Straightforward Parallax

Classical Mechanics Level 3

You are trying to measure the distance to a star. You are able to measure that over 6 months the position of the star shifts by an angle $(2a$ —see the diagram $)$ of 0.1 arcseconds.

Taking the distance from Earth to the sun to be $1.5\times 10^{11} \si{meter}$ , calculate the distance to the star in meters.

7.5 \times 10^{16}

1.5 \times 10^{17}

3.1 \times 10^{17}

6.2 \times 10^{17}

1 solution

Andrew Normand
Feb 1, 2017

The sun, the Earth and the star form a right angled triangle with an angle a = 0.05 arcseconds. The Earth-Sun distance (1AU), the distance to the star and the angle a are related by: $tan(a) = \frac{distance}{1 AU}$ . Since the angle is very small, tan(a) = a, so long as a is measured in radians.

Using this relationship, you can solve this in two ways. One is to convert the arcseconds into radians (divide by $\frac{3600*360}{2\pi}$ and then plug numbers into the equation. The second is to recognise that a parallax of 0.05 arcseconds tells you the star is 20 times more distant than a 1 arcsecond star, which would by definition be 1 parsec away. So you would then simply be converting 20 parsecs into metres.

Straightforward Parallax

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