To transmit a signal, if height of transmitting signal above surface of the earth is H, this signal can be received on surface of the earth upto distance from transmitter. Then

Options

A. `d prop H `
B. `d prop H^(2)`
C. `d prop H^((1)/(2))`
D. `d prop H^((3)/(2))`

Explanation

To solve the problem of determining the distance \( d \) up to which a signal can be received on the surface of the Earth from a transmitter at height \( H \), we can use the geometry of the Earth and the principles of right triangles.

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- Let \( R \) be the radius of the Earth.
- Let \( H \) be the height of the transmitter above the Earth's surface.
- The distance \( d \) from the transmitter to the point on the Earth's surface where the signal can be received can be visualized as the tangent line from the transmitter to the surface of the Earth.

2. **Setting Up the Right Triangle**:
- The center of the Earth can be denoted as point \( O \).
- The point directly below the transmitter on the surface of the Earth is point \( A \).
- The transmitter is at point \( B \) (height \( H \) above point \( A \)).
- The point where the signal just touches the Earth's surface is point \( C \).
- We can form a right triangle \( OBC \) where:
- \( OB = R + H \) (the distance from the center of the Earth to the transmitter),
- \( OA = R \) (the radius of the Earth),
- \( BC = d \) (the distance from the transmitter to the point on the surface).

3. **Applying the Pythagorean Theorem**:
- According to the Pythagorean theorem:
\[ OB^2 = OA^2 + BC^2 \]
- Substituting the lengths:
\[ (R + H)^2 = R^2 + d^2 \]

4. **Expanding and Rearranging**:
- Expanding the left side:
\[ R^2 + 2RH + H^2 = R^2 + d^2 \]
- Canceling \( R^2 \) from both sides:
\[ 2RH + H^2 = d^2 \]

5. **Solving for \( d \)**:
- Rearranging gives:
\[ d^2 = 2RH + H^2 \]
- For small heights \( H \) compared to the radius \( R \) of the Earth, \( H^2 \) becomes negligible, thus:
\[ d^2 \approx 2RH \]
- Taking the square root:
\[ d \approx \sqrt{2RH} \]

6. **Conclusion**:
- The distance \( d \) up to which the signal can be received is given by:
\[ d \approx \sqrt{2RH} \]
- This shows that the distance is directly proportional to the square root of the height \( H \) of the transmitter.