At a particular moment, a spacecraft of mass 15000 kg is at location (-6.4 × 107, 0, 0) m relative to an origin at the center of the Earth. At this instant the momentum of the spacecraft is (0, 3.7 × 10', 0) kg • m/s, and the force on the spacecraft by the Earth is (1472, 0, 0)N.

Assuming the force on the spacecraft by the Earth is approximately constant over a 1 hour interval, what will the momentum of the spacecraft be in 1 hour? (Remember to convert At to seconds.)

What will the spacecraft's new position be after 1 hour? Use the approximation vavg = pvector / m.





Let's break down the problem step by step.

### Given Information:
- Mass of the spacecraft: \( m = 15000 \) kg
- Initial position: \( \mathbf{r}_0 = \langle -6.4 \times 10^7, 0, 0 \rangle \) m
- Initial momentum: \( \mathbf{p}_0 = \langle 0, 3.7 \times 10^7, 0 \rangle \) kg·m/s
- Force on the spacecraft by Earth: \( \mathbf{F} = \langle 1472, 0, 0 \rangle \) N
- Time interval: \( \Delta t = 1 \) hour \( = 3600 \) s

### Step 1: Determine the Change in Momentum
The change in momentum \( \Delta \mathbf{p} \) over the time interval is given by:

\[
\Delta \mathbf{p} = \mathbf{F} \cdot \Delta t
\]

Substituting the given values:

\[
\Delta \mathbf{p} = \langle 1472, 0, 0 \rangle \text{ N} \times 3600 \text{ s} = \langle 1472 \times 3600, 0, 0 \rangle \text{ kg·m/s} = \langle 5.2992 \times 10^6, 0, 0 \rangle \text{ kg·m/s}
\]

### Step 2: Calculate the Final Momentum
The final momentum \( \mathbf{p}_f \) is the initial momentum plus the change in momentum:

\[
\mathbf{p}_f = \mathbf{p}_0 + \Delta \mathbf{p} = \langle 0, 3.7 \times 10^7, 0 \rangle \text{ kg·m/s} + \langle 5.2992 \times 10^6, 0, 0 \rangle \text{ kg·m/s}
\]
\[
\mathbf{p}_f = \langle 5.2992 \times 10^6, 3.7 \times 10^7, 0 \rangle \text{ kg·m/s}
\]

### Step 3: Calculate the Average Velocity
Using the approximation \( \mathbf{v}_{\text{avg}} = \frac{\mathbf{p}}{m} \), we calculate the average velocity over the 1-hour interval:

\[
\mathbf{v}_{\text{avg}} = \frac{\mathbf{p}_f}{m} = \frac{\langle 5.2992 \times 10^6, 3.7 \times 10^7, 0 \rangle \text{ kg·m/s}}{15000 \text{ kg}}
\]
\[
\mathbf{v}_{\text{avg}} = \langle \frac{5.2992 \times 10^6}{15000}, \frac{3.7 \times 10^7}{15000}, 0 \rangle \text{ m/s} = \langle 353.28, 2466.67, 0 \rangle \text{ m/s}
\]

### Step 4: Calculate the Displacement Over 1 Hour
The displacement \( \Delta \mathbf{r} \) over the 1-hour interval is given by:

\[
\Delta \mathbf{r} = \mathbf{v}_{\text{avg}} \times \Delta t = \langle 353.28, 2466.67, 0 \rangle \text{ m/s} \times 3600 \text{ s}
\]
\[
\Delta \mathbf{r} = \langle 353.28 \times 3600, 2466.67 \times 3600, 0 \rangle \text{ m} = \langle 1.272 \times 10^6, 8.880 \times 10^6, 0 \rangle \text{ m}
\]

### Step 5: Calculate the Final Position
The final position \( \mathbf{r}_f \) is the initial position plus the displacement:

\[
\mathbf{r}_f = \mathbf{r}_0 + \Delta \mathbf{r} = \langle -6.4 \times 10^7, 0, 0 \rangle \text{ m} + \langle 1.272 \times 10^6, 8.880 \times 10^6, 0 \rangle \text{ m}
\]
\[
\mathbf{r}_f = \langle -6.2728 \times 10^7, 8.880 \times 10^6, 0 \rangle \text{ m}
\]

### Final Answer
- **Momentum after 1 hour**: \( \mathbf{p}_f = \langle 5.2992 \times 10^6, 3.7 \times 10^7, 0 \rangle \) kg·m/s.
- **Position after 1 hour**: \( \mathbf{r}_f = \langle -6.2728 \times 10^7, 8.880 \times 10^6, 0 \rangle \) m.

### Follow-up Questions
1. How would the momentum change if the force applied was not constant?
2. What would be the effect of other forces (e.g., from the Sun) on the spacecraft's trajectory?
3. How would the calculation differ if the mass of the spacecraft changed over time (e.g., fuel consumption)?
4. What if the initial momentum had components in all three directions?
5. How would the result change if the time interval was different, say 2 hours instead of 1 hour?

### Tip
Always check the consistency of units when performing calculations in physics. Converting time, distance, and force into compatible units is crucial for accurate results.

Learn how to calculate the momentum and position change of a spacecraft over a 1-hour interval due to a constant force by Earth. Detailed step-by-step solution with calculations using Newton's second law of motion and average velocity formula. Suitable for high school physics students.

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