To find the period of revolution of a satellite orbiting the Earth in a circular path, we can use the formula:
T = 2π√(r³/GM)
Where:
T is the period of revolution,
r is the radius of the orbit,
G is the gravitational constant (approximately 6.67430 × 10^(-11) m³/(kg·s²)),
M is the mass of the Earth (approximately 5.972 × 10^24 kg).
Given:
Radius of the orbit, r = 9480 km = 9480 × 10^3 m
Radius of the Earth, R = 6371 km = 6371 × 10^3 m (approximately)
Height above the surface of the Earth = r - R = (9480 × 10^3) - (6371 × 10^3) m
We can calculate the total distance between the satellite and the center of the Earth as the sum of the radius of the Earth and the height above the surface:
Total distance = R + (9480 × 10^3 - 6371 × 10^3) m
Now, substituting the values into the formula for the period of revolution:
T = 2π√((R + (9480 × 10^3 - 6371 × 10^3))^3 / (G * M))
Calculating the period of revolution:
T = 2π√(((6371 × 10^3) + (9480 × 10^3 - 6371 × 10^3))^3 / (6.67430 × 10^(-11) * (5.972 × 10^24)))
Simplifying the expression:
T = 2π√((9480 × 10^3)^3 / (6.67430 × 10^(-11) * (5.972 × 10^24)))
Calculating the value:
T ≈ 2π√(8.982144 × 10^19 / (4.4649917 × 10^14))
Simplifying further:
T ≈ 2π√(2015.75 × 10^5)
Calculating the square root:
T ≈ 2π * 1422.374108
Calculating the value:
T ≈ 8952.778661 s
Therefore, the period of revolution of the satellite is approximately 8952.778661 seconds.