Halley’s comet orbits the Sun with a period of 76.2 years.
a) Find the semi-major axis of the orbit of Halley’s comet in astronomical units (1 AU is equal to the semi-major axis of the Earth’s orbit
Physics - drwls, Thursday, May 5, 2011 at 8:51am
There is a simple form of Kepler's Third law for objects orbiting the sun.
If the period P is in years and the semimajor axis a is in a.u.,
P^2 = a^3
You know that P = 76.2 yr
Solve for a
Thanks a lot, I'm confused with part b as well...
If Halley’s comet is 0.56 AU from the Sun at perihelion, what is the maximum distance from the Sun, and what is the eccentricity of its orbit?
.59 AU
To find the maximum distance from the Sun and the eccentricity of Halley's comet's orbit, we can use the information given:
1. Maximum distance from the Sun:
The maximum distance from the Sun, also known as the aphelion, can be calculated using the information that the comet is 0.56 AU from the Sun at perihelion.
Perihelion distance = 0.56 AU
Aphelion distance = maximum distance from the Sun
Since the orbit of Halley's comet is an ellipse, we can use the relationship between the perihelion distance and the aphelion distance given by:
Aphelion distance = 2 * semimajor axis - perihelion distance
Since we know the perihelion distance (0.56 AU) and we want to find the aphelion distance, we can rearrange the equation to solve for the aphelion distance:
Aphelion distance = perihelion distance + 2 * semimajor axis
Now, let's use the equation we derived from Kepler's Third Law:
P^2 = a^3
We know the period of Halley's comet, which is 76.2 years, and we want to find the semimajor axis (a) in AU. Rearranging the equation, we can solve for the semimajor axis:
semimajor axis = (P^2)^(1/3)
Plugging in the values:
semimajor axis = (76.2^2)^(1/3) = 17.8 AU (approx)
Now, we can find the aphelion distance:
Aphelion distance = 0.56 AU + 2 * 17.8 AU = 0.56 AU + 35.6 AU = 36.16 AU (approx)
Therefore, the maximum distance from the Sun (aphelion) is approximately 36.16 AU.
2. Eccentricity of the orbit:
The eccentricity of an ellipse can be calculated using the formula:
eccentricity = (aphelion distance - perihelion distance) / (aphelion distance + perihelion distance)
Plugging in the values we found:
eccentricity = (36.16 AU - 0.56 AU) / (36.16 AU + 0.56 AU) = 35.6 AU / 36.72 AU = 0.970 (approx)
Therefore, the eccentricity of Halley's comet's orbit is approximately 0.970.