Timeline for What is the typical relative impact velocity of orbital debris in low Earth orbit?
Current License: CC BY-SA 3.0
10 events
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Jun 12, 2017 at 20:33 | comment | added | HopDavid | @Uwe I was using 7.7 km/s in my spreadsheet. To look at yours, I changed to the first row to 7.8 but neglected to change the following rows. So the other numbers in my comment are also wrong with the exception of the first row. | |
Jun 12, 2017 at 19:48 | comment | added | Uwe | @HopDavid Yes, numbers were wrong. But your value for 90 ° is wrong too, 7.8 * 1.4142 is 11.03 | |
Jun 12, 2017 at 19:12 | history | edited | Uwe | CC BY-SA 3.0 |
A factor of 2 was missing.
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Jun 12, 2017 at 19:10 | comment | added | Uwe | Yes they were off by a factor of two. I edit the numbers now. Congratulations to all who found out that the values are wrong. | |
Jun 12, 2017 at 16:24 | comment | added | HopDavid | Numbers are wrong. Given two 7.8 km/s orbits: 5º .68 km/s, 45º 7.1 km/s, 60º 7.8 km/s (In this case both velocity vectors and delta V vector form the sides on an equilateral triangle), 90º 10.9º (In this case an isosceles right triangle, hypotenuse is sqrt(2) times each side). | |
Jun 12, 2017 at 13:35 | comment | added | Blake Walsh | Two objects both at 30 degrees inclination, but with opposite LANs, should collide at exactly orbital velocity as the velocity component that doesn't cancel out should be 2 x sin(30) | |
Jun 12, 2017 at 12:05 | comment | added | uhoh | Better double check the math. I did it wrong the first time also. At 90 degrees the relative velocity is $\sqrt{2} \times v_0$ or 1.414 times 7.8 km/s See this link. | |
Jun 12, 2017 at 11:15 | comment | added | Uwe | @Antzi The difference between 7.8 and 10 km/s is not that large. You may add a calculation for elliptical orbits. | |
Jun 12, 2017 at 11:06 | comment | added | Antzi | I think you are missing out on elliptical debris. In cases such as a Molniya orbit, it could be around 10km in LEO. | |
Jun 12, 2017 at 11:01 | history | answered | Uwe | CC BY-SA 3.0 |