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Distance by Parallax NULL

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Problem

Derive the formula for parallax. Then use the formula to calculate the distance, angular separation, and actual separation in parsecs and astronomical units of Proxima Centauri and Alpha Centauri (A). (Data are from Cox (2000).)

Hints:

Derive the parallax formula by converting radians to seconds of arc.

Use the parallax formula and the given data to find the distance and separation of the two stars.

 

Data

Parallax of Proxima Centauri

pp := .77233

 

Number of Metres in 1 parsec

pm := 3.0856776*10^16

 

Number of arcseconds in 1 radian

rd := 2.062648062*10^5

further*data*given*below

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Useful Equations:

dd := bb/tan(alpha)

deci := proc (h, m, s) options operator, arrow; h+(1/60)*m+(1/3600)*s end proc

 

Solution:

 

Parallax is the angle in arcseconds that a star would appear to be displaced on the sky by an observer moving one astronomical unit.

 

dd := bb/tan(alpha)

bb/tan(alpha)

(1)

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rd

206264.8062

(2)

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One radian equals approximately 206,265 seconds of arc. Therefore, one second of arc equals

 

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Unit('rad')/rd

0.4848136812e-5*Units:-Unit('rad')

(3)

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dd = AU/(0.4848136812e-5*pc*alpha)

bb/tan(alpha) = 206264.8062*AU/(pc*alpha)

(4)

 

Since 1 pc = 206,265 seconds of arc, this formula reduces to dd = 1/alpha as the distance in parsecs, where one parsec is defined as the distance at which one A.U. would subtend an angle of one second of arc.

 

In the triple-star α Centauri system, Proxima Centauri (α Centauri C) is the nearest star to the Sun. Its parallax is 0.77233 arcsec. Its distance, therefore, is

 

r = 1/.77233

r = 1.294783318

(5)

or approximately 1.29 parsecs. This corresponds to

(1.29*3.0856776)*10^16

0.3980524104e17

(6)

or approximately 3.98 * 1016 metres.

The right ascension (α) of Proxima Centauri for epoch J2000.0 is 14h 29m 42.95s, and its declination (δ) is -62o 40' 46.1". The brightest star in the α Centauri system. α Centauri (A), has coordinates (α, δ) of 14h 39m 36.50s, -60o 50' 02.3". Create a function that converts hours, minutes, and seconds of arc to degrees:

 

deci := proc (h, m, s) options operator, arrow; h+(1/60)*m+(1/3600)*s end proc

The two stars are separated on the sky by

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`Δα` = (360*(1/24))*(deci(14, 39, 36.50)-deci(14, 29, 42.95))

`Δα` = 2.4731250

(7)

"(->)"

2.4731250

(8)

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in right ascension, and

 

`Δδ` = deci(62, 40, 46.1)-deci(60, 50, 2.3)

`Δδ` = 1.84550001

(9)

"(->)"

1.84550001

(10)

in declination. The declination of α Centauri (A) in radians is

delta = (2*Pi*(1/360))*deci(-60, 50, 2.3)

delta = -.3287001543*Pi

(11)

"(->)"

-.3287001543*Pi

(12)

and the cosine of this angle is

cos(-.3287001543*Pi)

cos(.3287001543*Pi)

(13)

"(->)"

.51259

(14)

`Δθ` := (2.4731250^2*.51259^2+1.84550001^2)^(1/2)

2.238957661

(15)

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The separation of the two stars in the sky is approximately 2.239 degrees. (More accurate measurements give 2.18 degrees.)

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2.239*(2*Pi*(1/360))

0.1243888889e-1*Pi

(16)

"(->)"

0.39078e-1

(17)

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or 0.039 radians. Their physical separation is

 

d = 0.39078e-1*rhs(r = 1.294783318)

d = 0.5059754250e-1

(18)

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rd

206264.8062

(19)

rd*rhs(d = 0.5059754250e-1)

10436.49230

(20)

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five hundredths of a parsec, or 10,436 A.U. (The accepted value is about 13,000 A.U.)

 

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Reference

 

Cox, A. (Ed.). (2000). Allen's Astrophysical Quantities. (4th ed.). New York: Springer.

 

 

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