Calculate the change in the value of g at altitude `45^(@)`. Take radius of earth `R = 6.37 xx 1-0^(3)` km`.

Explanation

To calculate the change in the value of \( g \) at an altitude of \( 45^\circ \), we will follow these steps:

### Step 1: Understand the formula for the change in \( g \)
The change in the value of gravitational acceleration \( g \) at a latitude \( \theta \) is given by the formula:
\[ g' = g - R \omega^2 \cos^2 \theta \]
where:
- \( g' \) is the new value of gravitational acceleration,
- \( g \) is the standard value of gravitational acceleration (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( R \) is the radius of the Earth,
- \( \omega \) is the angular speed of the Earth,
- \( \theta \) is the latitude.

### Step 2: Convert the radius of the Earth to meters
Given:
\[ R = 6.37 \times 10^3 \, \text{km} = 6.37 \times 10^6 \, \text{m} \]

### Step 3: Calculate the angular speed \( \omega \)
The angular speed \( \omega \) of the Earth can be calculated using:
\[ \omega = \frac{2\pi}{T} \]
where \( T \) is the time period of rotation of the Earth, which is \( 24 \, \text{hours} = 24 \times 3600 \, \text{seconds} \).

Calculating \( T \):
\[ T = 24 \times 3600 = 86400 \, \text{s} \]
Now substituting \( T \) into the formula for \( \omega \):
\[ \omega = \frac{2\pi}{86400} \approx 7.27 \times 10^{-5} \, \text{rad/s} \]

### Step 4: Calculate \( \cos^2 \theta \)
For \( \theta = 45^\circ \):
\[ \cos 45^\circ = \frac{1}{\sqrt{2}} \quad \Rightarrow \quad \cos^2 45^\circ = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \]

### Step 5: Substitute values into the formula
Now we can substitute \( R \), \( \omega \), and \( \cos^2 \theta \) into the formula:
\[ g' - g = -R \omega^2 \cos^2 \theta \]
Substituting the values:
\[ g' - g = - (6.37 \times 10^6) \times (7.27 \times 10^{-5})^2 \times \frac{1}{2} \]

Calculating \( \omega^2 \):
\[ \omega^2 = (7.27 \times 10^{-5})^2 \approx 5.29 \times 10^{-9} \, \text{rad}^2/\text{s}^2 \]

Now substituting this back into the equation:
\[ g' - g = - (6.37 \times 10^6) \times (5.29 \times 10^{-9}) \times \frac{1}{2} \]

Calculating:
\[ g' - g \approx - (6.37 \times 10^6) \times (2.645 \times 10^{-9}) \approx -0.168 \, \text{m/s}^2 \]

### Step 6: Conclusion
Thus, the change in the value of \( g \) at an altitude of \( 45^\circ \) is approximately:
\[ \Delta g \approx -0.168 \, \text{m/s}^2 \]