4 ms·
The shutter speed was 1/250th of a second, so the earth rotated about 4 miles or 6 kms while the shutter was open. Not enough to blur the photo obviously, but c
by jvm___ 2y ago
The shutter speed was 1/250th of a second, so the earth rotated about 4 miles or 6 kms while the shutter was open. Not enough to blur the photo obviously, but crazy to think about.
- danielecook 2y agoI’m confused. Does this math check out? Circumference is 24,901 - so about 1000 mph at equator. 1000 mph / 3600 s / h = 0.27 mps 0.27 * (1/250) = 0.001 miles? Doing this math on my phone but am I missing something here?
- thsksbd 2y agoI got the same answer
- Choco31415 2y agoEven if we stretch out and look at how fast the Earth orbits the sun, it still doesn't explain the 4 mile figure. Earth's orbital speed: 66,200 mph 66,200 mph / 3600 s/h = 18.38 mps 18.38 * (1/250) = 0.07 miles
- mrb 2y agoYou are correct. Using GNU units: $ units '2 * pi * earthradius / day * (second / 250)' Definition: 1.8532517 m Which is 0.0011515572 miles...
- tomrod 2y agoI think it's relative and phrased poorly, since the orbiter has to circle the earth at a certain faster speed. But quick googling shows Apollo 8 was traveling about a mile a second?
- re 2y agoI think the parent comment was confusing/misremembering the rotational speed value in miles per hour as miles per second.
- serf 2y agorotation: 360/86,400=0.0041667deg/sec 0.0041667 * 0.004sec = 000016667degR circumference: 2pi * 6371km = 40,030km land covered by rotation: 40,030 / 360 = 111.194km/deg 0.000016667deg × 111.194 km/degree = 0.001854 km km to m: 0.001854km × 1,000 meters/km = 1.854 meters more like 0.001 miles. ... oh, woops. I see your answer is in kms. it was something like 0.46km/s.
- Cerium 2y agoI have no idea how to calculate it, but I interpreted this to mean not that the earth rotated (which everyone is trying to calculate) but that the earth was crossing the horizon of the moon such that four miles of earth crosses the horizon during the shot causing earth blur for a moon-stable reference frame.
- Thorrez 2y agoRevolved I think would be the correct term for that.
- mrb 2y agoFrom the moon's reference, the Earth orbits around it, traveling 2 × π × distance_moon_earth per orbit. Divide this by 27.3 days (sidereal orbital period) to get the Earth's speed. Multiply by 1/250th of a second. And we find this is, again, much less than 4 miles. Using GNU units: $ units "2 * pi * moondist / (27.3 day) * (second / 250)" Definition: 4.0958765 m which is only 0.0025450597 miles.
- lamontcg 2y agoIf the orbital period was 80 minutes then that is 1/1,200,000th of a period and with Earth's circumference being ~25,000 miles that should only be about 0.02 miles. Or if the orbital velocity was 17,000 mph and neglecting the height of the orbit, 17000 / 3600 / 250 = 0.018 miles. So either way, about 100 feet.