# How do you calculate the aphelion and perihelion of a comet's orbit if it has an orbital period of 83 years and an eccentricity of .875?

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

6
Nov 24, 2015

Using Kepler's 3rd law the perihelion distance is about 2.375AU and the aphelion distance is about 35.625AU.

#### Explanation:

Kepler's 3rd law enables an approximate calculation of the size of the comet's orbit. The actual obit will not be a true ellipse due to the gravitational effect of other bodies in the solar system.

Kepler's 3rd law states that the square of the orbital period divided by the cube of the cube of the semi-major axis of the orbit is a constant. For our solar system if the period T is in years and the semi-major a axis is in Astronomical Units, then the constant is one. Then ${T}^{2} = {a}^{3}$.

The Astronomical Unit is based on the distance of the Earth from the Sun. Earths period $T = 1$ year which means that Earth's semi major axis is $a = 1$ AU.

For the comet with a period of $t = 83$ years then the semi major axis ${a}^{3} = {83}^{2}$. Squaring and taking the cube root gives approximately $a = 19$ AU.

The perihelion distance $P = a \left(1 - e\right)$ and the aphelion distance $A = a \left(1 + e\right)$ where $e = 0.875$ is the eccentricity. This gives a perihelion distance of 2.375AU and an aphelion distance of 35.625AU.

• 16 minutes ago
• 16 minutes ago
• 16 minutes ago
• 16 minutes ago
• 4 minutes ago
• 6 minutes ago
• 8 minutes ago
• 11 minutes ago
• 14 minutes ago
• 16 minutes ago
• 16 minutes ago
• 16 minutes ago
• 16 minutes ago
• 16 minutes ago