An astronaut whose pulse frequency remains constant at 72 beats \( / \mathrm{min} \) is sent on a voyage. What would her pulse beat be, relative to Earth, when the ship is moving relative to Earth at (a) 0.10 c and (b) 0.90 c ? (Note: In this case, we are asking for the period, which is the time in the spaceship, 72 beats \( / \mathrm{min} \) is the frequency, the formula of period \( =1 / \mathrm{f} \) )

Let the astronaut’s proper pulse frequency be \[ f = 72\, \text{beats/min}. \] Thus, her proper period (time between beats) in the spaceship is \[ T = \frac{1}{f} = \frac{1}{72}\, \text{min/beat}. \] For an observer on Earth, time dilation must be taken into account. The time dilation factor is given by the Lorentz factor \[ \gamma = \frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^2}}. \] In the Earth frame, the period of the astronaut’s pulse appears as \[ T' = \gamma T. \] The corresponding observed frequency is then \[ f' = \frac{1}{T'} = \frac{1}{\gamma T} = \frac{f}{\gamma}. \] We now consider the two cases: **(a) For \( v = 0.10c \):** First, calculate the Lorentz factor: \[ \gamma = \frac{1}{\sqrt{1-(0.10)^2}} = \frac{1}{\sqrt{1-0.01}} = \frac{1}{\sqrt{0.99}}. \] Approximating, \[ \sqrt{0.99} \approx 0.9950,\quad \text{so}\quad \gamma \approx 1.0050. \] Thus, the observed frequency is \[ f' = \frac{72}{1.0050} \approx 71.64\, \text{beats/min}. \] **(b) For \( v = 0.90c \):** First, calculate the Lorentz factor: \[ \gamma = \frac{1}{\sqrt{1-(0.90)^2}} = \frac{1}{\sqrt{1-0.81}} = \frac{1}{\sqrt{0.19}}. \] Approximating, \[ \sqrt{0.19} \approx 0.4359,\quad \text{so}\quad \gamma \approx 2.294. \] Thus, the observed frequency is \[ f' = \frac{72}{2.294} \approx 31.38\, \text{beats/min}. \] **Final Answers:** - (a) At \( v = 0.10c \), the pulse is observed to be about \( 71.64\, \text{beats/min} \) on Earth. - (b) At \( v = 0.90c \), the pulse is observed to be about \( 31.38\, \text{beats/min} \) on Earth.