RickB

Is it just me, or did Y!A make a change a few days ago in how it lists questions in a given forum? Seems it used to list the most recent questions first. Now I get questions from a week ago on top; then maybe a question from a day ago; then 2 days ago; random time order. I typically have to scroll way down if I want to read a question that somebody's posted in the last few hours, and that I haven't already read (maybe days ago).

I thought this might have been an account setting that I (accidentally?) set, but if so, I can't find it anywhere. Any help?

A friend is seeking photo(s) of the total lunar eclipse of April 4, 2015 as seen from Zion National Park, Utah. (She was there at the time, and wishes to commemorate the event.) Her preference is for photos that include park terrain as well as the moon; but she says that's not absolutely necessary. Any tips on where I might look? Thanks!

I've found a bug (actually there are many!) in the Yahoo!Answers software; but I can't find any link or other mechanism by which to report it. Any suggestions?

Some members of my astronomy club and I witnessed Iridium 7 flaring on Saturday, June 15, at 11:02 PM CDT, from Lincoln, Nebraska. This was a predicted flare on heavens-above.com. But the weird thing was, it was trailed by what seemed to be a second Iridium satellite, in the same trajectory, separated by not more than 2 or 3 degrees from the first one. The second satellite flared to about the same brightness as the first (about mag -6), but a few seconds later.

The "mystery" satellite showed all the characteristics of an Iridium flare, but it was not listed on heavens-above. (Come to think of it, it's possible that the trailing one was Iridium 7 and the leading one was the "mystery" satellite. That's how close together they were.)

Does anybody know whether Iridium 7 is temporarily traveling in tandem with another Iridium?

I know that the rate of change of the velocity vector is called "acceleration". Is there a name for the rate of change of SPEED? (that is, the rate of change of the MAGNITUDE of the velocity vector)?

For example: Consider a particle moving in a circle of radius R, with a speed given by: |v| = kt. It's easy to show that the magnitude of its acceleration is:

|a| = sqrt((kt)^4/R² + k²)

While on the other hand:

rate of change of speed = d|v|/dt = k

Is there a name (other than just "rate of change of speed") for this second quantity?

Considering that the sun's pull on the moon is much greater than the earth's pull on the moon (by a factor of about 183), how do you explain the fact that the moon stays in orbit around the earth, rather than getting ripped away by the sun's gravity?

How do you calculate the average distance (over TIME) of an orbiting body from its primary, given the periapsis and apoapsis (or equivalently, the orbit's semimajor axis and its eccentricity)?

I know in abstract that the solution should be:

R_avg = ∫R(t)dt / T

where R(t) is the body's distance as a function of time; and T is its period. However, I don't think there's a closed-form solution for R(t), is there? Failing that, is there a (short) formula that gives a fair numerical approximation?

Also, my intuition is that "T" should cancel out of the final equation (i.e. R_avg should depend just on the geometry of the orbit), since the geometry uniquely determines the fraction of time that the body spends in any particular segment of the orbit.

How do you calculate the average distance (over TIME) of an orbiting body from its primary, given the periapsis and apoapsis (or equivalently, the orbit's semimajor axis and its eccentricity)?

I know in abstract that the solution should be:

R_avg = ∫R(t)dt / T

where R(t) is the body's distance as a function of time; and T is its period. However, I don't think there's a closed-form solution for R(t), is there? Failing that, is there a (short) formula that gives a fair numerical approximation?

Also, my intuition is that "T" should cancel out of the final equation (i.e. R_avg should depend just on the geometry of the orbit), since the geometry uniquely determines the fraction of time that the body spends in any particular segment of the orbit.

Last night I watched the shuttle Discovery and the ISS pass over my town, at about 8:40 PM CDT. Discovery led the ISS by about a minute.

