A rocket is fired from the earth towards the sun. At what distance from the earth's centre is the gravitational force on the rocket zero? Mass of the sun`=2xx10^(30)kg`, mass of the earth`=6xx10^(24)kg`. Neglect the effect of other planets etc. (orbital radius=`1.5xx10^(11)m`).

Options

A. `2.6xx10^(8)m`
B. `3.2xx10^(8)m`
C. `3.9xx10^(9)m`
D. `2.3xx10^(9)`m

Answer

To solve the problem of finding the distance from the Earth's center where the gravitational force on the rocket is zero, we can follow these steps:

### Step 1: Understand the Forces Acting on the Rocket
The gravitational force acting on the rocket due to the Earth and the Sun must be equal for the net gravitational force to be zero.

### Step 2: Set Up the Equation
Let:
- \( M_e = 6 \times 10^{24} \, \text{kg} \) (mass of the Earth)
- \( M_s = 2 \times 10^{30} \, \text{kg} \) (mass of the Sun)
- \( r = 1.5 \times 10^{11} \, \text{m} \) (distance from the Earth to the Sun)
- \( x \) = distance from the center of the Earth to the point where the gravitational force is zero.

The gravitational force due to the Earth at distance \( x \):
\[ F_e = \frac{G M_e m}{x^2} \]
The gravitational force due to the Sun at distance \( r - x \):
\[ F_s = \frac{G M_s m}{(r - x)^2} \]
Where \( G \) is the gravitational constant and \( m \) is the mass of the rocket.

### Step 3: Set the Forces Equal
Since we want the net force to be zero:
\[ F_e = F_s \]
This gives us:
\[ \frac{G M_e m}{x^2} = \frac{G M_s m}{(r - x)^2} \]
We can cancel \( G \) and \( m \) from both sides:
\[ \frac{M_e}{x^2} = \frac{M_s}{(r - x)^2} \]

### Step 4: Cross-Multiply to Solve for \( x \)
Cross-multiplying gives:
\[ M_e (r - x)^2 = M_s x^2 \]

### Step 5: Substitute Known Values
Substituting the values of \( M_e \) and \( M_s \):
\[ (6 \times 10^{24}) (1.5 \times 10^{11} - x)^2 = (2 \times 10^{30}) x^2 \]

### Step 6: Expand and Rearrange
Expanding the left side:
\[ 6 \times 10^{24} (1.5 \times 10^{11} - x)^2 = 2 \times 10^{30} x^2 \]
Let’s expand \( (1.5 \times 10^{11} - x)^2 \):
\[ (1.5 \times 10^{11})^2 - 2(1.5 \times 10^{11})x + x^2 \]
Substituting back:
\[ 6 \times 10^{24} \left( (1.5 \times 10^{11})^2 - 2(1.5 \times 10^{11})x + x^2 \right) = 2 \times 10^{30} x^2 \]

### Step 7: Collect Like Terms
This will yield a quadratic equation in \( x \):
\[ (6 \times 10^{24}) (1.5 \times 10^{11})^2 - 12 \times 10^{24} (1.5 \times 10^{11}) x + 6 \times 10^{24} x^2 - 2 \times 10^{30} x^2 = 0 \]
Combine the \( x^2 \) terms:
\[ (6 \times 10^{24} - 2 \times 10^{30}) x^2 - 12 \times 10^{24} (1.5 \times 10^{11}) x + (6 \times 10^{24}) (1.5 \times 10^{11})^2 = 0 \]

### Step 8: Solve the Quadratic Equation
Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = (6 \times 10^{24} - 2 \times 10^{30}) \), \( b = -12 \times 10^{24} (1.5 \times 10^{11}) \), and \( c = (6 \times 10^{24}) (1.5 \times 10^{11})^2 \).

### Step 9: Calculate the Values
After performing the calculations, we find:
\[ x \approx 2.6 \times 10^8 \, \text{m} \]

### Final Answer
The distance from the Earth's center where the gravitational force on the rocket is zero is approximately \( 2.6 \times 10^8 \, \text{m} \).

---