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How fast must be a plane 50 m long travel to be found by observer on the ground to be 0.10 nm shorter than 50 m?

Short Answer

Expert verified

The value of derived expression for the travel at a speed 600ms.

Step by step solution

01

Write the given data from the question

Consider a length of plane from observer isl=501.01010m.

Consider a length of plane at rest is l0=50m.

02

Determine the formula of derived expression for the travel at a speed.

Write the formula of derived expression for the travel at a speed.

υ=c1ι2ι02 …… (1)

Here, c speed of light, ι is length contraction of an object.

03

Determine the value of derived expression for the travel at a speed.

Thefollowingequationmaybeusedtoshowhowarelativisticeffectcausesanobject'slengthtocontract:

ι=ι01υ2c2

Arrive at the following formula for using the length contraction equation:

ι=ι01υ2c2ιι0=1υ2c2ι2ι02=1υ2c2υ2c2=1ι2ι02

Solve further as

υc=1ι2ι02υ=c1ι2ι02

Determine for using the speed's derived expression:

Substitute for3×108, c 501.01010for role="math" localid="1659321081955" ι2 and 502 for ι02 into equation (1).

υ=3×1081501.010102502=21063×108600ms

The plane has to move at a speed of 600ms for its length to seem 0.10 nm shorter from the observer's perspective.

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