A comet orbits the Sun with a period of 72.8 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 0.55 AU from the Sun at perihelion, what is its maximum distance from the Sun and what is the eccentricity of its orbit?
To find the semimajor axis of the comet's orbit, we can use Kepler's third law, which states that the square of the orbital period (T) is proportional to the cube of the semimajor axis (a) of the orbit:
T^2 = a^3
(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):
Given: T = 72.8 years
Using Kepler's third law, we can solve for a:
T^2 = a^3
(72.8)^2 = a^3
a^3 = 5314.24
a ≈ 18.35 AU
Therefore, the semimajor axis of the comet's orbit is approximately 18.35 astronomical units.
(b) If the comet is 0.55 AU from the Sun at perihelion, we can find its maximum distance from the Sun and the eccentricity of its orbit.
Given: perihelion distance (r_peri) = 0.55 AU
The maximum distance from the Sun, also known as the aphelion distance (r_aphe), can be calculated using the following formula:
r_aphe = 2 * a - r_peri
Substituting the known values:
r_aphe = 2 * 18.35 AU - 0.55 AU
= 36.7 AU - 0.55 AU
= 36.15 AU
Therefore, the maximum distance from the Sun is approximately 36.15 astronomical units.
The eccentricity (e) of the orbit can be calculated using the perihelion distance (r_peri) and aphelion distance (r_aphe) as follows:
e = (r_aphe - r_peri) / (r_aphe + r_peri)
Substituting the known values:
e = (36.15 AU - 0.55 AU) / (36.15 AU + 0.55 AU)
= 35.6 AU / 36.7 AU
≈ 0.97
Therefore, the eccentricity of the comet's orbit is approximately 0.97.
To find the answers to these questions, we can use Kepler's third law and the formula for eccentricity.
(a) The period, T, and the semimajor axis, a, are related by Kepler's third law equation:
T^2 = (4π^2 * a^3) / (G * M)
Where G is the gravitational constant (6.67430 x 10^-11 m^3 kg^-1 s^-2) and M is the mass of the Sun (1.989 x 10^30 kg).
We can rearrange this equation to solve for the semimajor axis:
a = [(G * M * T^2) / (4π^2)]^(1/3)
Plugging in the given values:
T = 72.8 years
G = 6.67430 x 10^-11 m^3 kg^-1 s^-2
M = 1.989 x 10^30 kg
π ≈ 3.14159
a = [((6.67430 x 10^-11) * (1.989 x 10^30) * (72.8)^2) / (4π^2)]^(1/3)
Calculating this expression will give us the semimajor axis in meters. To convert it to astronomical units (AU), we need to divide by the semimajor axis of the Earth's orbit, which is approximately 1 AU.
(b) To find the maximum distance from the Sun and the eccentricity, we need to use the formula for eccentricity:
e = (r_apo - r_peri) / (r_apo + r_peri)
Where e is the eccentricity, r_apo is the maximum distance from the Sun (apoapsis), and r_peri is the distance at perihelion.
We are given r_peri = 0.55 AU.
To find r_apo, we need to use the fact that the sum of the distances at apoapsis and perihelion is equal to twice the semimajor axis:
r_apo + r_peri = 2a
Solving for r_apo:
r_apo = 2a - r_peri
Plugging in the calculated value of a, we can find r_apo. Once we have r_apo and r_peri, we can calculate the eccentricity using the formula above.