pavelcherepan Posted April 6, 2015 Posted April 6, 2015 Ok, MigL, I understand where you're coming from but let's imagine we throw one ball high enough that it reaches the orbit of the Moon and the second one - 10000 kilometers higher and then start falling down. How big would be the difference in velocity then? Secondly, we weren't talking about starting at 0 because initially we were playing from a paper that described collision simulation and there the initial speed is estimated at 4 km/s. Wikipedia article has the same estimate. Now how much change due to Earth's gravitational influence (percentage) with initial velocity of 4 km/s and starting points at 150 and 300 million kilometers away? The difference in escape velocities ( relative to Earth) between these points are some 100 m/s.
Robittybob1 Posted April 6, 2015 Author Posted April 6, 2015 (edited) ..... let's imagine we throw one ball high enough that it reaches the orbit of the Moon and the second one - 10000 kilometers higher and then start falling down. How big would be the difference in velocity then? .... The differences in velocities are proportional to the ratio Escape velocity of the Earth minus the sqrt(2*M*G/r), r being the new starting distance. But remember the 2*M*G parts of the equation remains constant. So at those two specific distances the impact speeds won't be much different as you calculated (as they are relatively similar) but once you compare the distance to the Moon compared to infinity there would be a much greater of difference. Secondly, we weren't talking about starting at 0 because initially we were playing from a paper that described collision simulation and there the initial speed is estimated at 4 km/s. Wikipedia article has the same estimate. Now how much change due to Earth's gravitational influence (percentage) with initial velocity of 4 km/s and starting points at 150 and 300 million kilometers away? From memory that paper said the initial velocity was "<4 km/s" which could include 0, couldn't it? As I have previously said any initial velocity above that which is gained from falling toward the Earth would represent the amount of velocity it had gained from also falling toward the Sun. Well that is my prediction at this stage. The L4 position is 60 degrees ahead of the Earth on its orbit around the Sun so as rough calculation we could say it is 1/6 0f the orbit circumference away from the Earth or 1/3 * pi() * r r = 150 million km or from geometry it is exactly the length of r directly. So the difference in Earth Escape velocities from both positions: 11185.8 and 2305.3 so difference in Earth Escape velocities from both positions, when using current Earth Mass: = 8880.5 m/sec 9687.2 and 1996.4 so difference in Earth Escape velocities from both positions, when using Earth mass 0.75 current mass: = 7690.7 m/sec. So from infinity a mass will strike the Earth at 11185.8 m/sec but from the L4 position at 8880.5 m/sec (current mass) Edited April 6, 2015 by Robittybob1
pavelcherepan Posted April 6, 2015 Posted April 6, 2015 Can you show me the formula you used for these calculations?
Robittybob1 Posted April 6, 2015 Author Posted April 6, 2015 (edited) Can you show me the formula you used for these calculations? It is just the difference in the escape velocities from the two locations Terminal Impact velocity (Vt) Vt = sqrt(2*M*G/r1) - sqrt(2*M*G/r2) M current mass of Earth r1 - radius of the Earth (current) R2 distance out to center point of the L4 lagrangian Point which by geometry = radius of the Earth's orbit. Edited April 6, 2015 by Robittybob1
pavelcherepan Posted April 6, 2015 Posted April 6, 2015 (edited) I'll 're-check your calculations when I get back to my laptop but they don't seem right. At 150 or 300 million kilometers from Earth escape velocities can't be in order of kilometers per second, it's some meters per second. You've lost some zeros so it seems Edited April 6, 2015 by pavelcherepan
Robittybob1 Posted April 7, 2015 Author Posted April 7, 2015 I'll 're-check your calculations when I get back to my laptop but they don't seem right. At 150 or 300 million kilometers from Earth escape velocities can't be in order of kilometers per second, it's some meters per second. You've lost some zeros so it seems I'm using Excel and formulated cells so the same information is going into all the calculations. It is hard to make an error with Excel.
