The radius of the orbit of a geosynchronous satellite is 36000 km. then the period of revolution of a satellite with its
orbital radius 9000 km would be ?
Kepler's third law:
The ratio of the squares of the periods of any two planets is equal to the ratio of the cubes of their average distances from the sun. (The Law of Harmonies)
(periodsecond/24hrs)^2= (36000/9000)^3
period second= 24 sqrt (36/9)^3
To determine the period of revolution of a satellite with a given orbital radius, we need to use Kepler's Third Law of Planetary Motion, which states that the square of the period of revolution (T) of a satellite is proportional to the cube of the mean orbital radius (r) of the satellite.
The formula for Kepler's Third Law is:
T^2 = k * r^3
where T is the period of revolution, r is the orbital radius, and k is the proportionality constant.
To find the value of k, we can use the given information for the geosynchronous satellite. The radius of the orbit is 36000 km. This means that the period of revolution is 24 hours or 86400 seconds. We can substitute these values into the equation to solve for k:
(86400)^2 = k * (36000)^3
Simplifying the equation:
k = (86400)^2 / (36000)^3
Now we can use this value of k to find the period of revolution for the satellite with an orbital radius of 9000 km. Substituting the values into the formula:
T^2 = (86400)^2 / (36000)^3 * (9000)^3
Simplifying further:
T^2 = (86400)^2 / (36000)^3 * 72900000000
Taking the square root of both sides, we get:
T = square root[(86400)^2 / (36000)^3 * 72900000000]
Evaluating this equation using a calculator, we find that the period of revolution of the satellite with an orbital radius of 9000 km is approximately 3 hours and 48 minutes.