# How to calculate diameter of moon's shadow, it's speed and duration during eclipse?

To make calculations simple, an asteroid eclipsed a star(Regulus) for 14 seconds and was viewed in New York City. For simplification consider star very far, asteroid about 200 million miles away from earth and 12 miles in diameter.
How to calculate the diameter of its shadow, speed and duration(14 secs in NY) on earth?
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Commented:
Assuming a circular orbit, somewhere between Mars and Jupiter,
its speed relative to the Earth would be between about 6 and 17 km/s
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Commented:
Assuming the star was directly overhead, the diameter of the shadow would be the same as the diameter of the asteroid, and its speed would be diameter/duration.
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DeveloperAuthor Commented:
What if the shadow was on equator? You simplified by discounting speed of earth (rotation).
And how 14 sec was predicted?
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Commented:
Was it over one specific point in NYC for 14 seconds or over any part of NYC for 14 seconds (such that the size of NYC would matter)? I'm guessing the former?
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DeveloperAuthor Commented:
Yes, NYC size would not matter, In fact you can simply assume shadow at 45 degrees N Latitude.
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Commented:
Speed = distance/time applies regardless of speed of  earth (rotation), as long as you are consistent about how you measure speed, distance and time.
14 sec was predicted based on your question statement.
45 degrees N Latitude allows us to take into account speed of  earth (rotation)
but does not tell us much about the altitude angle of the star other than that it can be no more than about 57 degrees, since the declination of Regulus is about 12 degrees.
(although since NYC is closer to 40 N, that that could allow an altitude of up to about 63 degrees)
We also  don't know the direction the asteroid is traveling with respect to the direction of
earth (rotation) or with respect to the azimuth of Regulus.
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DeveloperAuthor Commented:
If everything is in ecliptic plane and we watch from equator for 14 seconds(the eclipse)
What is diameter of asteroid (we don't know if it is 12 miles), if it is also revolving around sun at half the speed of earth's revolution?
So factors are:
Earth's Rotation
Star is distant and stationary.
All happens in ecliptic plane.
Eclipse lasted 14 seconds.
We are on equator.
We can ignore Earth's revolution as well as asteroid's for further simplicity.

Have I simplified enough to calculate? Can you show the calculation?
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Commented:
You are changing the conditions of the problem.
You gave the diameter of asteroid as 12 miles, Now you say
" (we don't know if it is 12 miles)"
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" if it is also revolving around sun at half the speed of earth's revolution?  "
This is a very improbable assumption and not needed to first approximation.)
In any case the size of the shadow, the time of eclipse, and the distance of asteroid are not independent.
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"We can ignore Earth's revolution as well as asteroid's for further simplicity."
With this degree of reasonable simplification Ozo is correct.
(If the star is very far away, the rays coming from it are essentially parallel.)
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DeveloperAuthor Commented:
Agreed, but if the star is over North Pole as ozo assumed, then we simplified a lot  because earth's rotation is zero and it took 14  seconds for eclipse. With magnitude 12 miles we can know the speed of asteroid. There is no challenge or realism in question here.
The one which was seen over NYC for 14 secs was 80 km long did not fit in my calculations. I am a newbie.
So I tweaked the problem and made it simple too. I just want to know diameter of asteroid with everything in ecliptic plane and observation on equator. Why asteroid going half the earth's speed is unreasonable assumption?
There is asteroid belt between Mars and Jupiter and asteroids would be slower than earth and are on ecliptic plane. Mars, itself is at half the speed. I want to know the formula by which you calculate magnitude of asteroid. I refined my question, I admit. Sorry about that. I  refined my question by making magnitude unknown, as all other variables are known and some can be ignored.

9
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Commented:
I never assumed that the star was over the North Pole, I assumed that the star was directly overhead, so that the shadow would be projected vertically.  If the shadow is cast obliquely, then the notion of diameter becomes more ambiguous.
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Commented:
If all we know is that the shadow lasts for 14 seconds at the equator, then we have no way to distinguish between a large shadow moving quickly, or a small shadow moving slowly.

