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Distances to the nearest stars (up to 500 ly away) can be measured by a technique called parallax, as shown in Figure $34.26 .$ What are the angles $\theta_{1}$ and $\theta_{2}$ relative to the planeof the Earth's orbit for a star 4.0 ly directly above the Sun?
$\theta_{1}=\theta_{2}=89.999773^{\circ}$
Physics 103
Chapter 34
Frontiers of Physics
Wave Optics
Particle Physics
Rutgers, The State University of New Jersey
University of Washington
Simon Fraser University
McMaster University
Lectures
02:51
In physics, wave optics is…
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Interference is a phenomen…
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(I) The parallax angle of …
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Parallax To find the dista…
02:16
Two stars that are $10^{9…
01:14
The 300-m-diameter Arecibo…
00:53
(Refer to the following fi…
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(I) The Sun subtends an an…
01:24
01:00
(I) A star is 36 pe away. …
01:16
Star A and star B are two …
00:41
ASTRONOMY The comet Hale-B…
02:10
An Earth-observing satelli…
02:30
Consider a satellite orbit…
01:19
(I) A star exhibits a para…
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The 305 -m-diameter Arecib…
01:57
Astronomy Use the followin…
01:29
Solve each problem.Sca…
02:49
Round each answer to one d…
02:22
Parallax One method of mea…
question Number five is asking us to find the parallax angle for a star that is 55 light years away. Now there's a conversion factor between light years and parsecs. It is one par sick is equal to 3.26 light years. So when we do the math, this equates to be 16.87 parsecs for the distance to our star. Now what a parse IQ is one over the peril X angle for a star parallax angle being the difference in the angle between stars, Line of sight with sun and that stars line of sight with her. That's our angle phi there. So since we know that 1/5 is equal to our 16.87 parsecs distance weaken, just take 1/16 0.87 parsecs to find the distance and arc seconds, and that distance turns out to be 0.6 arc seconds. As for finding the angle in degrees, we just take 0.6 arc seconds and divided by 3 600 which is how maney arc seconds there are in one degree that gets us an angular distance of 1.65 times, 10 to the minus fifth degrees
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