Chapter 39, Problem 57P

(a)

To determine

The minimum amount of energy that the acceleration require from the spaceship’s fuel.

Answer to Problem 57P

The minimum amount of energy that the acceleration require from the spaceship’s fuel is $\underset{\_}{8.63\times {10}^{22}\text{J}}$.

Explanation of Solution

Write the expression for the initial relativistic kinetic energy of the spaceship.

  $K_i=\left({\gamma }_i-1\right)m{c}^2$                                                                                                         (I)

Here, $K_i$ is the initial relativistic kinetic energy of the spaceship, ${\gamma }_i$ is the Lorentz factor corresponding to the initial state, $m$ is the mass of the spaceship, and $c$ is the speed of light.

Write the expression for the final relativistic energy of the spaceship.

  $K_f=\left({\gamma }_f-1\right)m{c}^2$                                                                                                      (II)

Here, $K_f$ is the final relativistic kinetic energy of the spaceship, and ${\gamma }_f$ is the Lorentz factor corresponding to the final state.

Write the expression for the change in kinetic energy of the spaceship.

  $\Delta K=K_f-K_i$                                                                                                          (III)

Here, $\Delta K$ is the change in kinetic energy of the spaceship.

Use expressions (I) and (II) in (III).

  $\begin{array}{c}\Delta K=\left({\gamma }_f-1\right)m{c}^2-\left({\gamma }_i-1\right)m{c}^2\\ =\left({\gamma }_f-{\gamma }_i\right)m{c}^2\end{array}$                                                                              (IV)

The minimum amount of energy required to accelerate the spaceship is equal to the change in kinetic energy of the spaceship.

Write the expression for Lorentz factor corresponding to the initial state.

  ${\gamma }_i=\frac{1}{\sqrt{1-\frac{u_i^2}{c^2}}}$                                                                                                             (V)

Here, $u_i$ is the initial velocity of spaceship.

Write the expression for the Lorentz factor corresponding to the final state.

  ${\gamma }_f=\frac{1}{\sqrt{1-\frac{u_f^2}{c^2}}}$                                                                                                          (VI)

Here, $u_f$ is the final velocity of spaceship.

Use expressions (V) and (VI) in expression (IV).

  $\Delta K=\left(\frac{1}{\sqrt{1-\frac{u_f^2}{c^2}}}-\frac{1}{\sqrt{1-\frac{u_i^2}{c^2}}}\right)m{c}^2$                                                                             (VII)

Conclusion:

Substitute $0.700\text{c}$ for $u_f$, and $0$ for $u_i$, $2.40\times {10}^6\text{kg}$ for $m$, and $2.998\times {10}^8\text{m}/\text{s}$ for $c$ in equation (VII) to find $\Delta K$.

  $\begin{array}{c}\Delta K=\left(\frac{1}{\sqrt{1-\frac{{\left(0.700\text{c}\right)}^2}{c^2}}}-\frac{1}{\sqrt{1-\frac{0}{c^2}}}\right)\left(2.40\times {10}^6\text{kg}\right){\left(2.998\times {10}^8\text{m}/\text{s}\right)}^2\\ =\left(1.40-1\right)\left(2.40\times {10}^6\text{kg}\right){\left(2.998\times {10}^8\text{m}/\text{s}\right)}^2\\ =8.63\times {10}^{22}\text{J}\end{array}$

Therefore, The minimum amount of energy that the acceleration require from the spaceship’s fuel is $\underset{\_}{8.63\times {10}^{22}\text{J}}$.

(b)

To determine

The amount of fuel required to provide the required energy.

Answer to Problem 57P

The amount of fuel required to provide the required energy is $\underset{\_}{9.61\times {10}^5\text{kg}}$.

Explanation of Solution

Write the expression for the rest energy of the fuel.

  $E=m_f{c}^2$                                                                                                              (VIII)

Here, $E$ is the rest energy of the fuel, and $m_f$ is the rest mass of the fuel.

Rest energy of the fuel is equal to the change in kinetic energy of the fuel.

  $\Delta K=m_f{c}^2$                                                                                                              (IX)

Use expression (IX) for $m$.

  $m_f=\frac{\Delta K}{c^2}$                                                                                                                  (X)

Use expression (VII) in (X).

  $\begin{array}{c}{m}_f=\frac{\left(\frac{1}{\sqrt{1-\frac{u_f^2}{c^2}}}-\frac{1}{\sqrt{1-\frac{u_i^2}{c^2}}}\right)m{c}^2}{c^2}\\ =\left(\frac{1}{\sqrt{1-\frac{u_f^2}{c^2}}}-\frac{1}{\sqrt{1-\frac{u_i^2}{c^2}}}\right)m\end{array}$                                                                         (XI)

Conclusion:

Substitute $0.700\text{c}$ for $u_f$, and $0$ for $u_i$, $2.40\times {10}^6\text{kg}$ for $m$, and $2.998\times {10}^8\text{m}/\text{s}$ for $c$ in equation (XI) to find $m_f$.

  $\begin{array}{c}{m}_f=\left(\frac{1}{\sqrt{1-\frac{{\left(0.700\text{c}\right)}^2}{c^2}}}-\frac{1}{\sqrt{1-\frac{0}{c^2}}}\right)\left(2.40\times {10}^6\text{kg}\right)\\ =\left(\frac{1}{\sqrt{1-{\left(0.700\right)}^2}}-1\right)\left(2.40\times {10}^6\text{kg}\right)\\ =9.61\times {10}^5\text{kg}\end{array}$

Therefore, the amount of fuel required to provide this energy is $\underset{\_}{9.61\times {10}^5\text{kg}}$.