Chapter 15, Problem 33A

Joe Burpy is 30 years old and has a daughter who is 6 years old. Joe leaves on a space bus and takes a 5-year (space-bus time) round-trip at 0.99c. How old will he and his daughter be when he returns?

To determine

The age of person J and his daughter.

Answer to Problem 33A

  $t_J\text{= 35}\text{years}$

  $t_{daughter}=\text{41}\text{.7 years}$

Explanation of Solution

Given:

Before going to round trip with space bus,

Age of person J is 30 years.

Age of daughter is 6 years.

Velocity of the space bus is $v=0.99c$

Space bus time is $t_s=5\text{years}$

Formula used:

Timedilation formula is given by,

  $t=\frac{t_s}{\sqrt{1-\frac{v^2}{c^2}}}$ $\left(1\right)$

Where,

  $t$ is observed time

  $t_s$ is the space bus time

  $v$ is the velocity of the object moving with relativistic velocity

  $c$ is the velocity of light, $c=3\times {10}^8\text{m/s}$

Calculation:

Substituting the numerical values in equation $\left(1\right)$ ,

  $t=\frac{5\text{years}}{\sqrt{1-\frac{{\left(0.99c\right)}^2}{c^2}}}$

  $=\frac{5\text{years}}{\sqrt{1-0.98}}$

  $=\frac{5\text{}\text{years}}{\sqrt{1-0.98}}$

  $=\frac{5\text{years}}{\sqrt{0.02}}$

  $=35.4\text{years}$

Therefore, after returning from 5 years of round trip with space bus,

Age of person J is

  $\begin{array}{c}{t}_J=30\text{years}+5\text{years}\\ \text{ = 35}\text{years}\end{array}$

Age of his daughter is

  $\begin{array}{c}{t}_{daughter}=6\text{years}+35.4\text{years }\\ \text{= 41}\text{.4}\text{years}\end{array}$

Conclusion:

Therefore, age of person J is 35 years while his daughter is 41.4 years.