(perihelion distance) + (aphelion distance) = 2 a
You should have got a = 17.97 au in the first part
The aphelion (greatest) distance is therefore 36.95 - 0.56 = 35.4 au
There is an equation that relates eccentricity to the perihelion/semimajor ratio.The eccentricity (e) equals 1 - (perihelion)/(semi-major axis).
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... could you please help me with this?
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?
2 answers
e = 1 - (0.56/17.97) = 0.9688
e = 0 would be a circular orbit
e = 1 is the maximum possible eccentricity for an ellipse.
e = 0 would be a circular orbit
e = 1 is the maximum possible eccentricity for an ellipse.