Measuring distance with parallax

[clandarkfire]

Homework Statement

a) Two observers are separated by a baseline equal to Earth's diameter. If the differences in their measurements of Mars' position at opposition (i think this means when it is closest to earth) is 33.6" (arcsec). What is the distance between Earth and Mars at opposition?

b)If the distance to Mars is to be measured within 10%, how closely must the clocks used by the two observers be synchronized? (Ignore the rotation of the earth. THe orbital velocities of Earth and Mars are 29.79 km/s and 24.13km/s respectively.)

Homework Equations

$$tan\frac{\theta}{2}=\frac{R_{Earth}}{d}$$

The Attempt at a Solution

a)I think I know this part.
$$tan\frac{\theta}{2}=\frac{R_{Earth}}{d}$$

$$d=\frac{R_{Earth}}{tan\frac{\theta}{2}}$$
Converting theta back into radians I get 7.82x10^10 m.

b) This I'm having trouble with. Well I want 10% error; this means that my error should be 7.82x10^9 m. Observers on Earth will see that Mars is moving at 5.66 km/s (the difference in the orbital velocity between Earth and Mars). I know that the distance Mars seems to travel will be 5660*Δt meters. Somehow, I need to figure out angular momentum. And this is how far I got.

It might make the problem easier to point out that when computing parallax, $$d=\frac{2R_{Earth}}{\theta}$$ if theta is measured in radians. This means part a) could also be done like this.
$$\theta=33.6''*\frac{1 arcmin}{60 arcsec}*\frac{1 degree}{60 arcsec}*\frac{2\pi radians}{360 degrees} = 1.63*10^{-4} radians$$
so $$d=\frac{2R_{Earth}}{1.63*10^{-4}}=7.82*10^{10} m$$

 

[Simon Bridge]

(a) the measurements form an equi-lateral triangle with the diameter of the Earth as a base. The distance d is the height of the triangle, and the difference in positions is the angle at the apex ... you figure it out :)

comment on your figure - how does it compare to more routinely accepted figures?

(b) In the time between the measurements, Mars and Earth move in their orbits ... so the two observers are pointing their telescopes at different positions. A quick sketch will show you what this does to the measurement.

 

[clandarkfire]

I know how to do part a using parallax... though I believe its an isosceles triangle, not an equal-lateral one, right?

For part b, I've spent the last hour doing sketches... but I haven't really gotten very far. Basically, I know that from Earth's perspective, Mars is moving clockwise at 5.66*10^3 m/s. This means that the angle will no longer be 33.6''; because one vertex of the triangle will have the same angle as before, and the other will have that angle plus some other angle α. However, I don't know how to express α in terms of time or distance, and don't know at what rate its changing. Could you be a bit more specific, please?

 

[Simon Bridge]

I know how to do part a using parallax... though I believe its an isosceles triangle, not an equal-lateral one, right?

Good: you spotted the um deliberate mistake :/

For part b, I've spent the last hour doing sketches... but I haven't really gotten very far. Basically, I know that from Earth's perspective, Mars is moving clockwise at 5.66*10^3 m/s. This means that the angle will no longer be 33.6''; because one vertex of the triangle will have the same angle as before, and the other will have that angle plus some other angle α. However, I don't know how to express α in terms of time or distance, and don't know at what rate its changing. Could you be a bit more specific, please?

Are the orbits big enough that they can be considered flat in the timeframe of the measurements?

 

[clandarkfire]

Good: you spotted the um deliberate mistake :/

Are the orbits big enough that they can be considered flat in the timeframe of the measurements?

Well, they are pretty big, but obviously not flat. I imagine any curvature will be minimal in the time-frame available. My professor did not, however, tell us that it would be OK to treat them as if they were flat.

 

[Simon Bridge]

If the systematic error introduced by ignoring the curvature is less than 10% then cannot you treat them as flat?

Anyway - treating them as curved: you'll have a lot of triangles to draw. Do you think they are sufficiently low eccentricity orbits to treat as circular over the time frame?

 

[clandarkfire]

If the systematic error introduced by ignoring the curvature is less than 10% then cannot you treat them as flat?

Anyway - treating them as curved: you'll have a lot of triangles to draw. Do you think they are sufficiently low eccentricity orbits to treat as circular over the time frame?

Yes. Perhaps its also best to just treat them as flat.

 

[clandarkfire]

Does anyone else know how to do part b? I haven't gotten anywhere.

 

[voko]

Assume that the observers have time discrepancy D. That is to say, they measure the angles not at the same time, one measures it D seconds after another. I think you can assume that the relative motion of Mars is just a straight line. That will result in a slightly different parallax. Express it in terms of the distance to Mars, discrepancy D, the relative velocity of Mars, and the true parallax.

 

[Simon Bridge]

Yeh - what voko said is what I'm trying to get you to do. Have you tried that?
Cannot properly advise you without knowing where you are getting stuck.

 

[clandarkfire]

I think I've got it. Can someone check my work?

First, refer to the attached diagram.

Since d=7.82*10^10, the margin for error is 7.82*10^9. We need them to calculate a distance between 7.04*10^10 and 8.60*10^10 so that $$\frac{|d-d_{0}|}{d}=\frac{1}{10}$$
Suppose person B measures and records angle β. Person A measures Δt seconds later. Angle α will be bigger than it should be, because x has gotten larger. Specifically, if A measures Δt seconds after B, then $$\alpha=\arctan(\frac{x}{d})=\arctan{(\frac{R_{E}+5660*\Delta{t}}{d})}$$
We know R and d; we just need to figure out what α will give a d=7.04*10^10 and a d=8.60*10^10. We can do this using the original equation we used to find parallax, $$d=\frac{R_{E}}{\tan(\frac{\alpha+\beta}{2})}$$.
So, let's solve for θ=α+β, and from there we'll solve for α.
$$7.04*10^{10}=\frac{6371000}{\tan(\frac{\theta}{2})}$$
$$\theta=2\arctan(\frac{6371000}{7.04*10^{10}})=0.0103°=37.33''$$
To keep things simple, let's assume α and β were each 33.6/2=16.8 when time was measured correctly. Then β=16.8'' and $$α=37.33''-16.8''=20.53''=5.7*(10^{-3})°$$
Ok, now we've found α. Now we use it in the original equation to solve for Δt.
$$\tan(\alpha)=\frac{R_{E}+5660*\Delta{t}}{d}$$
$$\Delta{t}=\frac{d*\tan(\alpha)-R_{E}}{5660}=252.8 s$$

Now, we could do the whole thing over again using d=8.60*10^10 in our equation to solve for α, and we would get Δt = -206.8 seconds. However, since we're assuming the first measurement is made at the time of opposition and the second one Δt seconds later, Δt can't be negative.

Thus, the clocks must be synchronized to at least 252.8 seconds
And I have a feeling there's an easier way to do this...