A: If the force of gravitation in inversely proportional to the distance r rather than `r^(2)` given by Newton , then orbital velocity of the satellite around the earth is independent of r .
R : `(GMm)/r = (mv^(2))/r ` so , ` v = sqrt(GM)` hence independent of r .

Options

A. If both Assertion & Reason are true . And the reason is the correct explanation of the assertion , then mark (1)
B. If both Assertion & Reason are true but the reason is not the correct explanation of the assertion , then mark (2)
C. If Assertion is true statement but Reason is false , then mark (3)
D. It will move the same speed , tangentially to the spacecraft

Answer

To solve the problem, we need to analyze the gravitational force and the orbital velocity of a satellite under the assumption that the gravitational force is inversely proportional to the distance \( r \) instead of \( r^2 \).

### Step-by-Step Solution:

1. **Understanding the Gravitational Force**:
The gravitational force \( F_g \) acting on a satellite of mass \( m \) in orbit around the Earth (mass \( M \)) is given by the new assumption:
\[ F_g = \frac{k}{r} \]
where \( k \) is a constant.

2. **Centripetal Force Requirement**:
For the satellite to maintain circular motion, the gravitational force must provide the necessary centripetal force. The centripetal force \( F_c \) required for circular motion is given by:
\[ F_c = \frac{mv^2}{r} \]
where \( v \) is the orbital velocity of the satellite.

3. **Setting Up the Equation**:
Since the gravitational force provides the centripetal force, we can equate the two forces:
\[ \frac{mv^2}{r} = \frac{k}{r} \]

4. **Cancelling \( r \)**:
We can multiply both sides by \( r \) (assuming \( r \neq 0 \)):
\[ mv^2 = k \]

5. **Solving for Orbital Velocity**:
Rearranging the equation gives us:
\[ v^2 = \frac{k}{m} \]
Taking the square root of both sides, we find:
\[ v = \sqrt{\frac{k}{m}} \]
Here, \( k \) and \( m \) are constants, meaning that \( v \) is independent of \( r \).

6. **Conclusion**:
Thus, under the assumption that the gravitational force is inversely proportional to \( r \), the orbital velocity \( v \) of the satellite does not depend on the radius \( r \) of its orbit.