I have seen the ISS and the shuttle(s) pass overhead many times, but this one was a bit different. There was a faint, hazy "plume" trailing the shuttle; very dim, best seen using averted vision, but definitely there. It reminded me of the look of a comet's tail. I didn't have the presence of mind to stick my thumb out; otherwise I could give you a good estimate of the plume's angular size; however, my impression from memory is that when the shuttle was at maximum altitude, the plume was about 1 degree in length; and maybe 1/3 of that in width. This would have made it about 6 or 7 km in length. (Based on info from Heavens-above.com, which says the shuttle was 382 km distant at the time.)

I thought it might have been an artifact of hazy skies (or dirty eyeglasses); except the ISS came by just 1 minute later, and it definitely did NOT have any such plume.

What was this? Could it have been a water dump or something?

I have been trying in vain to use the ASPError object in my ".asp" web scripts (We have IIS6 installed). I have seen numerous examples online and though I copy and paste them exactly, the ASPError object always returns blank values. Here's something I tried:

<%

on error resume next

x = 4/0 'this should crash it

response.clear

set aspErr = server.getLastError

response.write aspErr.Number & "<br>" & aspErr.Description & "<br>" & aspErr.Line

on error goto 0

%>

I expected to see information displayed relating to the division by zero error: but the aspErr.Number and AspErr.Line are zero; and the aspErr.Description is blank. Do you get the same result on your IIS server? Can somebody show me some sample code that actually prints the error description? Thanks!

I'm running Vista Home Edition and have a wireless router for my Internet connection. If I go to "Start -> Connect to," then Vista always sees my network and I can easily (manually) connect. But if I shut down and restart, I have to manually connect to the network again.

How do I tell Vista, "ALWAYS find and connect to this network every time I start up?" I can't seem to find any Control Panel that lets me set my wireless connection as the "default".

(BTW, I have no such trouble if I use a "hard" connection (ethernet cable), nor do I have the same problem w/my Mac.)

Two astronauts are each in a circular orbit around an unknown planet. Astronaut "A" is displaced 100 meters "vertically" from Astronaut "B" (that is, the two astronauts are 100 meters apart, on a line that intersects the planet's center).

How far apart will the astronauts be from one another, after they have completed one orbit?

(Surprisingly, the answer does not depend on the planet's mass or on the radius of the orbit. You may assume, however, that the radius of the orbit is much, much greater than 100 meters (and use any resulting mathematical approximations).)

This is similar to a question I posted previously; but hopefully this version is stated more clearly. I know the answer, but the best response deserves 10 pts. anyway.

Two spaceships pass each other at a relative speed of 1/2 c. At the moment they cross paths, they each synchronize their on-board clocks to read "12:00 PM."

The captain of Ship "A" knows that when his own clock reads "1:00", it's only 12:52 on Ship "B", due to time dilation. He can prove this by sending a signal to Ship "B" asking what time it is (and then accounting for the time it takes for the signal to return).

But the captain of Ship "B" knows that when his own clock reads "12:52", it's only 12:45 on Ship "A", due to time dilation. He can prove this by sending a signal of his own.

So here's the question: When it's 12:52 on Ship "B", what time is it on Ship "A"?

First of all, let me say that I already know the answer to this one. I'm asking it just to generate some thought and discussion.

Here's the scenario:

The captain of Spaceship "A" watches as Spaceship "B" flies past at a (constant) speed of 1/2 c. According to the theory of relativity, moving clocks tick slowly; this means the time on Spaceship "B" is passing only 87% as fast as the time on Spaceship "A".

But now look at it from the point of view of the other captain. According to him, his own Spaceship (B) is standing still, while Spaceship "A" is moving at 1/2 c. So Captain "B" believes that Spaceship "A" is the one with the slow clock.

Can each "moving" clock actually be slower than the other ("stationary") clock? If not, are they both "really" ticking at the same speed? If not, which one is "really" slower?

This question is a little like the famous "Twin Paradox," except I've set this one up so that nobody accelerates and nobody changes reference frames.

There has been a _continuous_ human presence in space for a number of years now. On what date did that continuous presence start?

Note: I'm NOT asking, "when did people start going into space." I know that was 1959. I'm asking, "when was the last time outer space was completely free of humans?"