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?
To find the semimajor axis of the comet's orbit in astronomical units (AU), you can use Kepler's third law of planetary motion, which states that the square of the period of revolution of a planet is proportional to the cube of the semimajor axis of its orbit.
(a) Find the semimajor axis of the orbit:
1. Convert the period of revolution (T) to years: 71.8 years
2. Apply Kepler's third law:
T^2 = a^3
Where T is the period and a is the semimajor axis.
3. Rearrange the equation to solve for a:
a = (T^2)^(1/3)
Substitute the value of T:
a = (71.8^2)^(1/3) = 27.87 AU
Therefore, the semimajor axis of the comet's orbit is approximately 27.87 AU.
(b) To find the maximum distance from the Sun and the eccentricity of the orbit, we need the perihelion distance and the semimajor axis.
Given:
Perihelion distance (r_min) = 0.50 AU
Semimajor axis (a) = 27.87 AU
1. The maximum distance from the Sun (r_max) can be found using the relationship between the perihelion distance and the eccentricity (e) of the orbit:
r_max = a(1 + e)
2. To find the eccentricity (e), we use the equation:
r_min = a(1 - e)
Rearrange the equation to solve for e:
e = 1 - (r_min / a)
Substitute the given values:
e = 1 - (0.50 / 27.87) = 1 - 0.01794 = 0.9821 (rounded to four decimal places)
Therefore, the eccentricity of the orbit is approximately 0.9821.
Now, substitute the values of a and e into the equation for r_max:
r_max = 27.87(1 + 0.9821) = 27.87(1.9821) = 55.23 AU
Therefore, the maximum distance from the Sun is approximately 55.23 AU.