Two solid spherical planets of equal radii R having masses 4M and 9M their centre are separated by a distance 6R. A projectile of mass m is sent from theplanet of mass 4 M towards the heavier planet. What is the distance r of the point from the lighter planet where the gravitational force on the projectille is zero ?

Options

A. 1.4 R
B. 1.8 R
C. 1.5 R
D. 2.4 R

Answer

To solve the problem, we need to find the distance \( r \) from the lighter planet (mass \( 4M \)) where the gravitational force on the projectile (mass \( m \)) is zero.

### Step-by-Step Solution:

1. **Identify the Setup**:
- We have two planets:
- Planet 1 (lighter) with mass \( 4M \)
- Planet 2 (heavier) with mass \( 9M \)
- The distance between the centers of the two planets is \( 6R \).

2. **Define the Variables**:
- Let \( r \) be the distance from the lighter planet (mass \( 4M \)) to the point where the gravitational force on the projectile is zero.
- Therefore, the distance from the heavier planet (mass \( 9M \)) to this point will be \( 6R - r \).

3. **Write the Gravitational Forces**:
- The gravitational force \( F_1 \) exerted by the lighter planet on the projectile is given by:
\[ F_1 = \frac{G \cdot 4M \cdot m}{r^2} \]
- The gravitational force \( F_2 \) exerted by the heavier planet on the projectile is given by:
\[ F_2 = \frac{G \cdot 9M \cdot m}{(6R - r)^2} \]

4. **Set the Forces Equal**:
- For the gravitational forces to balance (i.e., the net force on the projectile is zero), we set \( F_1 = F_2 \):
\[ \frac{G \cdot 4M \cdot m}{r^2} = \frac{G \cdot 9M \cdot m}{(6R - r)^2} \]
- We can cancel \( G \), \( M \), and \( m \) from both sides:
\[ \frac{4}{r^2} = \frac{9}{(6R - r)^2} \]

5. **Cross-Multiply**:
- Cross-multiplying gives:
\[ 4(6R - r)^2 = 9r^2 \]

6. **Expand and Rearrange**:
- Expanding the left side:
\[ 4(36R^2 - 12Rr + r^2) = 9r^2 \]
\[ 144R^2 - 48Rr + 4r^2 = 9r^2 \]
- Rearranging gives:
\[ 144R^2 - 48Rr - 5r^2 = 0 \]

7. **Use the Quadratic Formula**:
- This is a quadratic equation in terms of \( r \):
\[ 5r^2 + 48Rr - 144R^2 = 0 \]
- Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 5 \), \( b = 48R \), and \( c = -144R^2 \).
\[ r = \frac{-48R \pm \sqrt{(48R)^2 - 4 \cdot 5 \cdot (-144R^2)}}{2 \cdot 5} \]
\[ r = \frac{-48R \pm \sqrt{2304R^2 + 2880R^2}}{10} \]
\[ r = \frac{-48R \pm \sqrt{5184R^2}}{10} \]
\[ r = \frac{-48R \pm 72R}{10} \]

8. **Calculate the Roots**:
- This gives two potential solutions:
- \( r = \frac{24R}{10} = 2.4R \) (valid solution since it is positive)
- \( r = \frac{-120R}{10} = -12R \) (not valid since distance cannot be negative)

9. **Final Answer**:
- Therefore, the distance \( r \) from the lighter planet where the gravitational force on the projectile is zero is:
\[ r = 2.4R \]