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To solve this problem, we need to break it down into parts. Let’s analyze each step based on the information provided.

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### Given Data:
- **Potential energy function** \( U(r) \) of a projectile is plotted as a function of \( r \), the radial distance from the planet's surface.
- The **mechanical energy** \( E \) of the projectile is given as \( E = -2.0 \times 10^9 \, \text{J} \).
- Radius of the planet, \( R_s = 8800 \, \text{km} = 8.8 \times 10^6 \, \text{m} \).
- We need to find:
  - (a) The **kinetic energy** of the projectile at \( r = 1.25 R_s \).
  - (b) The **turning point** of the projectile.

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### Step-by-Step Solution:

#### Part (a): Kinetic Energy at \( r = 1.25 R_s \)

1. **Calculate \( r \) in meters**:
   \[
   r = 1.25 R_s = 1.25 \times 8.8 \times 10^6 \, \text{m} = 1.1 \times 10^7 \, \text{m}
   \]

2. **Determine the potential energy \( U(r) \) at \( r = 1.25 R_s \)**:
   From the plot, you would need to read the value of \( U(r) \) at \( r = 1.25 R_s \). Since we don't have numerical data on the graph, let's assume that the value can be visually approximated or given in context. Let's denote this value as \( U(r = 1.25 R_s) \).

3. **Use the mechanical energy equation**:
   The mechanical energy \( E \) is the sum of potential and kinetic energies:
   \[
   E = K + U(r)
   \]
   Rearranging for \( K \), we get:
   \[
   K = E - U(r)
   \]
   Substitute \( E = -2.0 \times 10^9 \, \text{J} \) and the value of \( U(r = 1.25 R_s) \) from the graph.

4. **Calculate \( K \)**:
   Once \( U(r = 1.25 R_s) \) is identified, plug it into the equation to find \( K \).

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#### Part (b): The Turning Point

1. The turning point is the radial distance \( r \) where the projectile momentarily comes to rest, meaning its **kinetic energy \( K = 0 \)**.

2. At the turning point, the mechanical energy \( E \) is equal to the potential energy \( U(r_{\text{turning}}) \):
   \[
   E = U(r_{\text{turning}})
   \]
   Since \( E = -2.0 \times 10^9 \, \text{J} \), we need to find the point on the \( U(r) \) curve where \( U(r) = -2.0 \times 10^9 \, \text{J} \).

3. **Read \( r_{\text{turning}} \) from the graph**:
   Locate the point on the graph where \( U(r) = -2.0 \times 10^9 \, \text{J} \) and find the corresponding value of \( r \). This will be the turning point.

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### Summary of Steps:
1. Use the mechanical energy equation to calculate kinetic energy at \( r = 1.25 R_s \).
2. Identify the point on the potential energy graph that corresponds to the given mechanical energy to determine the turning point.

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Do you need further help interpreting the graph, or would you like a more detailed breakdown of the calculations? 

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### Related Questions to Explore:
1. How does the potential energy function \( U(r) \) relate to gravitational force?
2. Why is the kinetic energy zero at the turning point?
3. How would the result change if the mechanical energy were positive?
4. Can this method be applied to find escape velocity from a planet's surface?
5. What does a negative mechanical energy signify about the projectile's motion?

### Tip:
When dealing with potential energy graphs, always remember that the turning points occur where the mechanical energy equals the potential energy, as kinetic energy becomes zero at these points.

The figure gives the potential energy function U(r) of a projectile, plotted outward from the surface of a planet of radius Rs. If the projectile is launched radially outward from the surface with a mechanical energy of -2.0 x 10^9 J, what are (a) its kinetic energy at radius r = 1.25Rs and (b) its turning point if Rs = 8800 km?

Explore how to determine the kinetic energy and turning point of a projectile launched radially outward from a planet's surface. This problem involves calculating kinetic energy at a specific radius and finding the turning point using the conservation of mechanical energy. Suitable for undergraduate physics students.

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