Solar energy is incident normally on the earths surface at the rate of about 1.4 kW `m^(-2)`. The distance between the earth and the sun is `1.5 10^(11)` m. Energy (E) and mass (m) are related by Einstein equation `E=mc^2` where c `(3 xx 10^8 ms^1)` is the speed of light in free space. The decrease in the mass of the sun is

Options

A. `10^9 kg s^(-1)`
B. `10^(30) kg s^(-1)`
C. `10^(28) kg s^(-1)`
D. `10^(11) kg s^(-1)`

Explanation

To find the decrease in the mass of the Sun due to the solar energy emitted, we can follow these steps:

### Step 1: Calculate the total power output of the Sun reaching the Earth
Given that the solar energy incident on the Earth's surface is \(1.4 \, \text{kW/m}^2\), we need to calculate the total power (P) received by the Earth.

First, we calculate the surface area of the sphere with a radius equal to the distance from the Earth to the Sun, which is \(1.5 \times 10^{11} \, \text{m}\).

The surface area \(A\) of a sphere is given by the formula:
\[ A = 4\pi r^2 \]
Substituting the value of \(r\):
\[ A = 4\pi (1.5 \times 10^{11})^2 \]

### Step 2: Calculate the total power received by the Earth
Now, we can calculate the total power received by the Earth using the area calculated in Step 1:
\[ P = \text{Power per unit area} \times \text{Area} \]
\[ P = 1.4 \times 10^3 \, \text{W/m}^2 \times A \]

### Step 3: Use Einstein's equation to find the mass loss
According to Einstein's equation, the relationship between energy (E) and mass (m) is given by:
\[ E = mc^2 \]
We can rearrange this to find the mass loss per second:
\[ m = \frac{E}{c^2} \]
Where \(E\) is the total power output (which is energy per second), and \(c\) is the speed of light, approximately \(3 \times 10^8 \, \text{m/s}\).

### Step 4: Substitute values to find the mass loss
Now we substitute the values into the equation:
\[ m = \frac{P}{c^2} \]
Substituting \(P\) from Step 2 and \(c = 3 \times 10^8 \, \text{m/s}\):
\[ m = \frac{1.4 \times 10^3 \times 4\pi (1.5 \times 10^{11})^2}{(3 \times 10^8)^2} \]

### Step 5: Calculate the numerical value
Now we can calculate the numerical value for the mass loss per second.

1. Calculate the area:
\[ A = 4\pi (1.5 \times 10^{11})^2 \approx 4\pi \times 2.25 \times 10^{22} \approx 28.26 \times 10^{22} \, \text{m}^2 \]

2. Calculate the total power:
\[ P \approx 1.4 \times 10^3 \times 28.26 \times 10^{22} \approx 39.56 \times 10^{25} \, \text{W} \]

3. Finally, calculate the mass loss:
\[ m = \frac{39.56 \times 10^{25}}{(3 \times 10^8)^2} = \frac{39.56 \times 10^{25}}{9 \times 10^{16}} \approx 4.40 \times 10^9 \, \text{kg/s} \]

### Final Answer
The decrease in the mass of the Sun per second is approximately \(4.40 \times 10^9 \, \text{kg/s}\).

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