(2-4) ① u = velocity of the star in its orbit = circumference / orbit time = 2r / T
T = 27.2 days = (27.2 x 24 x 60 x 60) = 2 350 000 s
2 x π x 3.0 x 1010 / 2 350 000 = 80 000 ms-1
② t(1) = distance / velocity = D / (c + u)
t(2) = D / (c – u)
③
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which (using a similar rearrangement to the one you used with the merry-go-round), simplifies to:
④ In this case, c >>u so we can simplify this further, to a very good approximation, as:
⑤ D = 70 light years = 70 x 9.46 x 1015 = 6.6 x 1017 m
so t2 - t1 = (2 x 6.6 x 1017 x 80 000) / (3 x 108)2 = 1.2 x 106 s = 13.6 days
This suggests that it should take 13.6 days longer for light to reach the Earth from the orbiting star when it is moving away from us. The question is, during this extra time, how far round would this star get in its orbit?
Orbital period, T = 27.2 days. So in 15 days, it completes 13.6 / 27.2 = 0.5 orbits.
In other words, light from star B should reach the Earth simultaneously from two different positions, half an orbit apart. Therefore two images of this star would simultaneously be visible, on opposite sides of star A, during part of its orbit. Which is amazing and exciting, except that it doesn’t happen ………
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