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An alien spacecraft is flying overhead at a great distance as you stand in your backyard. You see its searchlight blink on for The first officer on the spacecraft measures that the searchlight is on for (a) Which of these two measured times is the proper time? (b) What is the speed of the spacecraft relative to the earth expressed as a fraction of the speed of light c?

$a ) t=12 m s$

$b ) u=0.998 c$

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So here we have the diagram, the space cop as it cools past, that's be you return to be up the proper time, remember, is the time measure in the frame with the two events occur at the same point, and hence the proper time in this situation, which we need to calculate, is the time measured by the first often officer on the spacecraft. But this proper time enabled by doubt he not. And this is equal to 12 maybe sick. It's the time, major, but the light pulse as an observer in the spacecraft. So next up, can we calculate the speed off the space cop you? Well, we know an expression that's a T. The time measured by the observable up is equal to gamma delta t not, And that's down to Do you know what the proper time do you mind it by the square root both one minus use crib a C scream. And so if we do the mathematics and review in this equation, we can solve the speed you and you to see and the square roots off one minus out, not divided by doubts. T, both of which have known all squared And so now we can substitute our values. It's crazy. So we leave. See, as is quantify. Per square root off one minus The proper time is 12 many second. So that's 12 just into the minus three. Divided buying the time measured by the observable on but 0.19 seconds. And this is squared. Indians Uribe Cancer off zero point nine nine eight time. See? So the speed of the spacecraft is actually 99.8 times the speed of light. 99.8 sinned felons.

University of Kwazulu-Natal

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