To solve this problem, we'll break it down into a few steps, considering the physics principles involved.

Step 1: Understand the Initial Conditions

• The spacecraft has an initial mass $m = 2.7 \times 10^4$ kg.
• Its initial velocity $\mathbf{v}_0 = \langle 0, 28, 0 \rangle$ km/s.
• The initial position is $\mathbf{r}_0 = \langle 2, 6, 0 \rangle$ km.

Step 2: Determine the Change in Velocity Due to the Thrusters

The thrusters apply a force for a given duration, which will change the velocity of the spacecraft. We'll use the impulse-momentum theorem, which states:

$\Delta \mathbf{v} = \frac{\mathbf{F} \cdot \Delta t}{m}$

Where:

• $\mathbf{F} = \langle 5 \times 10^5, 0, 0 \rangle$ N is the force applied.
• $\Delta t = 23.5$ s is the duration the force is applied.
• $m = 2.7 \times 10^4$ kg is the mass of the spacecraft.

First, convert the force to the appropriate units (N to kg·km/s²):

$\mathbf{F} = \langle 5 \times 10^5, 0, 0 \rangle \text{ N} = \langle 5 \times 10^5 \times 10^{-3}, 0, 0 \rangle \text{ km/s²} = \langle 5 \times 10^2, 0, 0 \rangle \text{ km/s²}$

Now, calculate the change in velocity:

$\Delta \mathbf{v} = \frac{\langle 5 \times 10^2, 0, 0 \rangle \times 23.5 \text{ s}}{2.7 \times 10^4 \text{ kg}} = \frac{\langle 1.175 \times 10^4, 0, 0 \rangle \text{ km/s}}{2.7 \times 10^4} = \langle 0.4352, 0, 0 \rangle \text{ km/s}$

Step 3: Calculate the New Velocity After the Thruster Fire

Add the change in velocity to the initial velocity:

$\mathbf{v}_f = \mathbf{v}_0 + \Delta \mathbf{v} = \langle 0, 28, 0 \rangle \text{ km/s} + \langle 0.4352, 0, 0 \rangle \text{ km/s} = \langle 0.4352, 28, 0 \rangle \text{ km/s}$

Step 4: Determine the Final Position After One Hour

After the thruster fire, the spacecraft continues coasting with constant velocity $\mathbf{v}_f$. We need to calculate the displacement over one hour (3600 seconds).

$\mathbf{r}_f = \mathbf{r}_0 + \mathbf{v}_f \times \Delta t$ $\mathbf{r}_f = \langle 2, 6, 0 \rangle \text{ km} + \langle 0.4352, 28, 0 \rangle \text{ km/s} \times 3600 \text{ s}$

Calculate the displacement:

$\mathbf{r}_f = \langle 2, 6, 0 \rangle + \langle 0.4352 \times 3600, 28 \times 3600, 0 \rangle$ $\mathbf{r}_f = \langle 2, 6, 0 \rangle + \langle 1566.72, 100800, 0 \rangle$ $\mathbf{r}_f = \langle 1568.72, 100806, 0 \rangle \text{ km}$

Step 5: Final Answer

One hour after the thruster fire, the spacecraft will be at the position $\langle 1568.72, 100806, 0 \rangle$ km.

Assumptions Made

1. The mass of the ejected gases is negligible, so the mass of the spacecraft remains constant.
2. The only force acting on the spacecraft during the 23.5 s interval is the thruster force; after this, no external forces are acting (assuming a vacuum with no gravitational effects).
3. Space is considered to be a vacuum with no resistance.

Follow-up Questions

1. How would the result change if the spacecraft's mass changed during the thruster fire?
2. What if the force applied by the thrusters had components in the $y$ or $z$ directions?
3. How would external gravitational forces affect the spacecraft's trajectory?
4. What if the thruster fired for a different duration, say 10 seconds instead of 23.5 seconds?
5. How would you calculate the position if the force applied wasn't constant?

Tip

Remember, in space, without any external forces, an object continues to move in a straight line at a constant speed due to Newton's first law of motion.