For circular orbits the potential energy of the companion star is constant throughout the orbit. if the radius of the orbit doubles, what is the new value of the velocity of the companion star?

Options

A. it is 1/2 of the old value
B. it is 1/`sqrt(2)` of the old value
C. it is the same as old value
D. it is double the old value

Explanation

To solve the problem, we need to analyze the relationship between the potential energy, kinetic energy, and velocity of the companion star in a circular orbit.

### Step-by-Step Solution:

1. **Understanding the Relationship Between Kinetic and Potential Energy**:
- For a star in a circular orbit, the kinetic energy (KE) is related to the potential energy (PE). The kinetic energy is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
- The potential energy in a gravitational field is given by:
\[ PE = -\frac{GMm}{r} \]
- Here, \( G \) is the gravitational constant, \( M \) is the mass of the central body, \( m \) is the mass of the companion star, and \( r \) is the radius of the orbit.

2. **Kinetic Energy and Potential Energy Relationship**:
- In circular orbits, the kinetic energy is related to the potential energy:
\[ KE = -\frac{1}{2} PE \]
- This means that if we know the potential energy, we can find the kinetic energy.

3. **Proportional Relationships**:
- The kinetic energy is proportional to the square of the velocity:
\[ KE \propto v^2 \]
- The potential energy is inversely proportional to the radius:
\[ PE \propto \frac{1}{r} \]

4. **Setting Up the Ratios**:
- Let \( v_1 \) be the initial velocity and \( v_2 \) be the final velocity after the radius is doubled (i.e., \( r_2 = 2r_1 \)).
- The relationship between the initial and final kinetic energies and potential energies can be expressed as:
\[ \frac{KE_1}{KE_2} = \frac{PE_1}{PE_2} \]
- Substituting the proportional relationships, we have:
\[ \frac{v_1^2}{v_2^2} = \frac{PE_1}{PE_2} = \frac{r_2}{r_1} \]

5. **Substituting the Radius**:
- Since \( r_2 = 2r_1 \), we can substitute this into our equation:
\[ \frac{v_1^2}{v_2^2} = \frac{2r_1}{r_1} = 2 \]
- This simplifies to:
\[ v_1^2 = 2v_2^2 \]

6. **Solving for Final Velocity**:
- Rearranging gives:
\[ v_2^2 = \frac{v_1^2}{2} \]
- Taking the square root of both sides:
\[ v_2 = \frac{v_1}{\sqrt{2}} \]

### Final Answer:
The new value of the velocity of the companion star when the radius of the orbit doubles is:
\[ v_2 = \frac{v_1}{\sqrt{2}} \]