In the distant future, a `5.40xx10^(5)-kg` intergalactic ship leaves Earth orbit and accelerates to a constant speed of 0.92c. Determine the difference, `p-p_(0)`, between the relativistic and classical momenta of the ship.

Options

A. `3.9xx10^(14)kg*m//s`
B. `8.0xx10^(13)kg*m//s`
C. `2.3xx10^(14)kg*m//s`
D. `5.8xx10^(13)kg*m//s`

Answer

To determine the difference between the relativistic momentum \( p \) and the classical momentum \( p_0 \) of the intergalactic ship, we will follow these steps:

### Step 1: Calculate Classical Momentum \( p_0 \)
The classical momentum \( p_0 \) is given by the formula:
\[ p_0 = m_0 v \]
where:
- \( m_0 = 5.4 \times 10^5 \, \text{kg} \) (rest mass of the ship)
- \( v = 0.92c \) (velocity of the ship, where \( c \) is the speed of light)

Substituting the values:
\[ p_0 = (5.4 \times 10^5 \, \text{kg}) \times (0.92 \times c) \]
Using \( c \approx 3 \times 10^8 \, \text{m/s} \):
\[ p_0 = (5.4 \times 10^5) \times (0.92 \times 3 \times 10^8) \]
Calculating this gives:
\[ p_0 = (5.4 \times 10^5) \times (2.76 \times 10^8) = 1.4904 \times 10^{14} \, \text{kg m/s} \]

### Step 2: Calculate Relativistic Momentum \( p \)
The relativistic momentum \( p \) is given by the formula:
\[ p = \frac{m_0 v}{\sqrt{1 - \frac{v^2}{c^2}}} \]
First, we need to calculate \( \sqrt{1 - \frac{v^2}{c^2}} \):
\[ \frac{v^2}{c^2} = (0.92)^2 = 0.8464 \]
Thus,
\[ 1 - \frac{v^2}{c^2} = 1 - 0.8464 = 0.1536 \]
Now, taking the square root:
\[ \sqrt{1 - \frac{v^2}{c^2}} = \sqrt{0.1536} \approx 0.39 \]
Now substituting this back into the relativistic momentum formula:
\[ p = \frac{(5.4 \times 10^5) \times (0.92 \times c)}{\sqrt{0.1536}} = \frac{(5.4 \times 10^5) \times (2.76 \times 10^8)}{0.39} \]
Calculating this gives:
\[ p \approx \frac{1.4904 \times 10^{14}}{0.39} \approx 3.8205 \times 10^{14} \, \text{kg m/s} \]

### Step 3: Calculate the Difference \( p - p_0 \)
Now we can find the difference between the relativistic momentum and the classical momentum:
\[ p - p_0 = 3.8205 \times 10^{14} - 1.4904 \times 10^{14} \]
Calculating this gives:
\[ p - p_0 \approx 2.3301 \times 10^{14} \, \text{kg m/s} \]

### Final Answer
The difference between the relativistic and classical momenta of the ship is approximately:
\[ \boxed{2.33 \times 10^{14} \, \text{kg m/s}} \]