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Falala
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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.
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.
none?
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?
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?