming from L3 would impact proto-Earth at higher velocity than object coming from L4 or L5 with same initial velocity.

[Robittybob1]

 

Sorry, mate, but you clearly have no idea what you're talking about. Both the planet coming from L3 and the planet coming from L4,5 will start at "the infinity" where gravitational influence of the Earth is negligible and will pass through entire Earth's gravitational well and get the same amount of acceleration. Your example is another logical fallacy where an analogy is used but such analogy is not applicable to the situation. Let's do the maths because you clearly can't.

 

Let's define 5 points - point 1 is the L3 Lagrangian point, point 2 is L4, point 3 is 50 million km away, point 4 is 10 million km away and point 5 is the impact with the Earth. And let's use some reasonable starting velocity ~ 1km/s.

 

Point 1.

r = 3*10¹¹m

escape velocity with regards to the Earth v_e = 44.62 m/s

orbital velocity [latex]v_{orbital}=\sqrt{v_e^2 + v_{\infty}^2} = \sqrt{44.62^2 + 1000^2} = 1000.99 m/s[/latex]

 

Point 2.

r~1.5*10¹¹m

v_e=63.10 m/s

[latex]v_{orbital} = 1001.99 m/s[/latex]

 

Point 3

r = 5*10¹⁰ m

v_e = 109.3 m/s

[latex]v_{orbital}=1005.96 m/s[/latex]

 

Point 4

r = 1*10¹⁰ m

v_e= 244.4 m/s

[latex]v_{orbital}=1029.43 m/s[/latex]

 

Point 5

r~6*10⁶m (from the center of the Earth)

v_e ~ 9977.4 m/s

[latex]v_{orbital}=10027.45 m/s[/latex]

 

 

First of all, can you see that even within 10 million km from the Earth it's gravitational influence is tiny - just adds 30 m/s to orbital speed and also the fact is that co-orbital object starting from L3 will have to go through L4 or L5 on the way to Earth and it doesn't matter if the object just starts from there or it's passing by with the same initial velocity the velocity at L4/5 will be the same.

 

Why are you calculating orbital velocity ? I was wanting terminal velocity.

If the Earth and Theia are L3 trojans we already know they have similar orbital velocities to orbit the Sun.

Edited April 5, 2015 by Robittybob1

[pavelcherepan]

Why are you calculating orbital velocity ? I was wanting terminal velocity.

If the Earth and Theia are L3 trojans we already know they have similar orbital velocities to orbit the Sun.

 

Those are velocities with respect to each other. So the velocity of Theia relative to the Earth. OK?

 

I gave you the terminal velocity. Point 5 it is.

Edited April 5, 2015 by pavelcherepan

[MigL]

I haven't gone back and read previous pages, but you and Robbity aren't arguing the same thing Pavel.

Robbity has clearly asked about terminal velocity at impact, while you're arguing orbital speeds.

 

Your calculated escape velocities at different radii ( different depths in the gravitational well ) clearly show the point he was making.

 

Or, I came in halfway through a conversation, haven't a clue what you guys were actually discussing and should keep my nose out of it.

[pavelcherepan]

Or, I came in halfway through a conversation, haven't a clue what you guys were actually discussing and should keep my nose out of it.

 

No, MigL, you're very much welcome and maybe you'll help clear the misunderstanding we're having

 

Robbity has clearly asked about terminal velocity at impact, while you're arguing orbital speeds.

 

I figured that what he really wanted to know was the impact velocity, because the term <Terminal velocity> is hardly even applicable in this situation? And the impact velocity would really (based on what I know) depend on excess velocity and strength of gravity field so I've done calculations for the Earth of 75% current mass and smaller radius. Maybe you know of the other way that can be done?

[Robittybob1]

I haven't gone back and read previous pages, but you and Robbity aren't arguing the same thing Pavel.

Robbity has clearly asked about terminal velocity at impact, while you're arguing orbital speeds.

 

Your calculated escape velocities at different radii ( different depths in the gravitational well ) clearly show the point he was making.

 

Or, I came in halfway through a conversation, haven't a clue what you guys were actually discussing and should keep my nose out of it.

No you are right on the money!

 

No, MigL, you're very much welcome and maybe you'll help clear the misunderstanding we're having

 

 

I figured that what he really wanted to know was the impact velocity, because the term <Terminal velocity> is hardly even applicable in this situation? And the impact velocity would really (based on what I know) depend on excess velocity and strength of gravity field so I've done calculations for the Earth of 75% current mass and smaller radius. Maybe you know of the other way that can be done?

It would be the combination of the Earth's escape velocity minus the escape velocity at distance X, y and Z. The further out the distances X, y and Z the greater the impact velocity (terminal velocity in my way of saying it).

<Terminal velocity> http://en.wikipedia.org/wiki/Terminal_velocity

 

Terminal velocity is the highest velocity attainable by an object in free fall. It occurs once the sum of the drag force (Fd) and buoyancy equals the downward force of gravity (FG) acting on the object. Since the net force on the object is zero, the object has zero acceleration.

