A space ship at rest on a launching pad has a mass of `1.00xx10^(5)` kg. How much will its energy have increased when the ship is moving at 0.600c?

Options

A. `1.12xx10^(21)J`
B. `2.25xx10^(21)J`
C. `1.62xx10^(21)J`
D. `6.00xx10^(21)J`

Answer

To solve the problem of how much the energy of a spaceship increases when it moves at a speed of 0.600c, we will use the principles of special relativity, specifically the concepts of rest mass and relativistic energy.

### Step-by-Step Solution:

1. **Identify the Rest Mass (m₀)**:
The rest mass of the spaceship is given as:
\[ m₀ = 1.00 \times 10^5 \, \text{kg} \]

2. **Determine the Speed as a Fraction of the Speed of Light (c)**:
The speed of the spaceship is given as:
\[ v = 0.600c \]

3. **Calculate the Lorentz Factor (γ)**:
The Lorentz factor is calculated using the formula:
\[ \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \]
Substituting \(v = 0.600c\):
\[ \gamma = \frac{1}{\sqrt{1 - (0.600)^2}} = \frac{1}{\sqrt{1 - 0.36}} = \frac{1}{\sqrt{0.64}} = \frac{1}{0.8} = 1.25 \]

4. **Calculate the Relativistic Mass (m)**:
The relativistic mass is given by:
\[ m = \gamma m₀ \]
Substituting the values:
\[ m = 1.25 \times (1.00 \times 10^5 \, \text{kg}) = 1.25 \times 10^5 \, \text{kg} \]

5. **Calculate the Total Energy (E)**:
The total energy of the spaceship when moving at speed \(v\) is given by:
\[ E = mc^2 \]
Substituting the relativistic mass:
\[ E = (1.25 \times 10^5 \, \text{kg}) \times (3 \times 10^8 \, \text{m/s})^2 \]
\[ E = (1.25 \times 10^5) \times (9 \times 10^{16}) = 1.125 \times 10^{22} \, \text{J} \]

6. **Calculate the Rest Energy (E₀)**:
The rest energy is calculated using:
\[ E₀ = m₀c^2 \]
\[ E₀ = (1.00 \times 10^5 \, \text{kg}) \times (3 \times 10^8 \, \text{m/s})^2 \]
\[ E₀ = (1.00 \times 10^5) \times (9 \times 10^{16}) = 9.00 \times 10^{21} \, \text{J} \]

7. **Calculate the Increase in Energy (ΔE)**:
The increase in energy is given by:
\[ \Delta E = E - E₀ \]
\[ \Delta E = (1.125 \times 10^{22} \, \text{J}) - (9.00 \times 10^{21} \, \text{J}) = 2.25 \times 10^{21} \, \text{J} \]

### Final Answer:
The increase in energy when the spaceship is moving at 0.600c is:
\[ \Delta E = 2.25 \times 10^{21} \, \text{J} \]