Imagine a light planet revolving around a very massive star in a circular orbit of radius r with a period of revolution T. On what power of r will the square of time period will depend if the gravitational force of attraction between the planet and the star is proportional to `r^(-5//2)`.

Explanation

To solve the problem, we need to find the relationship between the square of the time period \( T^2 \) and the radius \( r \) of the orbit, given that the gravitational force of attraction \( F \) is proportional to \( r^{-5/2} \).

### Step 1: Write the gravitational force equation
The gravitational force \( F \) between the planet and the star can be expressed as:
\[ F \propto \frac{1}{r^{5/2}} \]
This implies:
\[ F = k \cdot r^{-5/2} \]
where \( k \) is a proportionality constant.

### Step 2: Relate gravitational force to centripetal force
For a planet in circular motion, the gravitational force provides the necessary centripetal force. Therefore, we can equate the gravitational force to the centripetal force:
\[ F = m \cdot \frac{v^2}{r} \]
where \( m \) is the mass of the planet and \( v \) is its orbital speed.

### Step 3: Express orbital speed in terms of the period
The orbital speed \( v \) can be expressed in terms of the period \( T \):
\[ v = \frac{2\pi r}{T} \]
Substituting this into the centripetal force equation gives:
\[ F = m \cdot \frac{(2\pi r/T)^2}{r} = m \cdot \frac{4\pi^2 r}{T^2} \]

### Step 4: Set the two expressions for force equal
Now, we can set the two expressions for the gravitational force equal to each other:
\[ k \cdot r^{-5/2} = m \cdot \frac{4\pi^2 r}{T^2} \]

### Step 5: Rearranging for \( T^2 \)
Rearranging the equation to solve for \( T^2 \):
\[ T^2 = \frac{4\pi^2 m r^{7/2}}{k} \]

### Step 6: Identify the power of \( r \)
From the equation \( T^2 \propto r^{7/2} \), we can see that the square of the time period \( T^2 \) depends on \( r \) raised to the power of \( \frac{7}{2} \) or \( 3.5 \).

### Final Result
Thus, the square of the time period \( T^2 \) is proportional to \( r^{7/2} \).