Gravitation

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##### Christina K.

Rutgers, The State University of New Jersey

LB
##### Aspen F.

University of Sheffield

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### Video Transcript

we know the identity is equal to Ah, proper time. Delta Petey divided by square root off one minus. We squared, divided by C square. And this came to rate unease. Uh, Delta T It's equal to one minus, um, one minus. So we divided by See home square into minus one, divided by two multiplied by proper time. Now, using binomial tea room, we have Delta T is equal to one minus, uh, minus one divided by two times we divided by see whole square times proper time Delta Bi So be of his extra bracket. Let me do it. All right, so now let's up. Like in the values we have, Delta T is equal to one place, one divided by two into We have three and we equals, uh, 35 on zero 35 0.0, divided by speed, off light and speed of light ese tree multiplied by 10 to the power eat square. Multiply by appropriate averages 26 point zero minutes. I mean, it's further solving. Delta T is equal to, uh, 26 point Zito minutes. Bless 1.6 Multiply by 10 to the power minus 11 seconds minus 11 seconds.

#### Topics

Gravitation

##### Christina K.

Rutgers, The State University of New Jersey

LB
##### Aspen F.

University of Sheffield

Lectures

Join Bootcamp