# length of arcs

• Oct 16th 2007, 05:23 PM
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length of arcs
The Earth travels in a nearly circular orbit around the Sun. the radius of the orbit is about 149 000 000 km.

What is the measure in radians of the angle subtended at the sun by the positions of the Earth at two different times 24h apart?

thanks!
• Oct 16th 2007, 10:12 PM
ticbol
So let us say, the earth's orbit around the sun is circular.
Then its cirumference, or one complete orbit is 2pi(149Million) = 298pi Millions km

The earth takes 1 year to complete one orbit.
In hours, that is
(1yr)*(365.25days/1yr)(24hrs/day) = 8766 hrs

One orbit is 360 degrees or 2pi radians

So, by proportion,
x/(24 hr) = 2pi/(8766 hr)
x = [2pi / 8766]*24

(Umm, 360 degrees---365.25 days)

-----------------------
Edit:

Hey, what happened to the given R = 149 millions km?
Extraneous?
Or the question is not correct?
Or my solution is not correct?
• Oct 17th 2007, 03:49 AM
topsquark
Quote:

Originally Posted by ticbol
So let us say, the earth's orbit around the sun is circular.
Then its cirumference, or one complete orbit is 2pi(149Million) = 298pi Millions km

The earth takes 1 year to complete one orbit.
In hours, that is
(1yr)*(365.25days/1yr)(24hrs/day) = 8766 hrs

One orbit is 360 degrees or 2pi radians

So, by proportion,
x/(24 hr) = 2pi/(8766 hr)
x = [2pi / 8766]*24

(Umm, 360 degrees---365.25 days)

-----------------------
Edit:

Hey, what happened to the given R = 149 millions km?
Extraneous?
Or the question is not correct?
Or my solution is not correct?

Your answer is correct. If you look at it from the standpoint of the definition of an angle in radians:
$\theta = \frac{s}{r}$
where s is the arc length of the part of the orbit, and then note that s is a fraction of the circumference of the Earth's orbit, then we have $s = fC = 2 \pi r f$, so
$\theta = \frac{ 2 \pi rf}{r} = 2 \pi f$
and the r disappears.

-Dan