Question

Suppose the straight-line distance between New York and San Francisco is 4.1 × 106 m (neglecting the curvature of the earth). A UFO is flying between these two cities at a speed of 0.68c relative to the earth. What do the voyagers aboard the UFO measure for this distance?

The total energy of an object is 7.86 × 10¹² J, and its kinetic energy is 6.00 × 10¹² J. What is the mass m of the object?

A radar antenna is rotating and makes one revolution every 25 s, as measured on earth. However, instruments on a spaceship moving with respect to the earth at a speed v measure that the antenna makes one revolution every 45 s. What is the ratio v/c of the speed v to the speed c of light in a vacuum?

As observed on earth, a certain type of bacteria is known to double in number every 24 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 0.932c relative to the earth. At a time when the earthbound culture has grown to 16 bacteria, how many bacteria are in the culture on the rocket, according to an earth-based observer?

Answer 1

distance d between NewYork and SanFransisco as observed by voyager is given by

where is distance known to observer in earth

d = 4.110⁶ [ 1 - (0.68)² ]^1/2 = 3 10⁶ m

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rest energy = E - E_kin = mc²= (7.86 - 6.00)10¹² J = 1.86 10¹² J

mass = 1.86 10¹² / (3 10⁸)² = 2.06 10^-5 kg

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where is time measured by moving observer and is the time measured by observer at rest

( / ) = 25/45 = 5/9 = [ 1 - ( v² / c² ) ]^1/2

[ 1 - ( v² / c² ) ] = 25/81    or   v/c = 0.831

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16 bacteria grown and doubling rate is 24 hour,

then 2ⁿ = 16 or n =4 .

Hence time taken to get 16 bacteria = 424 = 96 hrs

time measured by observer at earth = 96 [1 - ( 0.932 )² ]^1/2 34.8 hr

Hence only 2 bacteria are in the culture of rocket according to earth based observer

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