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The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of parsecs? How much parallax would this star (named Alpha Centauri) show when viewed from two locations of the Earth six months apart in its orbit around the Sun?

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OK so I had done the first part but I am confused with the second one. 1.3333... pascal is the answer for 1st part. @hartnn @Algebraic! @sauravshakya
@hartnn any idea?

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Other answers:

nopes,seen such question for the first time.
well I can remind you that it is "parallax method"
@mukushla may help you...
\[4.26ly=(3*10^8)(365.25*24*60*60)m=9.467*10^{15}m\]\[1psc=3.09*10^{16}m\] \[4.26ly=\frac{9.467*10^{15}}{3.09*10^{16}}psc=0.306psc\]
\[\large{4.29 ly = 9.46*10^{15}m * 4.29 = 40.5834*10^{15} = 4 * 10^{16} m }\] \[\large{4.29 ly = \frac{4*10^{16}}{3*10^{16}}=\frac{4}{3}=1.3333...}\]
\(\large{1ly = 9.46*10^{15}m}\)
@henpen You have done a mistake there, you forgot to multiply 4.29 there and you by fault wrote 4.26 ly ..
Oh yes- just multiply the end result by 4.26
so the answer is 1.33... ?
*1.33... pascal ?
lol sorry forgot that but still m i correct with the solution given above?
Yes. As to the parallax, the earth sun distance *2 is 3*10^11 m, so just use trigonometry to work out the angle change
I tried that by applying the formula (parallax formula) : d = b / theta but I was unable to complete it... to the correct answer Can you show your work? or just answer you get so that I can check it with the correct answer... (I am not wanting a direct answer but here I want to check it so please provide your answer )
In short : what r u getting after doing that @henpen
|dw:1348407871723:dw| Multiply by 2 for the change
Actually, don't multiply by 2|dw:1348408141213:dw|. Take the answer away from 90, and multiply THAT answer by 2.
2*(90-arctan(8*10^5 /3))
If parallax is measured in degrees, that is.
Yes that is absolutely *correct* @henpen good work

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