Match the column :
A particle at a distance r from the centre of a uniform spherical planet of mass M radius `R(lt r)` has a velocity v magnitude of which is given in column I. Match trajectory from column II about possible nature of orbit.
`{:(,"Column I",,"Column II"),((A),0lt V lt sqrt((GM)/(r )),(P),"Straight line"),((B),sqrt((GM)/(r )),(Q),"Circle"),((C ),sqrt((2GM)/(r )),(R ),"Parabola"),((D),v gt sqrt((2GM)/(r )),(S ),"Ellipse"):}`

Answer

To solve the problem of matching the columns regarding the nature of the orbit of a particle at a distance \( r \) from the center of a uniform spherical planet of mass \( M \) and radius \( R \), we will analyze the velocity ranges given in Column I and determine their corresponding trajectories in Column II.

### Step-by-Step Solution:

1. **Understanding the Velocity Ranges:**
- We have four cases based on the velocity \( v \):
- (A) \( 0 < v < \sqrt{\frac{GM}{r}} \)
- (B) \( v = \sqrt{\frac{GM}{r}} \)
- (C) \( v = \sqrt{\frac{2GM}{r}} \)
- (D) \( v > \sqrt{\frac{2GM}{r}} \)

2. **Analyzing Case (A): \( 0 < v < \sqrt{\frac{GM}{r}} \)**
- In this case, the particle has a velocity less than the orbital velocity. It can either follow a straight line if projected directly away from the planet or could follow an elliptical path if projected at an angle.
- **Matching:** Possible trajectories are (P) "Straight line" and (S) "Ellipse".

3. **Analyzing Case (B): \( v = \sqrt{\frac{GM}{r}} \)**
- This velocity is the orbital velocity. If the particle is projected tangentially, it will move in a circular orbit. If projected at an angle, it will follow an elliptical path.
- **Matching:** Possible trajectories are (Q) "Circle" and (S) "Ellipse".

4. **Analyzing Case (C): \( v = \sqrt{\frac{2GM}{r}} \)**
- This velocity is the escape velocity. If projected tangentially, it will follow a parabolic trajectory. If projected directly away from the planet, it will follow a straight line.
- **Matching:** Possible trajectories are (P) "Straight line" and (R) "Parabola".

5. **Analyzing Case (D): \( v > \sqrt{\frac{2GM}{r}} \)**
- For velocities greater than the escape velocity, the particle will not return to the planet and will move in a straight line away from it.
- **Matching:** The only possible trajectory is (P) "Straight line".

### Final Matching:
- (A) \( 0 < v < \sqrt{\frac{GM}{r}} \) → (P) "Straight line", (S) "Ellipse"
- (B) \( v = \sqrt{\frac{GM}{r}} \) → (Q) "Circle", (S) "Ellipse"
- (C) \( v = \sqrt{\frac{2GM}{r}} \) → (P) "Straight line", (R) "Parabola"
- (D) \( v > \sqrt{\frac{2GM}{r}} \) → (P) "Straight line"

### Summary of Matches:
- A → P, S
- B → Q, S
- C → P, R
- D → P