To understand why the time period of a satellite orbiting a few hundred kilometers above the Earth's surface is approximately 2 hours, we need to consider the concept of geostationary satellites and the relationship between orbital radius and time period.

1. Geostationary Satellites:
A geostationary satellite is placed in a circular orbit above the Earth's equator, at an altitude of approximately 36,000 kilometers. It completes one orbit around the Earth in the same amount of time it takes for the Earth to complete one rotation, resulting in the satellite appearing stationary from the Earth's surface. This is useful for applications such as weather monitoring, communication, and broadcasting.

2. Orbital Radius and Time Period:
The time period of an orbiting satellite is the time it takes for the satellite to complete one full orbit around the Earth. It is influenced by the satellite's orbital radius, which is the distance between the center of the Earth and the satellite. The time period is directly proportional to the orbital radius, meaning that an increase in orbital radius leads to a longer time period.

3. Calculation:
Given that the radius of the Earth is 6,400 kilometers, and the satellite is orbiting a few hundred kilometers above the Earth's surface, we can approximate the satellite's orbital radius as 6,400 + 200 = 6,600 kilometers.

Using the relationship between orbital radius and time period, we can set up the following proportion:

(Orbital radius of geostationary satellite) / (Time period of geostationary satellite) = (Orbital radius of satellite above Earth's surface) / (Time period of satellite above Earth's surface)

Plugging in the values:

(36,000 km) / (Time period of geostationary satellite) = (6,600 km) / (Time period of satellite above Earth's surface)

Cross-multiplying and rearranging the equation, we get:

Time period of satellite above Earth's surface = (6,600 km) * (Time period of geostationary satellite) / (36,000 km)

Substituting the value of the geostationary satellite's time period as 24 hours (since it completes one orbit in the same time as Earth's rotation), we have:

Time period of satellite above Earth's surface = (6,600 km) * (24 hours) / (36,000 km)

Simplifying the expression:

Time period of satellite above Earth's surface = 2 hours

Therefore, the time period of a satellite orbiting a few hundred kilometers above the Earth's surface is approximately 2 hours, which corresponds to option A.

I'm sorry, but you have to specify how many kilometres exactly above the earth's surface.

To find the answer, you can use the Kepler's law for time period (T^2 ∝r^3).
Therefore, we can say :

T'/T" = (r'/r")^3/2
Here, T' will be 24 hrs (86400 seconds) and r' will be 36000 Km (36 × 10^ 6 m).
You'll need atleast one other value, it either may be T" or r", then only you can find the value of remaining quantity.

Please clarify the question with the value of the radius of the orbit.