Distance of Star: Solve Parallax Equation for km & Light Years

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Discussion Overview

The discussion revolves around solving the parallax equation to determine the distance to a star in both kilometers and light years. Participants explore the application of the equation and engage in calculations related to astronomical distances.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents the parallax equation D=(d/2)/tan(theta/2) and provides specific values for d and theta, seeking assistance with calculations.
  • Another participant suggests that the provided units are in kilometers and degrees, encouraging the original poster to attempt the calculations themselves.
  • A participant confirms that the calculated distance of approximately 343,774,677,078,406.66 km is correct and provides the exponential form as 34 x 10^13 km.
  • There is a discussion about the conversion of kilometers to light years, with one participant calculating the distance to be around 27 light years, while another suggests it should be approximately 36.3 light years, indicating a potential error in the calculations.
  • Participants share personal academic backgrounds, with one mentioning their coursework related to cosmological phenomena.

Areas of Agreement / Disagreement

Participants generally agree on the calculated distance in kilometers, but there is disagreement regarding the conversion to light years, with differing results presented. The discussion remains unresolved regarding the correct light year calculation.

Contextual Notes

There are unresolved aspects regarding the accuracy of the light year conversion and potential errors in calculations, as well as assumptions about the values used in the parallax equation.

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please can somebody help me with this parallax equation:

D=distance to star
theta=angle
using the rule D=(d/2)/tan(theta/2)
when d = 300*10tothe6
and theta = 5*10tothe-5

what is the distance to the star in km and light years?
 
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Okay - assuming that you've given the units in kilometres and degrees...

To get the answers in km you should just be able to plug the numbers into a calculator.

To get the answer in light years you have to first calculate how far a light year is (in km, obviously).

Rather than me just giving you the answer, why don't you have a go, then we can see if our answers agree...

Jess
 
Last edited:
hi there thankyou for your interest in my posting jess

d is in km
theta is in degrees
 
You'll be wanting to hit 'post reply' rather than 'new thread', then :wink:

Oooh look - we've been moved

Jess
 
Last edited:
Originally posted by Jess
You'll be wanting to hit 'post reply' rather than 'new thread', then :wink:

Jess

Presto! The threads are merged.
 
for this answer i get a result of:

343774677078406.6587981892079753

So, is this answer the number of km?

what is this answer in exponential form? ie: 34 * 10tothepower13?

1 light year is 9.467*10tothepower12, so in light years i think the distance will be:
27 light years

Please let me know if i am right
 
Yes, that's the answer in kilometres, and it's the same one that I've got.

34 x 10^13 km is correct as well.

If you take the speed of light to be 300,000 km/s, then a light year is 9.461 x 10^12 km - which looks roughly like what you put.

I get an answer of 36.3 l.y though - you might have typed a digit wrong somewhere, perhaps. Even a back-of-the-envelope type calculation gives an answer of at least 34, if you look closely.

Jess
 
re

thankyou very much jess!
I need the help for some uni coursework.
p.s. I am from edinburgh too!
 
This isn't Astronomy 1Ah or 1Ch work, is it?
 
  • #10
i study at the university of abertay dundee and travel from edinburgh.
My coursework is from "horizons of science 1".
this module is concerned with the big bang, black holes, telecommunications and cosmological phenomena
 
  • #11
Hehe - for a moment there I thought you might be one of my tutees (if that's the right word)

Good luck with the rest of your course...

Jess
 

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