Using Kepler's constant (9.84 X 10^-14), calculate the orbital radius of an artificial satellite whose period of revolution around the Earth is 1.43 X 10^4s(3.64 h).
To calculate the orbital radius of an artificial satellite, we can use Kepler's Third Law, which states that the square of the period of revolution (T) is proportional to the cube of the semi-major axis (a) of the satellite's orbit.
The equation related to Kepler's Third Law is:
T^2 = (4π^2 / GM) * a^3
Where:
T is the period of revolution in seconds
G is the gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2)
M is the mass of the Earth (approximately 5.972 × 10^24 kg)
a is the semi-major axis in meters
In this case, we are given:
T = 1.43 × 10^4 s
G = 6.67430 × 10^-11 m^3 kg^-1 s^-2
M = 5.972 × 10^24 kg
We need to solve for a, the semi-major axis.
Rearranging the equation, we get:
a = [(T^2 * GM) / (4π^2)]^(1/3)
Now we can substitute the given values into the equation and calculate the orbital radius.
a = [((1.43 × 10^4 s)^2) * (6.67430 × 10^-11 m^3 kg^-1 s^-2) * (5.972 × 10^24 kg)] / (4π^2)^(1/3)
Calculating this expression will give us the value of the semi-major axis, or orbital radius, in meters.