A satellite is placed 39875 m above the Moon's SURFACE. (M=7.35e22 kg, R=1.74e6 m) Find the Period of the satellite in seconds. (assume a circular orbit)
To find the period of the satellite, we can use Kepler's third law, which relates the period of an object in a circular orbit to the mass of the celestial body it is orbiting and the radius of its orbit.
Kepler's third law states that the square of the period (T) is directly proportional to the cube of the radius (R) of the orbit. Mathematically, this can be represented as:
T^2 = (4π^2 * R^3) / (G * M)
Where T is the period, R is the radius, G is the gravitational constant (6.67 x 10^-11 N m^2/kg^2), and M is the mass of the celestial object.
Let's calculate the period of the satellite using the given values:
R = 1.74 x 10^6 m
M = 7.35 x 10^22 kg
G = 6.67 x 10^-11 N m^2/kg^2
Substituting these values into the formula:
T^2 = (4π^2 * (1.74 x 10^6)^3) / (6.67 x 10^-11 * 7.35 x 10^22)
Now, we can solve for T by taking the square root of both sides of the equation:
T = sqrt((4π^2 * (1.74 x 10^6)^3) / (6.67 x 10^-11 * 7.35 x 10^22))
Calculating this equation will give us the period of the satellite in seconds.