- #1

clandarkfire

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## 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

[tex]tan\frac{\theta}{2}=\frac{R_{Earth}}{d}[/tex]

## The Attempt at a Solution

a)I think I know this part.

[tex]tan\frac{\theta}{2}=\frac{R_{Earth}}{d}[/tex]

[tex]d=\frac{R_{Earth}}{tan\frac{\theta}{2}}[/tex]

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, [tex]d=\frac{2R_{Earth}}{\theta}[/tex] if theta is measured in radians. This means part a) could also be done like this.

[tex]\theta=33.6''*\frac{1 arcmin}{60 arcsec}*\frac{1 degree}{60 arcsec}*\frac{2\pi radians}{360 degrees} = 1.63*10^{-4} radians[/tex]

so [tex]d=\frac{2R_{Earth}}{1.63*10^{-4}}=7.82*10^{10} m[/tex]

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