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 one year ago
A mechanism on Earth used to shoot down geosynchronous satellites that house laserbased weapons is
finally perfected and propels golf balls at 0.94c.
a) How far will a detector riding with the golf ball initially measure the distance to the satellite?
[Geosynchronous satellites are placed 3.58x104
km above the surface of the Earth.]
b) How much time will it take the golf ball to make the journey to the satellite in the Earth’s frame?
How much time will it take in the golf ball’s frame?
 one year ago
A mechanism on Earth used to shoot down geosynchronous satellites that house laserbased weapons is finally perfected and propels golf balls at 0.94c. a) How far will a detector riding with the golf ball initially measure the distance to the satellite? [Geosynchronous satellites are placed 3.58x104 km above the surface of the Earth.] b) How much time will it take the golf ball to make the journey to the satellite in the Earth’s frame? How much time will it take in the golf ball’s frame?

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modphysnoob
 one year ago
Best ResponseYou've already chosen the best response.0@Jemurray3 do I use length contraction on this? if so , how?

Jemurray3
 one year ago
Best ResponseYou've already chosen the best response.1Yes, for the first part. As straightforwardly as you possibly could... \[ L = L_0 \sqrt{1v^2/c^2} \]

modphysnoob
 one year ago
Best ResponseYou've already chosen the best response.0so L0 would be 3.58x10^4 ?

Jemurray3
 one year ago
Best ResponseYou've already chosen the best response.1Yes, the proper distance is measured in the earth frame because the geosynchronous satellites are at rest relative to the earth.

modphysnoob
 one year ago
Best ResponseYou've already chosen the best response.0for b) do I do just normal calcuation using t= l/v then lorentz transform that t?

Jemurray3
 one year ago
Best ResponseYou've already chosen the best response.1Yes. Alternatively, you could divide the contracted distance by the velocity.

modphysnoob
 one year ago
Best ResponseYou've already chosen the best response.0oh I see, that works too Thank you for your help
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