The gravitational force between two objects is proportional to `1//R` (and not as `1//R^(2)`) where `R` is separation between them, then a particle in circular orbit under such a force would have its orbital speed `v` proportional to

Options

A. `(1)/(R^(2))`
B. `R^(0)`
C. `R`
D. `(1)/(R)`

Answer

To solve the problem, we start by analyzing the gravitational force acting on a particle in a circular orbit under the assumption that the force is proportional to \( \frac{1}{R} \) rather than \( \frac{1}{R^2} \).

### Step-by-Step Solution:

1. **Understanding the Forces**:
The centripetal force required to keep a particle of mass \( m \) moving in a circular orbit of radius \( R \) is given by:
\[ F_{\text{centripetal}} = \frac{m v^2}{R} \]
where \( v \) is the orbital speed of the particle.

2. **Gravitational Force**:
According to the problem, the gravitational force \( F_{\text{gravity}} \) between two masses \( m_1 \) and \( m_2 \) is given by:
\[ F_{\text{gravity}} = \frac{G m_1 m_2}{R} \]
Here, \( G \) is the gravitational constant, and \( R \) is the distance between the two masses.

3. **Setting the Forces Equal**:
For a particle in circular motion, the gravitational force provides the necessary centripetal force. Therefore, we can set these two forces equal to each other:
\[ \frac{m v^2}{R} = \frac{G m_1 m_2}{R} \]

4. **Simplifying the Equation**:
We can cancel \( m \) and \( R \) from both sides (assuming \( R \neq 0 \)):
\[ v^2 = G m_1 m_2 \]

5. **Finding the Proportionality**:
Since \( G \), \( m_1 \), and \( m_2 \) are constants, we can express \( v \) as:
\[ v = \sqrt{G m_1 m_2} \]
This indicates that \( v \) is independent of \( R \) and can be considered as proportional to \( R^0 \).

6. **Conclusion**:
Therefore, the orbital speed \( v \) is proportional to \( R^0 \), which means:
\[ v \propto 1 \]

### Final Answer:
The orbital speed \( v \) is proportional to \( R^0 \) or simply a constant.