The earth (mass = `6 xx 10^(24)` kg) revolves round the sun with angular velocity `2 xx 10^(–7 )`rad/s in a circular orbit of radius `1.5 xx 10^(8)` km. The force exerted by the sun on the earth in newtons, is

Options

A. `18xx10^(25)`
B. Zero
C. `27xx10^(39)`
D. `36xx10^(21)`

Explanation

To find the force exerted by the sun on the earth, we can use the formula for centripetal force, which is given by:

\[ F_c = m \cdot a_c \]

where:
- \( F_c \) is the centripetal force,
- \( m \) is the mass of the earth,
- \( a_c \) is the centripetal acceleration.

The centripetal acceleration can be expressed in terms of angular velocity (\( \omega \)) and radius (\( R \)) of the circular orbit:

\[ a_c = R \cdot \omega^2 \]

### Step-by-Step Solution:

1. **Identify the given values:**
- Mass of the Earth, \( m = 6 \times 10^{24} \) kg
- Angular velocity, \( \omega = 2 \times 10^{-7} \) rad/s
- Radius of the orbit, \( R = 1.5 \times 10^{8} \) km = \( 1.5 \times 10^{11} \) m (since 1 km = 1000 m)

2. **Calculate the centripetal acceleration:**
\[ a_c = R \cdot \omega^2 \]
Substituting the values:
\[ a_c = (1.5 \times 10^{11} \, \text{m}) \cdot (2 \times 10^{-7} \, \text{rad/s})^2 \]
\[ a_c = (1.5 \times 10^{11}) \cdot (4 \times 10^{-14}) \, \text{m/s}^2 \]
\[ a_c = 6 \times 10^{-3} \, \text{m/s}^2 \]

3. **Calculate the centripetal force:**
\[ F_c = m \cdot a_c \]
Substituting the values:
\[ F_c = (6 \times 10^{24} \, \text{kg}) \cdot (6 \times 10^{-3} \, \text{m/s}^2) \]
\[ F_c = 36 \times 10^{21} \, \text{N} \]

4. **Express the final answer in scientific notation:**
\[ F_c = 3.6 \times 10^{22} \, \text{N} \]

### Final Answer:
The force exerted by the sun on the earth is \( 3.6 \times 10^{22} \) Newtons.