Solving Elliptical Orbit Homework: Comet's Semi-Major Axis & Sun Distance

[myoplex11]

Homework Statement

The period of a comet orbit is 100 years. As its closests approach is 0.37 AU from the sun. What is the semi-major axis of the comet's orbit? How far does it get from the sun?

Homework Equations

(T1/T2)^2 = (s1/s2)^3

The Attempt at a Solution

I don't even know where to begin with this problem any help would be really appreciated.

 

[alphysicist]

Hi myoplex,

Homework Statement

The period of a comet orbit is 100 years. As its closests approach is 0.37 AU from the sun. What is the semi-major axis of the comet's orbit? How far does it get from the sun?

Homework Equations

(T1/T2)^2 = (s1/s2)^3

The Attempt at a Solution

I don't even know where to begin with this problem any help would be really appreciated.

What does Kepler's laws have to say about periods and semi-major axis?

 

[thomate1]

I would suggest starting by reviewing the basic principles of Kepler's laws of planetary motion. These laws describe the relationship between the orbital period of a planet or comet, the semi-major axis of its orbit, and its distance from the sun.

In this case, we are given the orbital period (T) of the comet, which is 100 years. We also know that at its closest approach, the comet is 0.37 AU (astronomical units) from the sun. This information can be used to solve for the semi-major axis (a) of the comet's orbit using the following equation:

a = (s1 + s2)/2

Where s1 and s2 represent the distances from the sun at perihelion (closest approach) and aphelion (farthest distance), respectively. In this case, we are only given the distance at perihelion, so we can rewrite the equation as:

a = (0.37 AU + s2)/2

Next, we can use Kepler's third law (T1/T2)^2 = (a1/a2)^3 to solve for the distance at aphelion (s2). Rearranging the equation, we get:

s2 = s1 * (T2/T1)^2 * (a2/a1)^3

Plugging in the values we know, we get:

s2 = 0.37 AU * (100 years / 100 years)^2 * (a2/a1)^3

We can simplify this to:

s2 = 0.37 AU * (a2/a1)^3

Now, we can plug this value into our equation for the semi-major axis to get:

a = (0.37 AU + 0.37 AU * (a2/a1)^3)/2

We can further simplify this to:

a = 0.37 AU * (1 + (a2/a1)^3)/2

Finally, we can plug in a value for the semi-major axis (a) and solve for a2, the distance at aphelion. For example, if we assume a semi-major axis of 1 AU, we get:

1 AU = 0.37 AU * (1 + (a2/1 AU)^3)/2

Solving for a2, we get a distance of approximately 1.9 AU at aphelion.

In conclusion