Halley’s comet orbits the Sun with a period of 76.2 years.
a) Find the semi-major axis of the orbit of Halley’s comet in astronomical units (1 AU is equal to the semi-major axis of the Earth’s orbit
There is a simple form of Kepler's Third law for objects orbiting the sun.
If the period P is in years and the semimajor axis a is in a.u.,
P^2 = a^3
You know that P = 76.2 yr
Solve for a
To find the semi-major axis of Halley's comet's orbit in astronomical units (AU), we can use Kepler's third law of planetary motion, which states that the square of a planet's orbital period is directly proportional to the cube of the semi-major axis of its orbit.
The orbital period of Halley's comet is given as 76.2 years. Since the semi-major axis of Earth's orbit is considered 1 AU, we can rewrite this equation by substituting the values:
(Period of Halley's comet)^2 = (Semi-major axis of Halley's comet's orbit)^3 * (Period of Earth)^2
(76.2)^2 = (Semi-major axis of Halley's comet's orbit)^3 * (1)^2
Simplifying the equation:
(76.2)^2 = (Semi-major axis of Halley's comet's orbit)^3
5816.44 = (Semi-major axis of Halley's comet's orbit)^3
To find the semi-major axis of Halley's comet's orbit, we need to take the cube root of both sides of the equation:
Cube root of 5816.44 ≈ 22.910
Therefore, the semi-major axis of Halley's comet's orbit is approximately 22.910 AU.