That is exactly what we want. We may never get to a situation where there is no more acceleration. So we are just looking for the "highest velocity attainable by an object in free fall" from those distances starting from zero velocity in the normal direction.

Edited April 6, 2015 by Robittybob1

[pavelcherepan]

That is exactly what we want.

 

But what's the point in such a calculation? This is for an object moving in the atmosphere. How long before the impact would Theia move through the atmosphere? Couple seconds? At that velocity atmosphere wouldn't be able to slow it down appreciably. OK, let's do calculations based on current atmospheric characteristics, because I have no idea what the density of atmosphere was back then:

 

[latex]V_t = \sqrt{ \frac {2mg}{\rho AC_d}} = \sqrt{ \frac {2*6.39*10^{23}*9}{1.204* \left( 3.39*10^6 \right) ^2*3.14*0.47}} = 0.25*10^8 \,m/s [/latex]

 

Where I used current atmospheric density at surface of 1.2041, mass and radius of Mars and drag coefficient of sphere (0.47) and a lower g value of ~9 m/s² So the terminal velocity, if you're really wondering about it, is higher than the speed of light about 250 kilometers per second. Just as I said, atmosphere is not capable of slowing down such a massive body appreciably.

 

EDIT: Corrected a lame error in the calculation.

Edited April 6, 2015 by pavelcherepan

[Robittybob1]

 

But what's the point in such a calculation? This is for an object moving in the atmosphere. How long before the impact would Theia move through the atmosphere? Couple seconds? At that velocity atmosphere wouldn't be able to slow it down appreciably. OK, let's do calculations based on current atmospheric characteristics, because I have no idea what the density of atmosphere was back then:

 

[latex]V_t = \sqrt{ \frac {2mg}{\rho AC_d}} = \sqrt{ \frac {2*6.39*10^{23}*9}{1.204* \left( 3.39*10^6 \right) ^2*3.14*0.47}} = 0.25*10^8 \,m/s [/latex]

 

Where I used current atmospheric density at surface of 1.2041, mass and radius of Mars and drag coefficient of sphere (0.47) and a lower g value of ~9 m/s² So the terminal velocity, if you're really wondering about it, is higher than the speed of light about 250 kilometers per second. Just as I said, atmosphere is not capable of slowing down such a massive body appreciably.

 

EDIT: Corrected a lame error in the calculation.

Yes I didn't expect the atmosphere to slow it down much. Now do the calculation I suggested "the combination of the Earth's escape velocity minus the escape velocity at distance X, y and Z."

[pavelcherepan]

Yes I didn't expect the atmosphere to slow it down much. Now do the calculation I suggested "the combination of the Earth's escape velocity minus the escape velocity at distance X, y and Z."

 

Calculation for what?

[Robittybob1]

 

Calculation for what?

With a bit of luck I'm back to work tomorrow, and I'll do it as I promised.

You could have convinced yourself that planets further apart will collide at a higher impact speed.

You seem to be a wiz at the latex now. Does it do the calculation or is it just a display?

Edited April 6, 2015 by Robittybob1

[pavelcherepan]

You could have convinced yourself that planets further apart will collide at a higher impact speed.

 

Nope. I haven't.

[Robittybob1]

 

Nope. I haven't.

What sort of language is that? That wasn't even Aussie slang.

[pavelcherepan]

What sort of language is that? That wasn't even Aussie slang.

 

You said "You could have convinced yourself that planets further apart will collide at a higher impact speed." to which I replied that I haven't, which means that I haven't convinced myself. Your question was also very unclear, for example, why would I have to convince myself? That doesn't make much sense.

 

But anyway, I'll be waiting for your calculations.

[MigL]

I don't do LaTex either Pavel, so bear with me...

 

Take a ball and throw it up in the air with speed v. It will reach a certain height h, at which point all its kinetic energy has been swapped for gravitational potential and its speed is zero. It then falls back down and again swaps potential for kinetic, such that, if we disregard air resistance, it would land with the same speed.

Now take that same ball and throw it with speed v+x, it will now reach higher, say h+y, at which point its speed is zero. Note that now both balls are at rest compared to the planet, but the second ball is higher in the gravitational potential well than the one we first considered. And what happens ? As the second ball swaps its higher potential for higher kinetic, the impact speed is now v+x.

The conclusion we would reach is that a body, starting from rest, at a higher gravitational potential than another, will impact the ground with a higher speed than the other ( disregarding air resistance, of course )

 

The throw, or orbit, of a ball. easily and correctly scales to planetary distances ( that's why its called universal gravitation ), so this result also applies to planetary bodies.

Note however, that if a body has an initial speed towards the other body, but is not on a radial trajectory, the 'sling-shot' effect will boost it to above escape speed ( in a different direction ).