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$\bullet$ An alien spacecraft is flying overhead at a great distance asyou stand in your backyard. You see its searchlight blink on for0.190 s. The first officer on the craft measures the searchlightto be on for 12.0 ms. (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?

$=0.998 c$

Physics 101 Mechanics

Chapter 27

Relativity

Gravitation

Cornell University

University of Washington

University of Winnipeg

Lectures

03:55

In physics, orbital motion is the motion of an object around another object, which is often a star or planet. Orbital motion is affected by the gravity of the central object, as well as by the resistance of deep space (which is negligible at the distances of most orbits in the Solar System).

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Sir Isaac Newton described the law of universal gravitation in his work "Philosophiæ Naturalis Principia Mathematica" (1687). The law states that every point mass attracts every single other point mass by a force pointing along the line intersecting both points. The force is proportional to the product of the two masses and inversely proportional to the square of the distance between them.

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Okay, So in this problem, you see an alien spacecraft flying at a great distance of you. And you see the search light blink during a time that's called out G off zero point 19 seconds. But the captor in this brace craft watches the same searchlight to blink for a time. Let's go down to zero off 12 merely seconds. The first item of this problem, we need to answer which off this time is the proper time. Okay, so we know that the proper time is measured in the frame where the two events a car occurred at the same time. So the different reference where the search light is blinking is the reference frame off the captain. So we can say that for upper time needs to be Delta T zero equals 12 milli seconds. So this is a proper time. Now we have to calculate they speed off this spacecraft. Okay, so first of all, we know that the captor is under relativistic effects, so we can say that the time that we measure in the Earth is equal the time at the captain Measure die elated by gamma. So we can say that doubt t it. Close doubt T zero divided by the square roots off one miners. You're square C square. Okay, we can rearrange this, uh, putting square in both sides. Both sides. Let's rearrange the secretion. You're gonna have one miners. You're squared. Divided by C square equals doubt t zero divided by delta t square square. So we can say that they speed is equal to see square root one minus minus doubt t zero divided by delta t So this is the Squire route here. We have square here also have square. Okay, So institute in the values that we have, we can say that they speed is just see square it off one minus 12 times 10 minus street divided by zero point 19. Oh, this square. It's great rich. So after calculating this, we can see that the actual speed of the spacecraft is just zero point nine nine eight. See? And that's the final answer to just

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