We may be able to work out something if we have observations of the shadow of the same object from a number of different locations.
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DeveloperAuthor Commented:
OK. I misunderstood. But ecliptic plane will ensure shadow is round.
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Commented:
> ecliptic plane will ensure shadow is round
no, the star could still be anywhere from horizon to zenith
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DeveloperAuthor Commented:
Star is far at distant location. Rays are parallel.
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DeveloperAuthor Commented:
At equator the shadow is round. The rotational speed of earth is 1600 km/sec. So distance traveled in 14 sec at that speed is diameter of astroid. This is from your 1st reply. Am I right? Is so you get the credit.
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DeveloperAuthor Commented:
Sorry 1600 kmph
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Commented:
Rays are parallel, and they strike the terminator as well as the sub-astral point.
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DeveloperAuthor Commented:
Sorry, don't get it :-(
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Commented:
> At equator the shadow is round.
Let us take this as an assumption.
Let us also assume that the asteroid is round.
We can conclude from this that the star is at the zenith.
If we also knew the local time, we could determine the hour angle of the star, which could tell us something about how the Earth's orbital velocity would affect the motion of the shadow.
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DeveloperAuthor Commented:
We only need diameter of shadow, not speed.
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Commented:
Diameter of the shadow would be the same as the diameter of the asteroid.
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DeveloperAuthor Commented:
If eclipse lasts 14 sec what is diameter of asteroid?
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Commented:
If all we know is that the shadow lasts for 14 seconds, then we have no way to distinguish between a large shadow moving quickly, or a small shadow moving slowly.
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DeveloperAuthor Commented:
What are the main factors that affect speed of shadow?
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Commented:
The speed of the asteroid is all that will affect the speed of the shadow (since Earth's rotation has already been ignored).

The time the shadow is over a point is based on the speed of the asteroid and the size of the asteroid. If you could measure how fast the leading edge of the shadow was moving, then you could get the speed of the asteroid. Then the 14 seconds could be used to get the size.
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DeveloperAuthor Commented:
The asteroid is in asteroid belt between Mars & Jupiter, Can you know it's orbital velocity thus the speed using Kepler's Law? And using that determine the speed of shadow, again assuming asteroid and star are overhead at equator and star is eclipsed by asteroid for 14 secs.
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Commented:
If you want the transverse speed of the asteroid relative to you, it would be the same as speed of the shadow relative to you.
If you want the transverse speed of the asteroid relative to the centre of the earth, it would be the same as the speed of the shadow relative to the centre of the earth.
The total speed of the asteroid would be sqrt((radial speed)^2 + (transverse  speed)^2)
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Commented:
If you know the major and semimajor axes of the orbit, and the current distance from the sun, you can determine the velocity relative to the sun.
To determine the velocity relative to the earth, you would also need to know the position in the orbit relative to the earths position in its orbit.
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DeveloperAuthor Commented:
The asteroid is in line with Sun and Earth with closest distance from Earth directly above observer on earth on equator. We can assume equatorial plane same as ecliptic plane and ignore earth's tilt. See diagram below, top view:
asteroid.png
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DeveloperAuthor Commented:
Jupiter is 13 km/sec, Mars is 24 km/sec. Asteroid has to be in between?
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DeveloperAuthor Commented:
its speed relative to the Earth
I missed that, assuming 18 km/sec asteroid speed, and Earth 30 km/sec, it would be 12 km/sec relative to Earth. It took 14 sec to blink the star, so diameter is 14*12 =168 km?
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DeveloperAuthor Commented:
I've requested that this question be closed as follows:

Accepted answer: 0 points for vakils's comment #a39954916

for the following reason:

Thanks for providing various insights on the problem.
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DeveloperAuthor Commented:
Educational!
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