APOD: Star Trails over El Capitan (2014 Mar 21)

[APOD Robot]

Star Trails over El Capitan

Explanation: Towering 3,000 feet from base to summit, the famous granite face of El Capitan in Earth's Yosemite National Park just hides the planet's north celestial pole in this skyscape. Of course, the north celestial pole is at the center of all the star trails. Their short arcs reflecting the planet's daily rotation on its axis are traced in a digital stack of 36 sequential exposures. Linear trails of passing airplane navigation lights and a flare from car lights along the road below are also captured in the sequential stack. But the punctuated trail of light seen against the sheer El Capitan itself follows a climbing team on the night of November 8, 2013. The team is ascending toward the summit along The Nose, a historic rock climbing route.

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[Case]

... climbing team [at] night ...

Adding to the list of things that scare me if I were asked to do that : +Night rock climbing

[CURRAHEE CHRIS]

... climbing team [at] night ...

Adding to the list of things that scare me if I were asked to do that : +Night rock climbing

Absolutely!!! You were quicker to the punch than I. Couldnt imagine hanging off the side of El Cap at night!!! or day for that matter!!

[neufer]

Assuming that the 3000' granite face of El Capitan was perfectly vertical and there was absolutely no wind
how would these objects land to the East (from Coriolis forces) after being dropped from the very top:

1) a cannonball?
2) a feather with low terminal velocity V?

[Nitpicker]

Assuming that the 3000' granite face of El Capitan was perfectly vertical and there was absolutely no wind
how would these objects land to the East (from Coriolis forces) after being dropped from the very top:

1) a cannonball?
2) a feather with low terminal velocity V?

1) with a thud
2) with a whisper

I guess you meant how far, but I couldn't resist. Sorry.

[neufer]

Assuming that the 3000' granite face of El Capitan was perfectly vertical and there was absolutely no wind
how would these objects land to the East (from Coriolis forces) after being dropped from the very top:

1) a cannonball?
2) a feather with low terminal velocity V?

1) with a thud
2) with a whisper

I guess you meant how far, but I couldn't resist. Sorry.

I wonder if I could lift a small stone in China (while residing in Alexandria),
and then deposit it in Australia at some victim's head (with a bonk)
just to see if I can do so. (I better account for Coriolis forces.)

http://asterisk.apod.com/viewtopic.php? ... 50#p222450

[MarkBour]

Lovely picture of El Capitan and a beautiful fall evening. And it makes a great statement. The one light at the bottom of the "exclamation point" is what I would have expected. It looks like a climber that is tucked-in for the night. The other sounds scary. Trying to climb the nose in the dark. But with a good headlight, maybe it is actually less scary than during the day for some crazy folk? The main thing you wouldn't see is the distance below you. Personally, though, I would be worried that my headlight is not revealing everything and I'd be making dangerous mistakes. If they are in a position to fall about 3000 feet, that's the source of the adrenaline rush for the climber.

I have only looked at El Capitan from the valley below. It looks so large in front of you. But then, in this picture, it is the tiny points of light slowly curling across the sky that make you feel really small. And then a fall of 3000 feet seems pretty minor.

[Ron-Astro Pharmacist]

If you fall on a moonless night, do you still see your life flash before your eyes? Might stick to "day" rock climbing if you expect one last view. Of course if you could think fast enough to find a way to halt your reference frame, there's still hope. Focus on the apple falling next to you and convince yourself – nothing's moving, then grab on. If not, you've got a length contraction in your future. Talk about "special" relativity; afraid I qualify.

[Chris Peterson]

If you fall on a moonless night, do you still see your life flash before your eyes? Might stick to "day" rock climbing if you expect one last view. Of course if you could think fast enough to find a way to halt your reference frame, there's still hope. Focus on the apple falling next to you and convince yourself – nothing's moving, then grab on. If not, you've got a length contraction in your future. Talk about "special" relativity; afraid I qualify. :roll:

When you start falling, you're in a non-inertial frame, so special relativity doesn't apply. Once you reach terminal velocity, however, you can treat your reference frame as inertial and stationary. Unfortunately, the result will be the same when the now non-stationary ground hits you at 150 mph.

[neufer]

When you start falling, you're in a non-inertial frame, so special relativity doesn't apply. Once you reach terminal velocity, however, you can treat your reference frame as inertial and stationary. Unfortunately, the result will be the same when the now non-stationary ground hits you at 150 mph.

An accelerometer moving in an inertial frame of reference must detect zero acceleration so the guy starting free fall comes the closest to being in an inertial frame but technically it is not applicable:

<<An inertial frame of reference is a frame of reference that describes time and space homogeneously, isotropically, and in a time-independent manner. All inertial frames are in a state of constant, rectilinear motion with respect to one another; an accelerometer moving with any of them would detect zero acceleration.