pavelcherepan Posted April 7, 2015 Posted April 7, 2015 (edited) I'm using Excel and formulated cells so the same information is going into all the calculations. It is hard to make an error with Excel. 1. Check your numbers. You have done an arithmetic error in Excel. For example at L3 [latex] v_e = \sqrt \frac {2GM}{r} = \sqrt \frac {2*6.67*10^{-11}*6*10^{24}}{3*10^{11}} = \sqrt \frac {80.04*10^{13}}{3*10^{11}} = \sqrt {26.68*10^2}= 51.65 \, m/s[/latex] The rest of your numbers have the same error. 2. Explain physical reason for subtracting square roots of escape velocities at different distances? Can you show me where you got that formula from or explain what it's used for? 3. Have you invented the term "Terminal Impact velocity"? Edited April 7, 2015 by pavelcherepan
Robittybob1 Posted April 7, 2015 Author Posted April 7, 2015 (edited) 1. Check your numbers. You have done an arithmetic error in Excel. For example at L3 [latex] v_e = \sqrt \frac {2GM}{r} = \sqrt \frac {2*6.67*10^{-11}*6*10^{24}}{3*10^{11}} = \sqrt \frac {80.04*10^{13}}{3*10^{11}} = \sqrt {26.68*10^2}= 51.65 \, m/s[/latex] The rest of your numbers have the same error. 2. Explain physical reason for subtracting square roots of escape velocities at different distances? Can you show me where you got that formula from or explain what it's used for? 3. Have you invented the term "Terminal Impact velocity"? Where does the 3*10^11 come from in your calculation? Haven't you seen that been done before? That is pretty standard. You didn't like terminal velocity, and rightly so, so I described what we are looking for even better terminal impact velocity. I am just calculating escape velocity , not the square root of escape velocities. For a spherically symmetric massive body such as a (non-rotating) star or planet, the escape velocity at a given distance is calculated by the formula[1][latex] v_e = \sqrt{\frac{2GM}{r}} [/latex] [latex] v_e = \sqrt {\frac {2GM}{r}} [/latex] or you could say velocity at the point of impact. Edited April 7, 2015 by Robittybob1
pavelcherepan Posted April 7, 2015 Posted April 7, 2015 Where does the 3*10^11 come from in your calculation? Haven't you seen that been done before? That is pretty standard. You didn't like terminal velocity, and rightly so so I described what we are looking for even better terminal impact velocity. I am just calculating escape velocity , not the square root of escape velocities. or you could say velocity at the point of impact. 300 million kilometers = 300 billion meters = 300*10^9 = 3*10^11 Makes sense? Are your calculations correct or not?
Robittybob1 Posted April 7, 2015 Author Posted April 7, 2015 Where does the 3*10^11 come from in your calculation? Haven't you seen that been done before? That is pretty standard. You didn't like terminal velocity, and rightly so so I described what we are looking for even better terminal impact velocity. I am just calculating escape velocity , not the square root of escape velocities. or you could say velocity at the point of impact. Where does the 3*10^11 come from in your calculation? Haven't you seen that been done before? That is pretty standard. You didn't like terminal velocity, and rightly so so I described what we are looking for even better terminal impact velocity. I am just calculating escape velocity , not the square root of escape velocities. or you could say velocity at the point of impact. 1. Check your numbers. You have done an arithmetic error in Excel. For example at L3 [latex] v_e = \sqrt \frac {2GM}{r} = \sqrt \frac {2*6.67*10^{-11}*6*10^{24}}{3*10^{11}} = \sqrt \frac {80.04*10^{13}}{3*10^{11}} = \sqrt {26.68*10^2}= 51.65 \, m/s[/latex] The rest of your numbers have the same error. 2. Explain physical reason for subtracting square roots of escape velocities at different distances? Can you show me where you got that formula from or explain what it's used for? 3. Have you invented the term "Terminal Impact velocity"? 300 million kilometers = 300 billion meters = 300*10^9 = 3*10^11 Makes sense? Are your calculations correct or not? But what distance are you measuring? I still think I'm right.
