Astronomers observe two separate solar systems each consisting of a planet orbiting a sun. The two orbits are circular and have the same radius `R`. It is determined that the planets have angular momenta of the same magnitude `L` about their suns, and that the orbital periods are in the ratio of three to one, i.e. `T_(1)=3T_(2)`. The ratio `m_(1)//m_(2)` of the masses of the two planets is

Options

A. `1`
B. `sqrt(3)`
C. `2`
D. `3`

Explanation

To solve the problem, we need to find the ratio of the masses of two planets orbiting their respective suns, given that both planets have the same angular momentum and that their orbital periods are in the ratio of 3:1.

### Step-by-Step Solution:

1. **Understanding Angular Momentum**:
The angular momentum \( L \) of a planet in a circular orbit is given by the formula:
\[ L = m \cdot v \cdot r \]
where \( m \) is the mass of the planet, \( v \) is its orbital velocity, and \( r \) is the radius of the orbit.

2. **Given Conditions**:
- The radii of the orbits are the same: \( r_1 = r_2 = R \).
- The angular momenta are equal: \( L_1 = L_2 \).
- The periods are in the ratio \( T_1 = 3T_2 \).

3. **Expressing Velocity**:
The orbital velocity \( v \) can be expressed in terms of the radius and the period:
\[ v = \frac{2\pi r}{T} \]
Thus, for the two planets we have:
\[ v_1 = \frac{2\pi R}{T_1} \quad \text{and} \quad v_2 = \frac{2\pi R}{T_2} \]

4. **Substituting into Angular Momentum**:
Now substituting the expressions for velocity into the angular momentum equations:
\[ L_1 = m_1 \cdot v_1 \cdot R = m_1 \cdot \left(\frac{2\pi R}{T_1}\right) \cdot R = m_1 \cdot \frac{2\pi R^2}{T_1} \]
\[ L_2 = m_2 \cdot v_2 \cdot R = m_2 \cdot \left(\frac{2\pi R}{T_2}\right) \cdot R = m_2 \cdot \frac{2\pi R^2}{T_2} \]

5. **Setting Angular Momenta Equal**:
Since \( L_1 = L_2 \):
\[ m_1 \cdot \frac{2\pi R^2}{T_1} = m_2 \cdot \frac{2\pi R^2}{T_2} \]
The \( 2\pi R^2 \) cancels out:
\[ m_1 \cdot \frac{1}{T_1} = m_2 \cdot \frac{1}{T_2} \]

6. **Rearranging for Mass Ratio**:
Rearranging gives:
\[ \frac{m_1}{m_2} = \frac{T_1}{T_2} \]

7. **Substituting the Period Ratio**:
We know \( T_1 = 3T_2 \):
\[ \frac{m_1}{m_2} = \frac{3T_2}{T_2} = 3 \]

### Final Answer:
The ratio of the masses of the two planets is:
\[ \frac{m_1}{m_2} = 3 \]