# A comet orbits the Sun with a period of 71.8 yr. (a) Find the semimajor axis of the orbit of the...

## Question:

A comet orbits the Sun with a period of 71.8 yr.

(a) Find the semimajor axis of the orbit of the comet in astronomical units (1 AU is equal to the semimajor axis of the Earth's orbit).

(b) If the comet is 0.50 AU from the Sun at perihelion, what is its maximum distance from the Sun and what is the eccentricity of its orbit? maximum distance? eccentricity?

## Kepler's Laws

Based on the meticulous observations of planetary positions by Tycho Brahe, three empirical laws regarding planetary motion were constructed by Johannes Kepler. The first law or the law of ellipses states that the planets move around the sun in ellipses with the sun at a focus. The second law or the law of areas states that the areal velocity of the radial vector drawn from the sun to the planet is constant. The third law or the law of harmonies states that the square of the time period of the orbit is proportional to the cube of the semi-major axis of the ellipse.

a)

According to Kepler's third law, the time period of orbit {eq}\displaystyle {T} {/eq} and the length of the semi-major axis of the orbit {eq}\displaystyle {a} {/eq} are related according to,

{eq}\displaystyle {T^2\propto a^3} {/eq}.

Therefore for two different orbits, we have the proportionality,

{eq}\displaystyle {\frac{T_2^2}{T_1^2}=\frac{a_2^3}{a_1^3}}---------(1) {/eq}

The time period of orbit of the earth around the sun is {eq}\displaystyle {T_1=1\ year} {/eq} and the semimajor axis of earth's orbit has a length {eq}\displaystyle {a_1=1\ AU} {/eq}

For the comet {eq}\displaystyle {T_2=71.8\ years} {/eq}

Therefore using (1) we get,

{eq}\displaystyle { \frac{71.8^2}{1^2}=\frac{a_2^3}{1^3}} {/eq}

Therefore,

{eq}\displaystyle {a_2=71.8^{\frac{2}{3}}=17.28\ AU} {/eq}.

b)

The major axis must have a length of {eq}\displaystyle {2\times 17.28=34.56\ AU} {/eq}.

It is given that the perihelion distance is {eq}\displaystyle {0.5\ AU} {/eq}. Now the sum of the perihelion distance with the aphelion distance should equal the length of the major axis.

Hence the maximum distance of the planet from the sun is {eq}\displaystyle { 34.56-0.5= 34.06\ AU} {/eq}.

Thus,

Aphelion distance=34.06 AU.