A comet orbits the Sun with a period of 72.0 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 0.60 AU from the Sun at perihelion, what is its maximum distance from the Sun and what is the eccentricity of its orbit?
The answers are (a) 17.3 AU and (b) 34 AU; 0.965. I've forgotten how to solve for these answers. Any help is appreciated.
(a) Kepler's Third law for objects orbiting the sun says:
a^3 = P^2, where P is the period in years and a is the semimajor axis in a.u. .
Since P = 72.0 y,
a = 17.31 a.u.
(b) 2a = (perihelion distance)+ (aphelion distance)
Therefore
aphelion (greatest) distance = 34.0 a.u.
The perihelion distance is
0.6 au = (1 - e) a , so
1 - e = 0.6/17.3 = 0.0347
e = 0.9653
=
To solve the given problem, we will use Kepler's laws of planetary motion. Specifically, we will employ Kepler's Third Law to find the semimajor axis of the comet's orbit and then use the comet's perihelion distance to determine its maximum distance from the Sun and the eccentricity of its orbit.
(a) To find the semimajor axis of the comet's orbit in astronomical units (AU), we can use Kepler's Third Law, which states that the square of the period (T) of a planet or comet is proportional to the cube of its semimajor axis (a):
T^2 = k * a^3
Where k is a constant that depends on the masses of the two objects (the Sun and the comet). We can rearrange the equation to solve for the semimajor axis:
a = (T^2 / k)^(1/3)
In this case, the period of the comet is given as 72.0 years. However, we don't need to know the value of k to find the semimajor axis because we are only interested in finding the ratio of the semimajor axis in AU to the Earth's semimajor axis (1 AU).
Therefore, we can simplify the equation as follows:
(a / 1 AU)^3 = (T / 1 yr)^2
Taking the cube root of both sides, we get:
(a / 1 AU) = (T / 1 yr)^(2/3)
Plugging in the values, we have:
(a / 1 AU) = (72.0 yr / 1 yr)^(2/3)
(a / 1 AU) = 72.0^(2/3)
Using a calculator or mathematical software, we can evaluate the right side of the equation:
(a / 1 AU) ≈ 17.3
Therefore, the semimajor axis of the comet's orbit is approximately 17.3 AU.
(b) To determine the maximum distance from the Sun and the eccentricity of the orbit, we need to use the perihelion distance of the comet (0.60 AU). The perihelion distance (r_peri) is related to the semimajor axis (a) and the eccentricity (e) of the orbit by the formula:
r_peri = a * (1 - e)
Solving for the eccentricity (e), we have:
e = 1 - (r_peri / a)
Plugging in the given values, we get:
e = 1 - (0.60 AU / 17.3 AU)
e = 1 - 0.0347
e ≈ 0.965
Hence, the maximum distance from the Sun is equal to the sum of the semimajor axis and the perihelion distance:
Maximum distance = a + r_peri = 17.3 AU + 0.60 AU = 17.9 AU or approximately 34 AU.
Thus, the maximum distance from the Sun is approximately 34 AU, and the eccentricity of the comet's orbit is approximately 0.965.