As observed on earth, a certain type of bacterium is known to double in number e

Question

As observed on earth, a certain type of bacterium is known to double in number every 6 hours. Two cultures of these bacteria are prepared, each consisting initially of one bacterium. One culture is left on earth and the other placed on a rocket that travels at a speed of 0.980c relative to the earth. At a time when the earthbound culture has grown to 128 bacteria, how many bacteria are in the culture on the rocket, according to an earth-based observer? (Round your answer to the nearest whole number.)

Explanation / Answer

Doubling of population P every 26.4 h means P = 2^N , where N = 1 , 2 , 3, ... , with the total elapsed time (1) ..... Te = N*(26.4 h) Given P = 256 = 2^N, we solve for N in the following way: .......... ln(256) =ln(2^N) = N*[ln(2)] (2) ..... N = ln(256)/ln(2) = 8 The total elapsed time according to (1) is then (3) ..... Te = N*(26.4 h) = 8*(26.4 h) = 211.2 h Since there is no relative motion between the clock used in measuring the above time and the culture of bacteria whose population is being studied, that time is the proper time To (according to an earth-based observer), so that (4) ..... To = Te = 211.2 h Using the time dilation formula (5) ..... T = To/sqrt [ 1 – (V/C)^2 ] where V = 0.796*C, we get (6) ..... sqrt [1 – (V/C)^2 ] = sqrt [1 – (0.796)^2 ] = 0.6053 (7) ..... T = To/sqrt [ 1 – (V/C)^2 ] = (211.2 h)/(0.6053) = 349 h for the elapsed time on the rocket as measured by an observer on earth. According to (1), Te = N*(26.4 h) = 349 h , gives (8) ..... N = (349 h)/(26.4 h) = 13.22 or N = 13 only It follows that according to the earth-based observer, the number of bacteria in the culture on the rocket is (9) ..... P = 2^(13) = 8,192

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