The earth (mass `= 6 xx 10^(24)kg` ) revolves around the sun with angular velocity `2 xx 10^(-7)"rads"^(-1)` in a circular orbit of radius `1.5 xx 10^(11)` km. The force exerted by the sun on the earth in newton is

Options

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

\[ F = m \cdot a \]

where:
- \( F \) is the centripetal force,
- \( m \) is the mass of the Earth,
- \( a \) is the centripetal acceleration.

The centripetal acceleration \( a \) can be expressed in terms of angular velocity \( \omega \) and radius \( r \) as:

\[ a = \omega^2 \cdot r \]

### Step 1: Identify the given values
From the problem, we have:
- Mass of the Earth, \( m = 6 \times 10^{24} \, \text{kg} \)
- Angular velocity, \( \omega = 2 \times 10^{-7} \, \text{radians/second} \)
- Radius of the orbit, \( r = 1.5 \times 10^{11} \, \text{km} = 1.5 \times 10^{14} \, \text{m} \) (since \( 1 \, \text{km} = 1000 \, \text{m} \))

### Step 2: Calculate the centripetal acceleration
Using the formula for centripetal acceleration:

\[ a = \omega^2 \cdot r \]

Substituting the values:

\[ a = (2 \times 10^{-7})^2 \cdot (1.5 \times 10^{14}) \]

Calculating \( (2 \times 10^{-7})^2 \):

\[ (2 \times 10^{-7})^2 = 4 \times 10^{-14} \]

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

\[ a = 4 \times 10^{-14} \cdot 1.5 \times 10^{14} \]

Calculating this gives:

\[ a = 6 \times 10^{0} = 6 \, \text{m/s}^2 \]

### Step 3: Calculate the centripetal force
Now we can find the force using the mass and the centripetal acceleration:

\[ F = m \cdot a \]

Substituting the values:

\[ F = 6 \times 10^{24} \cdot 6 \]

Calculating this gives:

\[ F = 36 \times 10^{24} \, \text{N} \]

### Final Answer
Thus, the force exerted by the Sun on the Earth is:

\[ F = 36 \times 10^{24} \, \text{N} \]