Physics & Astrophysics?

Question

Hi, I have a particular question and I don't know how to go about it. Your help is appreciated.

Roughly how fast would you have had to move to follow the solar eclipse shadow in Europe in 1999?
Diameter (Sun)= 1.392*10^6km
Diameter (Moon)= 3476km
Diameter (Earth)= 12756km
Distance (Sun-Earth)= 1.5*10^8km
distance (Moon-Earth)= 384402km

Answer 1

1. Cirric's answer is good but don't forget to factor in the movement of the Earth also. When the Moon is in front of the Sun the Earth and Moon are moving in approximately opposite directions.

So if we use D (distance) to denote the circumference of the object's orbit, you get the equation:

(Dmoon / 28) + (Dearth / 365.25)

This will get you a speed per day. Divide by 24 to get a speed per hour.

2. For extra points, you could do some trigonometry. The moon's orbit is inclined at 5 degrees. This creates a very wide triangle shape. Side A would be the moon's speed, side B would be the Earth's speed, and the hypotenuse would represent the overall combined speed. The angle of AB would be 175 degrees.

It's a little tricky, because for trigonometry you need a square corner. We'll use line B as the "base" or bottom of the triangle.

So you take the Sin of 5 degrees and multiply it by A. That gets the height of the triangle. Then you take the Cos of 5 degrees, and multiply it by A. That, PLUS B, gets you the bottom of the triangle. Then the root of the sum of the squares gets you the hypotenuse, which will equal the total speed of the moon.

Here it is in equation form:

(Sin 5deg x Dmoon / 28)^2 x (Cos 5deg x Dmoon / 28 + Dearth / 365.25)^2

Take the square root of that.

3. Also, there's the rotation of the Earth to factor in. The Earth spins once per day, so you can use the circumference of the Earth as a rate per day. The trouble with this, though, is that this is the diameter at the equator, and the eclipse happened in Europe and Asia. Since we're specifically talking about Europe the eclipse happened between a lattitude of 45 and 42 degrees. The circumference of the 45th parallel is 70.71% that of the equator. Again, you use trigonometry to figure this out (it's the sine of the degree of lattitude). You would subtract this amount from the total you got above, since the Earth is spinning in the same direction that the moon orbits.

4. Oh, but it gets worse. The surface of the Earth is curved, while the distance to the moon is relatively constant. That means that the rate the eclipse passes over the surface of the earth *changes* depending where you are. If the moon is near the horizon, it will move very fast compared to the ground, while if it is directly overhead it will not move as fast.

That makes for a very tricky calculation. Your teacher will probably allow you to skip this, but if not the speed figure you got above has to be multiplied by secant of the angle of the ground's surface relative to the orbit of the moon.

The eclipse hit the coast of France just north of La Havre which is about 0 degrees longitude. The eclipse was at its greatest height somehwere over Romania, which is about 22 degrees East, so you can use that as your "zero point" (where the speed is as listed above). Since the secant of 22 degrees is 1.0785, when it was passing over La Havre the eclipse was moving 1.0785 times as fast as when it was passing over Romania, in terms of groundspeed.

5. Or you could cheat. You could use the diagram in the link below. It shows the path of the eclipse and the times when it appeared. It hit La Havre at 10:20, and passed out of Europe just north of Varna at around 11:03. Get the total distance between those two cities from an atlas and that's the amount of distance and multiply that by 60/43 to get an average speed in km per hour.

Answer 2

Hi. "Obi-wan... now that's a name I have not heard in a long time...". http://books.google.com/books?id=oYZ-0PUrjBcC&pg=PA3&lpg=PA3&dq=lunar+velocity&source=web&ots=r3oxlhryw7&sig=VNrklCyQ3na-i7tua2AAi1o0Hl0
Sorry, checking the link did not work, so here goes. The Lunar orbital diameter is 384402km*2. This number times pi is the circumference. Divide by the number of days it takes to orbit and you have velocity. This velocity is REAL close to the answer you seek. "Use the force" of a calculator.