As the earth revolves around the sun, distant objects can be seen from different locations. This idea can be used to measure the distance between earth and nearby stars and is called stellar parallax. To see how this process works, consider the simplified scenario below. A distant object is observed on earth from location 1 and again six months later at location 2. The angle at both location 1 and 2 is 89.9°. How many astronomical units is the distant object from location 1?
5 years ago
Answered By Leonardo F
Since the time elapsed from location 1 to 2 is 6 months, half of the time that the Earth takes to complete one lap around the sun, we will have an isosceles triangle with the base measuring the diameter of Earth's orbit, and base angles both 89.9°. Let's first find out the equal sides ofthe isosceles triangle. Calling them x and using sine law: remember that the average diameter of Earth's orbit is around 299 million km:
Smaller angle = 180° - 89.9° - 89.9° = 0.2°
x/sin(89.9°) = 299,000,000/sin(0.2°)
x = 1,406,100,000 km
We know that 1 astronomical unit (AU) is approximately 149,600,000 km. Dividing x by this value, we get:
1,406,100,000 km / 149,600,000 km = 9.40 AU
Hence, the distance of the distant object to location 1 is approximately 9.40 astronomical units.
5 years ago
Answered By Emily H
In this problem, as 6 months are between the two measurements, we know that we're dealing with an isocelese triangle with the diameter of earth's orbit as its base. Remembering that the distance between Earth and sun is 1AU, we can solve this problem in a few different ways:
if we consider that since we're dealing with an isosceles triangle, we can simplify the problem to one with two congruent right triangles. If we do this, then we need to solve for the hypotenuse of this triangle, which can be done by applying cosine (or secant). So making x the hypotenuse and distance between the earth and distant object, we get:
$\cos\left(89.9\right)=\frac{1}{x}$cos(89.9)=1x
$x=\frac{1}{\cos\left(89.9\right)}$x=1cos(89.9)
$x=\sec\left(89.9\right)$x=sec(89.9)
$x=579.6AU$x=579.6AU
Alternatively, we can use sine law on the entire triangle:
$\frac{\sin\left(\theta_1\right)}{x_2}=\frac{\sin\left(\theta_2\right)}{x_1}=\frac{\sin\left(180-\left(\theta_1+\theta_2\right)\right)}{2}$sin(θ1)x2 =sin(θ2)x1 =sin(180−(θ1+θ2))2
Where $x_1$x1 and $x_2$x2 are the distances from the Earth to the star at points 1 and 2, respectively and similarly $\theta_1$θ1 and $\theta_2$θ2 are the angles recorded at both observation points. Thus:
$\frac{x_1}{\sin\left(89.9\right)}=\frac{2}{\sin\left(180-\left(89.9+89.9\right)\right)}$x1sin(89.9) =2sin(180−(89.9+89.9))
$x_1=\frac{2\sin\left(89.9\right)}{\sin\left(0.2\right)}$x1=2sin(89.9)sin(0.2)
$x_1=\frac{1.999}{0.00349}$x1=1.9990.00349
$x_1=572.6AU$x1=572.6AU
So we can see that by both methods, the distance between Earth and the distant star is 572.6AU at location 1.