A space vehicle approaching a planet has a speed v when it is very far from the planet At that moment tangent of its trajectory would miss the centre of the planet by distance R if the planet has mass M and radius r what is the smallest value of R in order that the resulting orbit of the space vehicle will just miss the surface of the planet?

Options

A. a) `r/v[v^(2)+(2GM)/r]^(1/2)`
B. b) `vr[1+(2GM)/r]`
C. c) `r/v[v^(2)+(2GM)/r]`
D. d) `(2GMv)/r`

Explanation

To solve the problem, we need to determine the smallest value of R such that the space vehicle just misses the surface of the planet while approaching with speed \( v \). We will use the principles of conservation of momentum and conservation of energy.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- Let the mass of the planet be \( M \) and its radius be \( r \).
- The space vehicle has a speed \( v \) when it is very far from the planet.
- The trajectory of the vehicle will miss the center of the planet by a distance \( R \).

2. **Conservation of Momentum**:
- At point A (far away from the planet), the momentum of the vehicle is \( mv \).
- At point B (just missing the planet), the momentum is \( mv' \), where \( v' \) is the speed of the vehicle at that point.
- The perpendicular distance from the center of the planet to the trajectory is \( R \).
- From conservation of momentum, we can write:
\[ mv = mv' \cdot \frac{R}{r} \]
- Rearranging gives:
\[ v' = \frac{vR}{r} \]

3. **Conservation of Energy**:
- At point A, the total energy is purely kinetic:
\[ E_A = \frac{1}{2} mv^2 \]
- At point B, the total energy includes both kinetic and potential energy:
\[ E_B = \frac{1}{2} mv'^2 - \frac{GMm}{R} \]
- Setting \( E_A = E_B \):
\[ \frac{1}{2} mv^2 = \frac{1}{2} mv'^2 - \frac{GMm}{R} \]

4. **Substituting for \( v' \)**:
- Substitute \( v' = \frac{vR}{r} \) into the energy equation:
\[ \frac{1}{2} mv^2 = \frac{1}{2} m \left(\frac{vR}{r}\right)^2 - \frac{GMm}{R} \]
- Simplifying gives:
\[ \frac{1}{2} mv^2 = \frac{1}{2} m \frac{v^2 R^2}{r^2} - \frac{GMm}{R} \]

5. **Eliminating \( m \)**:
- Since \( m \) appears in every term, we can cancel it out:
\[ \frac{1}{2} v^2 = \frac{1}{2} \frac{v^2 R^2}{r^2} - \frac{GM}{R} \]

6. **Rearranging the Equation**:
- Rearranging gives:
\[ \frac{GM}{R} = \frac{1}{2} \frac{v^2 R^2}{r^2} - \frac{1}{2} v^2 \]
- Multiply through by \( 2R \):
\[ 2GM = \frac{v^2 R^3}{r^2} - v^2 R \]
- Rearranging leads to:
\[ \frac{v^2 R^3}{r^2} - v^2 R - 2GM = 0 \]

7. **Solving the Cubic Equation**:
- This is a cubic equation in \( R \). Using the cubic formula or numerical methods, we can find the smallest positive root.

8. **Final Expression for R**:
- The smallest value of \( R \) can be expressed as:
\[ R = \frac{2GM}{v^2} \left(1 + \sqrt{1 + \frac{4GM}{v^2 r^2}}\right) \]