Calculating distance (and percent error) using parallax

In summary, the conversation is about a homework problem that involves measuring the parallax angle of Mars at the time of opposition. The first part of the problem (a) asks for the distance between Earth and Mars using a baseline equal to Earth's diameter and a difference in measurements of 33.6". The second part (b) asks for the necessary synchronization of clocks in order to measure the distance with a 10% accuracy. The person seeking help is confident in their answer for part (a) but is struggling with part (b) and is seeking advice on how to approach it. They have consulted their professor and received conflicting advice, and the problem is due tomorrow.
  • #1
Falala
2
0
Hi all,

This is a homework problem, but I don't think asking for help here presents any ethical issues. I'm permitted to consult my classmates and compare solutions with them as long as I do my own work, and I plan to ask my professor for help if I still feel unsure after reading your responses.

Thanks for taking a look.

Homework Statement



In 1672, an international effort was made to measure the parallax angle of Mars at the time of opposition, when it was closest to the Earth.

(a) Consider two observers who are separated by a baseline equal to Earth's diameter. If the difference in their measurements of Mars's angular position is 33.6", what is the distance between Earth and Mars at the time of opposition?

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

Homework Equations



none?

The Attempt at a Solution



I'm relatively confident of my answer for (a) -- 7.82 x 10^10 m. To find the distance between Earth and Mars, I drew an isosceles triangle with base 1.28 x 10^7 m (the diameter of Earth). The distance between Earth and Mars is equal to 6.37 x 10^6 (half the diameter of Earth) divided by the tangent of 0.00467 degrees (half the parallax angle).

The problem is (b). I've drawn a diagram, but I'm not sure it's correct. The premise is that Observer 2 now observes Mars t seconds later than Observer 1 does. I started with the same triangle I drew for (a). Then I moved the original upper vertex (Mars 1) to the right a distance 5660t (the relative velocity of Mars times the time discrepancy t between the two observers) and called the new point Mars 2. Then I drew a line between Observer 2 and Mars 2, and I extended the original line between Observer 1 and Mars 1. (The rationale for this is that Observer 1's viewpoint hasn't changed, but Observer 2 is seeing Mars slightly later and, therefore, at a different location. In other words, Observer 1 is looking at Mars 1 and Observer 2 is looking at Mars 2. Right?) The intersection of the two lines O2-M2 and O1-M1 is the upper vertex of a new triangle, and the vertical altitude of that triangle is the new (incorrect) distance D.

Am I right so far?

Now I'm stuck. I need to relate t and D so that I can use the percent error to find the maximum allowed t. But I can't figure out what the relationship is. Help?
 
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  • #2
Sorry for the bump, but this is due tomorrow and I'm still stuck. I talked to the professor today, and she indicated that to do (b) I should run my calculations for (a) backwards, yielding an angle measurement, and then use the angular velocity of Mars to convert the angle into a time period. I did that, using the period of Mars's orbit, but it seems wrong to ignore Earth's movement (and ignore the hint given in the problem).

Twenty minutes after I left the professor's office, she sent me an email -- about something she'd thought about related to the problem -- that left me even more confused. I'm not sure if she was trying to restate what she'd already said or tell me that what she said this morning was not right.

HELP! This is due tomorrow, and I would appreciate any advice. You don't have to read my long paragraphs above, just tell me briefly how you would approach (b).

Thank you so much.
 

1. What is parallax and how is it used to calculate distance?

Parallax is the apparent shift in the position of an object when viewed from different angles. It is used in trigonometry to calculate the distance to an object by measuring the shift in its position from two different vantage points.

2. What is the formula for calculating distance using parallax?

The formula for calculating distance using parallax is d = (b / p) * 1 AU, where d is the distance to the object in parsecs, b is the baseline (distance between the two vantage points) in AU, and p is the parallax angle in arcseconds.

3. How accurate is the distance calculated using parallax?

The accuracy of the distance calculated using parallax depends on the precision of the measurements taken. Generally, parallax can be used to calculate distances up to a few hundred parsecs with an accuracy of about 10%. Beyond that, other methods such as spectroscopic parallax or standard candles are used.

4. What is percent error and how is it calculated in the context of parallax?

Percent error is a measure of how inaccurate a measurement is compared to the true value. In the context of parallax, percent error is calculated by taking the absolute value of the difference between the calculated distance and the actual distance, divided by the actual distance, and multiplied by 100.

5. What are some sources of error when calculating distance using parallax?

Some sources of error when calculating distance using parallax include measurement errors, atmospheric distortions, and the motion of the object being observed. These errors can be minimized by using precise instruments, taking multiple measurements, and accounting for any known motion of the object.

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