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how to solve this question

An airplane whose rest length is 40.0 m is moving at a uniform velocity with respect to the Earth at a speed of 630m/s. (a) By what fraction of its rest length will it appear to be shortened to an observer on Earth? (b) How long would it take by Earth clocks for the airplane’s clock to fall behind by 1 microsecond, assuming that only special relativity
applies?

even after using taylor series i get a very small value and 1 minus that small value gives me 1 again for part a
(edited 4 years ago)
you need to not use a calculator until the very end
Original post by studentzz123
An airplane whose rest length is 40.0 m is moving at a uniform velocity with respect to the Earth at a speed of 630m/s. (a) By what fraction of its rest length will it appear to be shortened to an observer on Earth? (b) How long would it take by Earth clocks for the airplane’s clock to fall behind by 1 microsecond, assuming that only special relativity
applies?

even after using taylor series i get a very small value and 1 minus that small value gives me 1 again for part a


Post your working. I don’t think you can use your “normal” school scientific calculator to compute the answer.
L/Lo = (1-(v^2)/(c^2))^1/2

using taylor series

(1-v^2/2c^2) = 1 - 2.205x10^-12 = 1 (from calculator) , what other way is there to compute the final answer?

Original post by Eimmanuel
Post your working. I don’t think you can use your “normal” school scientific calculator to compute the answer.
Original post by studentzz123
L/Lo = (1-(v^2)/(c^2))^1/2

using taylor series

(1-v^2/2c^2) = 1 - 2.205x10^-12 = 1 (from calculator) , what other way is there to compute the final answer?


In some courses, the lecturers allow the answer to be 1 - 2.205x10^-12 without further simplification.

If further simplification is a MUST, you can work out the pattern using your scientific calculator to derive the answer:

1 - 2.205x10^-3 = 0.997795
1 - 2.205x10^-4 = 0.9997795
1 - 2.205x10^-5 = 0.99997795
....

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