Final answer:
To calculate the orbital period of an object in space, such as Pluto or 1996 TL 66, we apply Kepler's third law, that the square of the orbital period is proportional to the cube of the semi-major axis of the orbit. The semi-major axis is given or converted to astronomical units (AU), and then the formula P² = a³ is used to find the period P in Earth years.
Step-by-step explanation:
Applying Kepler's Third Law to Calculate Orbital Period
To determine the orbital velocity and period of Pluto, we can use Kepler's laws of planetary motion. For the orbital period, Kepler's third law states that the square of a planet's orbital period (P) is directly proportional to the cube of the semi-major axis (a) of its orbit. The semi-major axis is half the longest diameter of an ellipse, which is the shape of Pluto's orbit. We can express Kepler's third law with the formula P² ≈ a³. For a semi-major axis, a, of 5.870x10⁹ km for Pluto, the units can be converted to astronomical units (AU), since 1 AU is the average distance from the Earth to the Sun (approximately 149.6 million km). After conversion, we apply the law to find the orbital period P in Earth years.
The calculation for orbital velocity involves the gravitational constant and the masses involved. However, to simplify and since the question primarily focuses on the period, we will adhere to the period calculation. The velocity would involve additional information such as the mass of the sun and Pluto, and is often extracted from more complex orbital mechanics equations. The orbital period for the object designated as 1996 TL 66 with a semi-major axis of 84 AU can be calculated by using the same law: P² = a³. This results in P being the square root of 84 cubed, which provides an approximate orbital period for this object in Earth years.