Acceleration of Communication Satellites in Earth's Orbit

Satellites in orbit around the Earth experience a centripetal force that keeps them moving in a circular path. This force is provided by the gravitational attraction between the satellite and the Earth. To find the acceleration of the satellite, we can use the equation for centripetal acceleration:

\[a = \frac{v^2}{r}\]

where:
- \(a\) is the acceleration
- \(v\) is the velocity of the satellite
- \(r\) is the radius of the orbit

Let's calculate the acceleration of the communication satellite using the given radius of 44,400 km.

1. Convert the radius to meters:
\[r = 44,400 \times 1000 \, \text{m} = 44,400,000 \, \text{m}\]

2. Determine the velocity of the satellite:
The velocity of a satellite in a circular orbit can be calculated using the formula:
\[v = \frac{2\pi r}{T}\]

where:
- \(v\) is the velocity
- \(r\) is the radius of the orbit
- \(T\) is the period of the satellite (time taken to complete one orbit)

Since the period of the satellite is not given in the question, we need to use Kepler's third law to find it.

Kepler's third law states that the square of the period of a satellite is proportional to the cube of its average distance from the center of the Earth:
\[T^2 = k \times r^3\]

where:
- \(T\) is the period
- \(r\) is the radius of the orbit
- \(k\) is a constant

To find the value of \(k\), we can use the known period and radius of the Earth's moon:
\[T_{\text{moon}}^2 = k \times r_{\text{moon}}^3\]

Substituting the values of \(T_{\text{moon}}\) (27.3 days) and \(r_{\text{moon}}\) (384,400 km) into the equation, we can solve for \(k\):
\[k = \frac{T_{\text{moon}}^2}{r_{\text{moon}}^3}\]

Now we can calculate the period of the communication satellite using the value of \(k\) we obtained:
\[T^2 = k \times r^3\]
\[T^2 = \frac{T_{\text{moon}}^2}{r_{\text{moon}}^3} \times r^3\]
\[T = \sqrt{\frac{T_{\text{moon}}^2}{r_{\text{moon}}^3} \times r^3}\]

3. Calculate the velocity:
\[v = \frac{2\pi r}{T}\]

4. Substitute the values into the equation for acceleration:
\[a = \frac{v^2}{r}\]

The final answer should be option 'C' which is 0.2 m/s².

Acceleration of0.2 metre per second..................