Timeline for Why do spacecraft enter the atmosphere violently instead of a smooth spiral?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 7, 2022 at 11:10 | history | edited | Uwe | CC BY-SA 4.0 |
removed the error, not dividing by 2
|
| S Jun 7, 2022 at 5:26 | history | suggested | No Nonsense | CC BY-SA 4.0 |
Add consistent formatting according to https://physics.nist.gov/cuu/Units/checklist.html and switch from "0.5" to "1/2" because it is much more commonly used as a fraction.
|
| Jun 7, 2022 at 1:18 | review | Suggested edits | |||
| S Jun 7, 2022 at 5:26 | |||||
| May 17, 2020 at 15:54 | comment | added | Uwe | You did not multiply by 0.5 in "Ek=0.5⋅5000kg⋅(7800m/s)2". The result is 1.52E11 J, not 3.04E11 J. So 7.6E6 s or 88 days. Almost a quarter of a year. | |
| May 17, 2020 at 15:10 | history | edited | SF. | CC BY-SA 4.0 |
deleted 4 characters in body
|
| Dec 25, 2015 at 20:39 | comment | added | SF. | @NathanTuggy: Definitely a ballpark - a 3kW heater in a closed small room will make it oppressively hot but survivable; now let's surround our craft with a heaters that add up to said 3kW emitting in each of 6 directions, 1/6th directed at the craft (down), the remainder away (back, front, left, right, up) - that would be 18kW out of which 1/6 is directed at the craft. Of course the mechanism of cooling of a small room on Earth is quite different, so a rather large error is possible, but - that's a ballpark of what would be survivable without extreme cooling mechanisms. | |
| Dec 25, 2015 at 20:07 | comment | added | Nathan Tuggy | Can you clarify the source of the 20kW ballpark? | |
| Dec 24, 2015 at 23:53 | history | answered | SF. | CC BY-SA 3.0 |