Question

A new asteroid is discovered in an orbit unusually close to the Earth’s

orbit. The semi-major axis of the orbit is 1.2 A.U. What is the period of

this asteroid’s orbit?

Answer 1

Hi, to calculate an asteroid's period which orbits around a central body, in this case that body being the sun (I'm assuming the central body is the sun because the Earth's orbit is mentioned), you use the astronomy orbital period equation:

$T=2\pi \sqrt{(a^{3}/\mu)}$

Where:

• T is the orbital period
• a is the length semi major-axis
• $\mu$ is the standard gravitational parameter of the central body

So if the central body is the Sun, because as I said, you're using the Earth as a reference, it's Standard Gravitational Parameter is $\mu = 132712440018 \frac{km^{3}}{s^{2}}$

This Standar Gravitational Parameters are established on a celestial body by calculating the product of its gravitational constant G and its mass M, so $\mu$=GM.

Given that 1 A.U.= 149597871 Km

Then 1.2 A.U. = 179517445.2 Km

So for the orbital period formula:

a = 179517445.2 Km

and

$\mu = 132712440018 \frac{km^{3}}{s^{2}}$

so for this problem, introducing the data in the formula, the answer would be

seconds

The base theory for this formula is the one used to make calculations over an elipse. Where the semi-major axis is one half of the major axis, this last one being the one that contains the longest diameter of the elipse.

The semi-major axis thus runs from the center of the elipse, through a focus and to the perimeter. It represents the radius of an orbit at its two most distant points.

Here's an image showing the semi-major and semi-minor axis of an elipse.

Semi-major axis Semi-minor axis focus focus

Thank you very much for asking, If you need any further information on this question, please feel free to ask again by posting another question and I will gladly answer it even more deeply for you.