How can the distance to nearby stars be calculated using the parallax method?

  • Thread starter Thread starter MathematicalPhysics
  • Start date Start date
  • Tags Tags
    Stars
AI Thread Summary
The parallax method calculates the distance to nearby stars using the formula d=1/(theta), where d is the distance in parsecs and theta is the parallax angle in arcseconds. Theta is indeed the parallax, determined by observing the star from two points in Earth's orbit, typically in January and June. The angular distance between these observations gives the value of theta. Alternatively, right ascension and declination can be used to compute the separation through spherical trigonometry. This method effectively allows astronomers to measure stellar distances accurately.
MathematicalPhysics
Messages
40
Reaction score
0
Just a little question, using the formula d=1/(theta), d is the distance in pc and theta is the angle in arsecs. Is theta simply the parallax? or if not, can it be calculated from right ascention and declination?

Thanks, Matt.
 
Astronomy news on Phys.org
Originally posted by MathematicalPhysics
Just a little question, using the formula d=1/(theta), d is the distance in pc and theta is the angle in arsecs. Is theta simply the parallax? or if not, can it be calculated from right ascention and declination?

Thanks, Matt.

You have to observe the star from two ends of the Earth's orbit, in January and in June (aphelion and perihelion, the two ends of the semimajor axis of the orbital ellipse). Then you compute the angular distance between the two observations; that's your theta. You could do the computation by recording your two observations as right ascension and declination and then doing spherical trig to calculate the separation.
 
This thread is dedicated to the beauty and awesomeness of our Universe. If you feel like it, please share video clips and photos (or nice animations) of space and objects in space in this thread. Your posts, clips and photos may by all means include scientific information; that does not make it less beautiful to me (n.b. the posts must of course comply with the PF guidelines, i.e. regarding science, only mainstream science is allowed, fringe/pseudoscience is not allowed). n.b. I start this...
Today at about 4:30 am I saw the conjunction of Venus and Jupiter, where they were about the width of the full moon, or one half degree apart. Did anyone else see it? Edit: The moon is 2,200 miles in diameter and at a distance of 240,000 miles. Thereby it subtends an angle in radians of 2,200/240,000=.01 (approximately). With pi radians being 180 degrees, one radian is 57.3 degrees, so that .01 radians is about .50 degrees (angle subtended by the moon). (.57 to be more exact, but with...
Back
Top