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This problem appears to involve gravitational interactions and the conservation of mechanical energy. Given that an object of mass \(370 \, \text{kg}\) is passing by a planet of mass \(5 \times 10^{22} \, \text{kg}\), we need to determine how far from the center of the planet the object will be when its speed decreases from \(1190 \, \text{m/s}\) to \(1000 \, \text{m/s}\).

### Step-by-step Solution

#### Step 1: Mechanical Energy Conservation
Since no external forces are doing work on the system, we can apply the conservation of mechanical energy:

\[
\text{Total Energy Initial} = \text{Total Energy Final}
\]

The mechanical energy at any point is the sum of the kinetic energy (KE) and the gravitational potential energy (PE).

#### Initial Point (at \( r = 5 \times 10^6 \, \text{m} \), \( v = 1190 \, \text{m/s} \)):
- **Kinetic Energy (KE):**

\[
KE_{\text{initial}} = \frac{1}{2} m v_{\text{initial}}^2
= \frac{1}{2} (370 \, \text{kg}) (1190 \, \text{m/s})^2
= 2.62115 \times 10^8 \, \text{J}
\]

- **Potential Energy (PE):**

\[
PE_{\text{initial}} = - \frac{G M m}{r_{\text{initial}}}
= - \frac{(6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2) (5 \times 10^{22} \, \text{kg}) (370 \, \text{kg})}{5 \times 10^6 \, \text{m}}
= -2.46838 \times 10^{10} \, \text{J}
\]

#### Final Point (at unknown \( r \), \( v = 1000 \, \text{m/s} \)):
- **Kinetic Energy (KE):**

\[
KE_{\text{final}} = \frac{1}{2} m v_{\text{final}}^2
= \frac{1}{2} (370 \, \text{kg}) (1000 \, \text{m/s})^2
= 1.85 \times 10^8 \, \text{J}
\]

- **Potential Energy (PE):**

\[
PE_{\text{final}} = - \frac{G M m}{r_{\text{final}}}
= - \frac{(6.674 \times 10^{-11}) (5 \times 10^{22}) (370)}{r_{\text{final}}}
= - \frac{1.235 \times 10^{12}}{r_{\text{final}}}
\]

#### Step 2: Applying Conservation of Mechanical Energy
Now, we apply energy conservation:

\[
KE_{\text{initial}} + PE_{\text{initial}} = KE_{\text{final}} + PE_{\text{final}}
\]

Substitute the values:

\[
2.62115 \times 10^8 - 2.46838 \times 10^{10} = 1.85 \times 10^8 - \frac{1.235 \times 10^{12}}{r_{\text{final}}}
\]

Simplifying:

\[
-2.44217 \times 10^{10} = - \frac{1.235 \times 10^{12}}{r_{\text{final}}}
\]

\[
r_{\text{final}} = \frac{1.235 \times 10^{12}}{2.44217 \times 10^{10}}
\]

\[
r_{\text{final}} \approx 50.6 \times 10^6 \, \text{m}
\]

Thus, the object will be approximately **50.6 million meters** from the center of the planet when its speed decreases to \(1000 \, \text{m/s}\).

---

Would you like more details on any step of the solution, or do you have any questions?

### Related Questions:
1. How does the mass of the planet affect the gravitational potential energy?
2. Can we apply the same principles if the object were on the surface of the planet?
3. How does energy conservation help in orbital mechanics?
4. What happens to the object if its velocity drops to zero?
5. How would the distance change if the planet had a higher mass?

### Tip:
When dealing with gravitational problems, always ensure to apply the correct formula for potential energy, which is negative due to the attractive nature of gravity.

An object of mass 370 kg goes past a planet of mass 5 × 10^22 kg at a distance of 5 × 10^6 m from the center of the planet, traveling at a speed of 1190 m/s. How far from the center of the planet will the object be when its speed is 1000 m/s?

Learn how to determine the distance of a 370 kg object from the center of a planet of mass 5 × 10^22 kg when its speed decreases from 1190 m/s to 1000 m/s. This problem uses the principles of gravitational potential energy and the conservation of mechanical energy.

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