Einstein’s general theory modifies the distinction between nominally "inertial" and "noninertial" effects by replacing special relativity's "flat" Minkowski Space with a metric that produces non-zero curvature. In general relativity, the principle of inertia is replaced with the principle of geodesic motion, whereby objects move in a way dictated by the curvature of spacetime. As a consequence of this curvature, it is not a given in general relativity that inertial objects moving at a particular rate with respect to each other will continue to do so. This phenomenon of geodesic deviation means that inertial frames of reference do not exist globally as they do in Newtonian mechanics and special relativity.>>

[Anthony Barreiro]

... climbing team [at] night ...

Adding to the list of things that scare me if I were asked to do that : +Night rock climbing

While certifiably insane speed climbers have climbed El Capitan in less than three hours, typical ascents take three or four days. Climbers sleep in hammocks hanging from wedges placed in the rock, and carry plastic bags to use as toilets.

[Chris Peterson]

An accelerometer moving in an inertial frame of reference must detect zero acceleration so the guy starting free fall comes the closest to being in an inertial frame but technically it is not applicable:

Actually, it's a little more complicated than that. When you're comparing two frames, you can use special relativity to perform transforms between them when an accelerometer in each frame reads the same. They do not need to read zero. So in the case of the falling climber and the mountain, special relativity is applicable once terminal velocity is reached, and each experiences 1G towards the center of the Earth. Even though the neither frame is an inertial one, they are inertial with respect to one another.

[neufer]

An accelerometer moving in an inertial frame of reference must detect zero acceleration so the guy starting free fall comes the closest to being in an inertial frame but technically it is not applicable:

Actually, it's a little more complicated than that. When you're comparing two frames, you can use special relativity to perform transforms between them when an accelerometer in each frame reads the same. They do not need to read zero. So in the case of the falling climber and the mountain, special relativity is applicable once terminal velocity is reached, and each experiences 1G towards the center of the Earth. Even though the neither frame is an inertial one, they are inertial with respect to one another.

• In an event, what I find interesting here is that
  we are dealing with a clockwise rotating frame of reference (vis-a-vis the stars)
  where: Ω = 2π cos(Lat)/86400 ~5.75 x 10^-5

Assuming that the 3000' granite face of El Capitan was perfectly vertical and there was absolutely no wind
how would these objects land to the East (from Coriolis forces) after being dropped from the very top:

1) a cannonball?
2) a feather with low terminal velocity V?

1) The cannonball case is a simple case where the top of El Capitan
is moving eastward with respect to the base by ΩH ~ 2 inches per second.

The cannonball drops for sqrt(2H/g) ~ 13.7 seconds
by which time the cannonball has traveled about 28 inches to the East.

2) The feather is even simpler in that the Coriolis acceleration to the east = 2ΩV
makes a constant angle θ to the vertical acceleration g: tan(θ) = 2ΩV/g

• 

The total eastward displacement (on average) is then H tan(θ) = 2ΩVH/g
~ 0.13 inches times V (where V the is feather's velocity in feet/second).

[rstevenson]

But the cannonball was dropped from a moving El Capitan, yes? The Earth did not pause while the ball was dropped, so the ball has the same velocity towards the east as El Capitan has, and, disregarding atmospheric drag, it will retain that velocity as it falls. Or is that covered in your equation?

Rob

[neufer]

But the cannonball was dropped from a moving El Capitan, yes? The Earth did not pause while the ball was dropped, so the ball has the same velocity towards the east as El Capitan has, and, disregarding atmospheric drag, it will retain that velocity as it falls. Or is that covered in your equation?

The top of El Capitan is moving (to the East) 2 in/s faster than the base; it is the relative motion that is important.

[Chris Peterson]

• In an event, what I find interesting here is that
  we are dealing with a clockwise rotating frame of reference (vis-a-vis the stars)
  where: Ω = 2π cos(Lat)/86400 ~5.75 x 10^-5

Yes. I remember in high school physics working out these problems with how far sideways tall waterfalls drift.

[rstevenson]

But the cannonball was dropped from a moving El Capitan, yes? The Earth did not pause while the ball was dropped, so the ball has the same velocity towards the east as El Capitan has, and, disregarding atmospheric drag, it will retain that velocity as it falls. Or is that covered in your equation?

The top of El Capitan is moving (to the East) 2 in/s faster than the base; it is the relative motion that is important.

Ah yes, got it! (I should read more carefully.)

Rob