Question

What is the orbital period (in years) of a planet with a semimajor axis of 15 AU?

Answer 1

T = 58.1 years

Step-by-step explanation:

To determine the orbital period (in years) of a planet with a semimajor axis of 15 Astronomical Units (AU), we will utilize Kepler's Third Law of Planetary Motion. This law relates the period of a planet's orbit around a star to its semimajor axis.

Given:

• a = 15 AU

`\hrulefill`

Kepler's Third Law states that the square of the period of any planet (T²) is proportional to the cube of the semimajor axis of its orbit (a³), mathematically given by:

`T^2 \propto a^3`

For a planet orbiting the Sun, the relationship can be expressed as:

`\Longrightarrow (T^2)/(a^3) =k`

Using Earth as a reference (with a semimajor axis of 1 AU and an orbital period of 1 year), the constant, 'k', is 1:

`\Longrightarrow (T^2)/(a^3) =1`

Solve for 'T':

`\Longrightarrow (T^2)/(a^3) =1\\\\\\\\\Longrightarrow T^2 =a^3\\\\\\\Longrightarrow T =√(a^3)`

Plugging in the value we have:

`\Longrightarrow T =√(a^3)\\\\\\\\\Longrightarrow T =√((15)^3)\\\\\\\\\Longrightarrow T =√(3375)\\\\\\\\\therefore \boxed{T=58.1 \ \text{years}}`

Thus, the orbital period 'T' is approximately 58.1 Earth years.