The radius of the orbit of a geosynchronous satellite is 36000km then the period of revolution of a satellite with its orbital radius9000km would be
The Law of Periods: The square of the period of any planet is proportional to the cube of the semimajor axis of its orbit.
(T/24hrs)^2=(9/36)^3
T^2=24^2(1/4)^3= 9 hr
t=3hr
To find the period of revolution of a satellite with its orbital radius, we can use Kepler's Third Law of Planetary Motion, which states that the square of the period is proportional to the cube of the orbital radius. The formula is as follows:
T^2 = k * r^3
Where T is the period of revolution, r is the orbital radius, and k is a constant.
Given that the radius of the geosynchronous satellite is 36000 km, let's assume that its period is T1. Using the formula above, we get:
T1^2 = k * (36000)^3
Now, we need to find the period of a satellite with an orbital radius of 9000 km. Let's call this period T2. Using the same formula, we have:
T2^2 = k * (9000)^3
To find T2, we can set up a proportion:
T1^2 / T2^2 = (36000)^3 / (9000)^3
Simplifying the equation:
T1^2 / T2^2 = 4^3
T1^2 / T2^2 = 64
Taking the square root of both sides:
T1 / T2 = √(64)
T1 / T2 = 8
Now, we can solve for T2:
T2 = T1 / 8
Therefore, the period of revolution of a satellite with an orbital radius of 9000 km would be T1 divided by 8.