pavelcherepan Posted April 7, 2015 Posted April 7, 2015 (edited) But what distance are you measuring? I still think I'm right. What do you mean? That's 300 million kilometers to L3 from the Earth. Your calculations are orders of magnitude off. 2. Explain physical reason for subtracting square roots of escape velocities at different distances? Can you show me where you got that formula from or explain what it's used for? Edited April 7, 2015 by pavelcherepan
Robittybob1 Posted April 7, 2015 Author Posted April 7, 2015 (edited) What do you mean? That's 300 million kilometers to L3 from the Earth. Your calculations are orders of magnitude off. 2. Explain physical reason for subtracting square roots of escape velocities at different distances? Can you show me where you got that formula from or explain what it's used for? It was you who has used the 300 million km figure not me. Well you tell me how far it is between the Earth and the L4 Lagrangian point? In this thread we are discussing the L4 or L5 points not the L3 point. I Have already told you we are not "subtracting the square roots of escape velocities at different distances". Get that right first please. Edited April 7, 2015 by Robittybob1 -1
pavelcherepan Posted April 7, 2015 Posted April 7, 2015 (edited) It was you who has used the 300 million km figure not me. Well you tell me how far it is between the Earth and the L4 Lagrangian point. In this thread we are discussing the L4 or L5 points not the L3 point. I Have already told you we are not "subtracting the square roots of escape velocities at different distances". Get that right first please. OK, so it wasn't you who said all that? To get a Giant Impact collision of course. I had already calculated the speed an object hit earth would get from infinity (hope I did it right) but how did that speed compare to a planet that just coming in from the L3 position (furthest point)? Well you agreed it was a trojan planet, so there are limited places it could have come from if that was the case, and none of them have anything to do with "infinity". The L3 is a postion co-orbiting with the Earth on opposites sides of the Sun (300 million km). OK it is a long way around the orbit but if it was perturbed by Venus and if it got out of the balanced situation it could begin a journey being pulled toward the Earth, slowly at first but after sometime it would impact. I don't believe the impact speed is the same for all starting positions especial from a Lagrangian point. Doesn't that prove my point for we are saying soon after forming at the L3 it became unstable and later impacted the Earth. Once I do the calculation you will see (and believe) that a planet coming from L3 will impact slower than from infinity. So where do you get that time limit from? I might have read them had you not been so condescending. How long do you think it would take to pull a planet around from the Earth's L3 position? OK? N one has even mentioned L3 until you started repeatedly bringing it up. You seem to have lost all your argumentation, you can't make a simple calculation without being off by orders of magnitude and you can't support your position with any relevant links or evidence. Now you're just making stuff up. I Have already told you we are not "subtracting the square roots of escape velocities at different distances". Get that right first please. Stop your straw manning. You said: Vt = sqrt(2*M*G/r1) - sqrt(2*M*G/r2) Here are we not subtracting square roots of escape velocities at two different points? EDIT: OK, my fault. We're just subtracting escape velocities. Anyway, what's the basis for this calculation? Edited April 7, 2015 by pavelcherepan
Robittybob1 Posted April 7, 2015 Author Posted April 7, 2015 OK, so it wasn't you who said all that? OK? N one has even mentioned L3 until you started repeatedly bringing it up. You seem to have lost all your argumentation, you can't make a simple calculation without being off by orders of magnitude and you can't support your position with any relevant links or evidence. Now you're just making stuff up. Stop your straw manning. You said: Here are we not subtracting square roots of escape velocities at two different points? OK we have talked about the L3 point as well but more recently you were telling me that Theia was more likely to come from the L4 or L5 position. Sqrt(2*M*G/r1) - sqrt(2*M*G/r2) is just one escape velocity minus another.
