In Einstein\'s theory of special relativity, the kinetic energy of an object Ls

Question

In Einstein's theory of special relativity, the kinetic energy of an object Ls given by the following formula Ek = mc2/1 - v/c)2 - m2 where m is the mass of the object when at rest, v Ls its velocity, and c is the speed of light. Using a first order Taylor expansion of 1/1 - (v/c)2, show that the above reduces to Ek = 1/2mv2. This is the kinetic energy from Newtonian mechanics. In a vacuum, c. = 3.0 108m/s. Using the error formula for Taylor polynomials. determine the maximum value of v for which the error is less than 10-2. Convert this to miles per hour. This shows how fast you need to be going before relativistic effects begin to make a difference.

Explanation / Answer

a) first order expansion of (1+x)^k = 1 + kx

so Ek = mc^2 ( 1 - (v/c)^2)^(-1/2) - mc^2

= mc^2 ( 1 - (-1/2) *(v/c)^2 ) - mc^2

= 1/2 m c^2 (v/c)^2 = 1/2 mv^2

b)
error/Ek = (Ek - 1/2 mv^2)/Ek

= (1/sqrt(1-(v/c)^2 - 1 - 1/2 v^2/c^2)/(1/sqrt(1-(v/c)^2 - 1) = 10^(-2)

(1/sqrt(1-(v/3E8)^2 - 1 - 1/2 v^2/3E8^2)/(1/sqrt(1-(v/3E8)^2 - 1) = 10^(-2)


v= 3.46E7 m/s

c) converting to mi/hr

v=7.74E7 mph

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