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$\cdot$ Inside a spaceship flying past the earth at three-fourths thespeed of light, a pendulum is swinging. (a) If each swing takes1.50 s as measured by an astronat performing an experimentinside the spaceship, how long will the swing take as measured by a person at mission control on earth who is watching the experiment? (b) If each swing takes 1.50 s measured by aperson at mission control on earth, how long will it take asmeasured by the astronaut in the spaceship?

$=0.9921 \mathrm{s}$

Physics 101 Mechanics

Chapter 27

Relativity

Gravitation

Rutgers, The State University of New Jersey

Simon Fraser University

University of Sheffield

McMaster University

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).

03:18

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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Inside a spaceship flying …

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Okay, This problem is asking about time dilation. So we're given a spaceship that's traveling at 3/4 C and mama pendulum on the spaceship is measured. Have a period of 1.5 seconds. So what time dilation says is that for um, observers outside the spaceship thiss time is going to be violated by a factor. Gamma, which comes from the speed of the ship. So we have 1.5 sometimes gamma equals 1.5 over square root of one minus 3/4 square and then for part B. Um, no, it's the people on earth that measure 1.5 seconds and there die elated in time with respect to the spaceships of the time lapse on the spaceship has to be shorter. And so that time will be given by 1.5 times square to one minus 34 squared. So when I think about these problems, tried to he track in my head whether the length or the time um, I should be getting larger or smaller if it's getting larger, Um, you multiply by gamma. It's getting smaller. You multiply by one over gamma

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