pavelcherepan Posted April 7, 2015 Posted April 7, 2015 (edited) Sqrt(2*M*G/r1) - sqrt(2*M*G/r2) is just one escape velocity minus another. Yeah, I've figured that out already. Anyway, what's the physical basis for calculating impact velocity by subtracting escape velocities at two points? Show me a link or physical explanation of the calculation. OK we have talked about the L3 point as well but more recently you were telling me that Theia was more likely to come from the L4 or L5 position. Yes, I was saying that from the very beginning , but you never agreed and have been trying to prove it came from L3. And then suddenly you agree with me? By the way, your calculation for escape velocity at L4 is still off by an order of magnitude. Edited April 7, 2015 by pavelcherepan
Robittybob1 Posted April 7, 2015 Author Posted April 7, 2015 (edited) Yes, I was saying that from the very beginning , but you never agreed and have been trying to prove it came from L3. And then suddenly you agree with me? By the way, your calculation for escape velocity at L4 is still off by an order of magnitude. Show me your figures and I'll compare them to mine tomorrow when I'm back in the office please.. Don't take the fact that I'm calculating the impact speed from the L4 or L5 is any indication that I agree with you. I'm Just trying to show you that you are wrong, really. Edited April 7, 2015 by Robittybob1
pavelcherepan Posted April 7, 2015 Posted April 7, 2015 Show me your figures and I'll compare them to mine tomorrow when I'm back in the office please.. Do you even read my posts? Or do you just start writing random stuff? Point 2. r~1.5*1011m ve=63.10 m/s
Robittybob1 Posted April 7, 2015 Author Posted April 7, 2015 Do you even read my posts? Or do you just start writing random stuff? You frustrate me. Of course I read what you write and notice you ask the same questions over and over again. So was that your Escape velocity from the Earth at 150 million km? My figures may be wrong but I'll check them tomorrow but be as clear as you can about what you have calculated please.
pavelcherepan Posted April 7, 2015 Posted April 7, 2015 (edited) You frustrate me. Of course I read what you write and notice you ask the same questions over and over again. So was that your Escape velocity from the Earth at 150 million km? My figures may be wrong but I'll check them tomorrow but be as clear as you can about what you have calculated please. [latex]v_e=\sqrt \frac{2GM}{r} = \sqrt \frac{2*6.67*10^{-11}*6*10^{24}}{1.5*10^{11}} = \sqrt {\frac{80.04*10^{13}}{1.5*10^{11}}}=\sqrt{53.36*10^2} = 73.05[/latex] My original calculation seems slightly off, still way lower than what you've got. Of course I read what you write and notice you ask the same questions over and over again. Would it be because you never answer those? Edited April 7, 2015 by pavelcherepan
imatfaal Posted April 7, 2015 Posted April 7, 2015 http://www.wolframalpha.com/input/?i=sqrt%28%282*6e24*6.67e-11%29%2F1.5e11%29 2
Strange Posted April 7, 2015 Posted April 7, 2015 (edited) It is hard to make an error with Excel. That is the funniest thing I have heard for a long time. "88% of spreadsheets have errors" http://www.marketwatch.com/story/88-of-spreadsheets-have-errors-2013-04-17 Edited April 7, 2015 by Strange
Robittybob1 Posted April 7, 2015 Author Posted April 7, 2015 (edited) http://www.wolframalpha.com/input/?i=sqrt%28%282*6e24*6.67e-11%29%2F1.5e11%29 Strangely I saw the graphs of the roots this time but I didn't see them on the other thread about Dark Matter. Thanks I might have entered the distance in wrongly to my Excel calculation. Edited April 7, 2015 by Robittybob1
Strange Posted April 7, 2015 Posted April 7, 2015 I might have entered the distance in wrongly to my Excel calculation. See post 96
Robittybob1 Posted April 7, 2015 Author Posted April 7, 2015 (edited) [latex]v_e=\sqrt \frac{2GM}{r} = \sqrt \frac{2*6.67*10^{-11}*6*10^{24}}{1.5*10^{11}} = \sqrt {\frac{80.04*10^{13}}{1.5*10^{11}}}=\sqrt{53.36*10^2} = 73.05[/latex] So in your understanding what is it that you have calculated here? Edited April 7, 2015 by Robittybob1
Strange Posted April 7, 2015 Posted April 7, 2015 You can confirm this with an online calculator: http://hyperphysics.phy-astr.gsu.edu/hbase/vesc.html Put in 1 for the mass (number of Earth masses) and 23500 for distance (the number of Earth radii) and you get a result of 73 m/s.
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