If the mass of the sun is `2 xx 10^(30)` kg , the distance of the earth from the sun is `1.5 xx 10^(11)` m and period of revolution of the earth around the sun is one year (`= 365.3` days ) , calculate the value of gravitational constant .

Answer

To calculate the gravitational constant \( G \) using the given data, we can follow these steps:

### Step 1: Understand the relationship between gravitational force and centripetal force
The gravitational force acting between the Earth and the Sun can be expressed as:
\[ F_g = \frac{G M m}{r^2} \]
where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Sun,
- \( m \) is the mass of the Earth (which will cancel out),
- \( r \) is the distance between the Earth and the Sun.

The centripetal force required to keep the Earth in orbit around the Sun is given by:
\[ F_c = m \omega^2 r \]
where \( \omega \) is the angular velocity.

### Step 2: Relate angular velocity to the period of revolution
The angular velocity \( \omega \) can be expressed in terms of the period \( T \):
\[ \omega = \frac{2\pi}{T} \]
where \( T \) is the period of revolution (in seconds).

### Step 3: Set the gravitational force equal to the centripetal force
Setting \( F_g = F_c \):
\[ \frac{G M m}{r^2} = m \omega^2 r \]
We can cancel \( m \) from both sides:
\[ \frac{G M}{r^2} = \omega^2 r \]

### Step 4: Substitute \( \omega \) in the equation
Substituting \( \omega = \frac{2\pi}{T} \):
\[ \frac{G M}{r^2} = \left(\frac{2\pi}{T}\right)^2 r \]
This simplifies to:
\[ \frac{G M}{r^2} = \frac{4\pi^2}{T^2} r \]

### Step 5: Rearranging the equation to solve for \( G \)
Rearranging gives us:
\[ G = \frac{4\pi^2 r^3}{M T^2} \]

### Step 6: Substitute the known values
Now, we can substitute the known values:
- Mass of the Sun, \( M = 2 \times 10^{30} \) kg
- Distance from the Earth to the Sun, \( r = 1.5 \times 10^{11} \) m
- Period of revolution, \( T = 365.3 \) days = \( 365.3 \times 24 \times 60 \times 60 \) seconds

Calculating \( T \):
\[ T = 365.3 \times 24 \times 60 \times 60 = 31,557,600 \text{ seconds} \]

### Step 7: Calculate \( G \)
Now substituting these values into the equation for \( G \):
\[ G = \frac{4\pi^2 (1.5 \times 10^{11})^3}{(2 \times 10^{30}) (31,557,600)^2} \]

Calculating \( (1.5 \times 10^{11})^3 \):
\[ (1.5 \times 10^{11})^3 = 3.375 \times 10^{33} \]

Calculating \( (31,557,600)^2 \):
\[ (31,557,600)^2 = 9.958 \times 10^{15} \]

Now substituting these values back into the equation for \( G \):
\[ G = \frac{4 \times (3.14)^2 \times (3.375 \times 10^{33})}{(2 \times 10^{30}) \times (9.958 \times 10^{15})} \]
Calculating the numerator:
\[ 4 \times (3.14)^2 \approx 39.4784 \]
Thus,
\[ G = \frac{39.4784 \times 3.375 \times 10^{33}}{(2 \times 10^{30}) \times (9.958 \times 10^{15})} \]

Calculating the final value:
\[ G \approx 6.69 \times 10^{-11} \text{ N m}^2/\text{kg}^2 \]

### Final Result
The value of the gravitational constant \( G \) is approximately:
\[ G \approx 6.69 \times 10^{-11} \text{ N m}^2/\text{kg}^2 \]