Mars has two moons, Phobos and Deimos. It is known that the larger moon, Phobos, has an orbital radius of 9.4 106 m and a mass of 1.1 1016 kg. Find its orbital period.
To find the orbital period of Phobos, we can use Kepler's Third Law of Planetary Motion, which states that the square of the orbital period (T) is directly proportional to the cube of the semi-major axis of the orbit (a^3) for all objects orbiting the same central body.
Kepler's Third Law equation:
T^2 = (4π^2 / GM) * a^3
Where:
T = Orbital period of the moon (unknown)
G = Gravitational constant = 6.67430 × 10^-11 m^3 kg^-1 s^-2
M = Mass of Mars = 6.39 × 10^23 kg
a = Orbital radius of Phobos = 9.4 × 10^6 m
Now, let's calculate the orbital period of Phobos using the equation:
T^2 = (4π^2 / GM) * a^3
Substituting the known values:
T^2 = (4π^2 / (6.67430 × 10^-11 * 6.39 × 10^23)) * (9.4 × 10^6)^3
Simplifying the equation:
T^2 = 4π^2 * (9.4 × 10^6)^3 / (6.67430 × 10^-11 * 6.39 × 10^23)
T^2 = (4 * (3.14159)^2 * 9.4^3 * 10^(6*3)) / (6.67430 * 6.39)
T^2 = (4 * 9.872 * 10^14 * 10^18) / (6.67430 * 6.39)
T^2 = 3.704163 * 10^34 /40.665987
T^2 ≈ 9.0975 * 10^32
Taking the square root of both sides:
T ≈ √(9.0975 * 10^32)
T ≈ 3.02 * 10^16 seconds
Hence, the orbital period of Phobos is approximately 3.02 * 10^16 seconds.