Chapter 26, Problem 70GP

To determine

The mass that is converted to energy to accelerate the spaceship and the time taken by trip.

Answer to Problem 70GP

The mass: $37,500kg$

The time: $\text{33}\text{y}$

Explanation of Solution

Given:

Total mass is $\text{150,000}\text{kg}$

Speed is $\text{0}\text{.60}\text{c}$

Distance is $\text{25}\text{ly}$

Formula used:

It is known that,

  $KE=\frac{m_0{c}^2}{\sqrt{1-\frac{v^2}{c^2}}}-m_0{c}^2$

Where,

  $m_0$ is rest mass

  $KE$ is Kinetic energy

  $v$ is velocity

  $c$ is speed of light, $3.00\times {10}^{8}m/s$

It is known that,

  $\Delta t=\frac{\Delta t_0}{{\left[1-\left(\frac{v^2}{c^2}\right)\right]}^{\frac{1}{2}}}$

Where,

  $t_0$ is rest time

  $t$ is time

  $v$ is velocity

  $c$ is speed of light, $3.00\times {10}^{8}m/s$

Calculation:

The kinetic energy that must be supplied is $KE=\frac{m_0{c}^2}{\sqrt{1-{\left(0.60\right)}^2}}-m_0{c}^2=0.25m_0{c}^2$

This is the amount of energy that would be liberated by converting $0.25m_0$ of mass into energy, or $0.25\left(150,000kg\right)=37,500kg$

From Earth’s point of view the trip will take

  $\Delta t=\frac{d}{v}=\frac{\left(25y\right)c}{0.60c}=41.7y$

According to the astronauts, it will take less time:

  $\Delta t=\frac{\Delta t_0}{{\left[1-\left(\frac{v^2}{c^2}\right)\right]}^{\frac{1}{2}}}$

  $41.7y=\frac{\Delta t_0}{\sqrt{1-{\left(0.60\right)}^2}}$

  $\Delta t_0=33y$