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`
B. `R^0`
C. `R^1`
D. `1/R^2`

Explanation

To solve the problem, we need to analyze the relationship between gravitational force and orbital speed when the gravitational force is given as proportional to \( \frac{1}{R} \) instead of the usual \( \frac{1}{R^2} \).

### Step-by-Step Solution:

1. **Understanding the Force**:
The gravitational force \( F \) between two objects is given by:
\[ F \propto \frac{1}{R} \]
This means:
\[ F = k \frac{m_1 m_2}{R} \]
where \( k \) is a constant, and \( m_1 \) and \( m_2 \) are the masses of the two objects.

2. **Centripetal Force Requirement**:
For a particle of mass \( m \) moving in a circular orbit of radius \( R \) with speed \( v \), the centripetal force \( F_c \) required to keep it in orbit is given by:
\[ F_c = \frac{mv^2}{R} \]

3. **Setting Forces Equal**:
In circular motion, the centripetal force is provided by the gravitational force. Therefore, we set the centripetal force equal to the gravitational force:
\[ \frac{mv^2}{R} = k \frac{m_1 m_2}{R} \]

4. **Canceling Common Terms**:
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)) and also \( R \) (assuming \( R \neq 0 \)):
\[ v^2 = k m_1 m_2 \]

5. **Analyzing the Result**:
Notice that \( k \), \( m_1 \), and \( m_2 \) are constants. Therefore, \( v^2 \) does not depend on \( R \). This implies:
\[ v^2 \propto R^0 \]
Hence, taking the square root gives:
\[ v \propto R^0 \]

6. **Conclusion**:
Therefore, the orbital speed \( v \) is proportional to \( R^0 \), which means:
\[ v \text{ is constant and does not depend on } R. \]

### Final Answer:
The orbital speed \( v \) is proportional to \( R^0 \), which means \( v \) is constant.

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