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91Ó°ÊÓ

The radius of Earth is 6370km, and its orbital speed about the Sun is 30m/s. Suppose Earth moves past an observer at this speed. To the observer, by how much does Earth’s diameter contract along the direction of motion?

Short Answer

Expert verified

In the rest frame of the observer, the diameter of earth contracts by8mm.

Step by step solution

01

Lorentz factor:

The result of the 2ndpostulate of special relativity is that the clock runs slower for a moving object when measured from a rest frame. The factor by which the clocks run differently is called the Lorentz factor.

The earth is moving at 30km/srelative to observer’s frame. The Lorentz factor for this speed is

γ=11-v2c2=11-30km/s23×105km/s2=1.000000006

02

Length contraction:

The length of an object measured in the object’s inertial reference frame is called proper length. When the length of this object is measured in any other inertial frame moving at relative speed is always shorter than the proper length. This is known as length contraction.

L=Lo1-β2=Loγ

Here, the diameter measured from the earth’s frame (proper length) is 12,740km, and the Lorentz factor is found to have a value of 1.000000006.

The distance measured from the observer’s reference(stationary) frame is

L=Loγ=12740km1.000000006=12739.99992km

In the rest frame of the observer, the diameter of earth contracts by,

12740km-12739.99992km=0.000008km=8mm

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