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Two events are observed in a frame of reference $S$ to occurat the same space point, the second occurring 1.80 s after thefirst. In a second frame $S^{\prime}$ moving relative to $S,$ the second eventis observed to occur 2.35 s after the first. What is the differencebetween the positions of the two events as measured in $S^{\prime \prime} ?$

$=4.53 \times 10^{8} \mathrm{m}$

Physics 101 Mechanics

Chapter 27

Relativity

Gravitation

University of Michigan - Ann Arbor

Simon Fraser University

University of Winnipeg

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.

05:36

Two events are observed in…

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Sam sees two events as sim…

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Two events in reference fr…

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An observer in reference f…

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03:53

Keilah, in reference frame…

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Okay, So in this problem, we have to reference frame. We have our initial reference frame. That is where we going called s. And we have a no only near show reference frame s prime, which is moving if the velocity v respect too. Reference s so in the inertial reference frame, we know that the distance between the two events is hero. Both events occur at the same point in space. But the difference in time off these events is just Let's go one point eight seconds. We know that in the norm inertial reference frame, the difference in time. Let's go without teeth grind dream. The two events is two point 35 seconds. But this time the difference in the position of the two events out My ex prime is different from zero. And the problem wants to calculate precisely what should be this difference in position relative to the initial reference frame. The norm inertial reference frame s Brian. So first of all, let's look so that dilation of time because we need to calculate the speed in his reference frame because we doubted speed. We can calculate we cannot calculated distance in this space. So First of all, we know that doubt deep prime equals gamma Doubter t because we see a dilation of time. So this is just to a point 35 equals 1.8 divided by one miners. Be square C square all days in the square root. If we rearrange this equation here we have that we have that fee equal. See square root for minors 1.8 that divides us two point 35. You square in the square average and we flick calculated this we're gonna have that developed. The speed between the two reference frame is just zero point 64 three. See? Okay, now that we have this speech, we can calculate the difference in position. Because if we think about that dear observer in the normal inertial reference frame s brain, it's watching the events. The events occur in different positions in the space. So this is just going to be Delta X spry. It grows V doubt. Cheap prime. So doubt Ex prime should be 0.643 c than multiplies two point 35. So this is just four. This is just four point 53 times 10 to the power off eight meters. And that's a separation between the two events. That observer in s crime cease. And that's the final answer. Thanks for watching.

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