A uniform ring of mass M and radius R is placed directly above a uniform sphere of mass 8M and of same radius R. The centre of the ring is at a distance of `d = sqrt(3)R` from the centre of the sphere. The gravitational attraction between the sphere and the ring is

Options

A. `(GM^(2))/(R^(2))`
B. `(3GM^(2))/(2R^(3))`
C. `(2GM^(2))/(sqrt(2)R^(2))`
D. `(sqrt(3)GM^(2))/(R^(2))`

Explanation

To find the gravitational attraction between the uniform ring and the uniform sphere, we can use the formula for gravitational force. The gravitational force \( F \) between two masses \( M_1 \) and \( M_2 \) separated by a distance \( d \) is given by:

\[ F = \frac{G M_1 M_2}{d^2} \]

Where:
- \( G \) is the gravitational constant,
- \( M_1 \) is the mass of the ring,
- \( M_2 \) is the mass of the sphere,
- \( d \) is the distance between the centers of the two objects.

### Step 1: Identify the masses and distance
- Mass of the ring, \( M_1 = M \)
- Mass of the sphere, \( M_2 = 8M \)
- Distance between the center of the ring and the center of the sphere, \( d = \sqrt{3}R \)

### Step 2: Substitute the values into the gravitational force formula
Now, substituting the values into the gravitational force formula:

\[ F = \frac{G (M) (8M)}{(\sqrt{3}R)^2} \]

### Step 3: Simplify the equation
Calculating \( (\sqrt{3}R)^2 \):

\[ (\sqrt{3}R)^2 = 3R^2 \]

Now substituting this back into the equation for \( F \):

\[ F = \frac{G (M) (8M)}{3R^2} \]

### Step 4: Further simplify the expression
This simplifies to:

\[ F = \frac{8GM^2}{3R^2} \]

### Final Answer
Thus, the gravitational attraction between the sphere and the ring is:

\[ F = \frac{8GM^2}{3R^2} \]

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