A satellite in a force - free space sweeps stationary interplanetary dust at a rate `dM//dt = alpha v`, where `M` is the mass , `v`is the velocity of the satellite and `alpha` is a constant. What is the deacceleration of the satellite ?

Options

A. ` - 2 alpha v^(2) //M`
B. `- alpha v^(2)//M`
C. `+ alpha v^(2)//M`
D. `-alpha v^(2)`

Explanation

To solve the problem, we need to determine the deceleration of the satellite as it captures interplanetary dust at a rate proportional to its velocity. Let's break this down step by step.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- The satellite is capturing dust, which means its mass is changing.
- The rate of change of mass is given by the equation:
\[ \frac{dM}{dt} = \alpha v \]
- Here, \( M \) is the mass of the satellite, \( v \) is its velocity, and \( \alpha \) is a constant.

2. **Define Momentum**:
- The momentum \( p \) of the satellite is given by:
\[ p = M v \]
- Since the mass \( M \) is changing, we need to consider the rate of change of momentum.

3. **Apply the Principle of Momentum Change**:
- The force \( F \) acting on the satellite can be expressed as the rate of change of momentum:
\[ F = \frac{dp}{dt} = \frac{d(Mv)}{dt} \]
- Using the product rule:
\[ F = v \frac{dM}{dt} + M \frac{dv}{dt} \]

4. **Substitute the Rate of Change of Mass**:
- We know from the problem that:
\[ \frac{dM}{dt} = \alpha v \]
- Substitute this into the force equation:
\[ F = v(\alpha v) + M \frac{dv}{dt} \]
- This simplifies to:
\[ F = \alpha v^2 + M \frac{dv}{dt} \]

5. **Set the Force Equal to the Mass Times Acceleration**:
- According to Newton's second law, the force can also be expressed as:
\[ F = M a \]
- Where \( a \) is the acceleration (or deceleration in this case). Thus, we have:
\[ M a = \alpha v^2 + M \frac{dv}{dt} \]

6. **Rearranging the Equation**:
- Rearranging gives us:
\[ M a - M \frac{dv}{dt} = \alpha v^2 \]
- Factoring out \( M \):
\[ M \left( a - \frac{dv}{dt} \right) = \alpha v^2 \]

7. **Solve for Acceleration**:
- Since \( a = \frac{dv}{dt} \) (deceleration), we can rewrite the equation:
\[ M a = \alpha v^2 \]
- Therefore, the deceleration \( a \) is given by:
\[ a = \frac{\alpha v^2}{M} \]

### Final Answer:
The deceleration of the satellite is:
\[ a = \frac{\alpha v^2}{M} \]