bcjochim07
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Novas and special relativity--check my work?
Suppose Earth astronomers see two novas occur simultaneously, one in constellation Orion and the other in Lyra. Both nova are the same distance from Earth (2.5e3 cy) and are in exactly opposite directions from Earth. Observers are on board an aircraft flying at 1000 km/h on a line from Orion toward Lyra.
a) for the observers on the aircraft, how much time separates the nova?
v=1000 km/hr = 277.78 m/s =9.26e7*c
Beta = 9.26e7
Gamma= approximately 1
a) I will say that the astronomers see the novas both occur at t1(Orion)=t2(Lyra)=0 and that the origin of the Earth's frame is on the Earth.
Applying the Lorentz transform:
t1'=1*(0-(9.26e7*c)(-2.5e3cy)/c^2) = 0.002315 y
t2'=1*(0-(9.26e7*c)(2.5e3cy)/c^2) = -0.002315 y
Therefore, the interval measured by the observers on the aircraft is 0.00463 y = 40.56 hr.
The observers on the aircraft see the nova on Lyra occur first.
Is this correct?
Homework Statement
Suppose Earth astronomers see two novas occur simultaneously, one in constellation Orion and the other in Lyra. Both nova are the same distance from Earth (2.5e3 cy) and are in exactly opposite directions from Earth. Observers are on board an aircraft flying at 1000 km/h on a line from Orion toward Lyra.
a) for the observers on the aircraft, how much time separates the nova?
Homework Equations
v=1000 km/hr = 277.78 m/s =9.26e7*c
Beta = 9.26e7
Gamma= approximately 1
The Attempt at a Solution
a) I will say that the astronomers see the novas both occur at t1(Orion)=t2(Lyra)=0 and that the origin of the Earth's frame is on the Earth.
Applying the Lorentz transform:
t1'=1*(0-(9.26e7*c)(-2.5e3cy)/c^2) = 0.002315 y
t2'=1*(0-(9.26e7*c)(2.5e3cy)/c^2) = -0.002315 y
Therefore, the interval measured by the observers on the aircraft is 0.00463 y = 40.56 hr.
The observers on the aircraft see the nova on Lyra occur first.
Is this correct?
I was thinking to complicated about it lol Thx