Halley\'s comet is the only short-period comet that is regularly visible to the
ID: 284278 • Letter: H
Question
Halley's comet is the only short-period comet that is regularly visible to the naked eye on Earth, making a return appearance every 76 years (last in 1986). The comet is in a highly elliptical elliptical orbit, and its distance of closest approach to the Sun is 0.6 A.U. where 1 A.U. equivalence 1 Astronomical Unit equivalence r_E, the average distance of the Earth from the Sun. How far is the comet from the Sun at the outer extreme of its orbit? (It is convenient and allowed to express your answer in A.U.'s) What is the ratio of its maximum orbit speed to its minimum orbit speed?Explanation / Answer
a) Halley's Comet is named after the astronomer Edmund Halley (1656 - 1742). He didn't really discover it but he supposed that comets actually orbit the Sun like planets and that the comet takes 75 to 76 years to complete an orbit around the Sun. Comets move in elliptical orbits, typically going from a great distance away from the Sun to comparatively close to it. It makes visible from the Earth, every 75 to 76 years, as Halley's Comet come near the Sun.
The largest distance (aphelion) from Halley's Comet from the Sun is 35 Astronomical Units (approximately 150 million kilometers). Halley's Comet is named after the astronomer Edmund Halley (1656 - 1742). He didn't really discover it but he supposed that comets actually orbit the Sun like planets and that the comet takes 75 to 76 years to complete an orbit around the Sun. Comets move in elliptical orbits, typically going from a great distance away from the Sun to comparatively close to it. It makes visible from the Earth, every 75 to 76 years, as Halley's Comet come near the Sun.
Th = time period for Halley's Comet = 76 years
Te = time period for earth = 1 year
Rh = distance of comet from sun
Re = distance of earth from sun using kepler's law
(Th/Te)2 = (Rh/Re)3
(76/1)2 = (Rh/1)3
Rh = 17.94 AU
Minimum distance given is 0.6 AU
Maximum distance = 2 Rh - minimum distance
Maximum distance = 2 (17.94) - 0.6 = 35.28 AU
b) The angular velocity = GM/D3,
Where
M is the mass of the sun,
D is the radius of the circular orbit around the sun.
Period, T is given by T2 = 42D3/GM
D3 =GMT2/42
When
T =1 year and
M=2x1030kg,
D=1.50x1011 m
So comet’s semi-major axes, a=2.69x1012m
The comet’s minimum distance, rmin =a (1-e),
Where
e is the eccentricity.
rmin has a value of 9x1010m, So
e = 1-(9.0x1010/2.69x1012)
= 0.9665
The maximum distance rmax from the sun is = a (1+e)
= 2.69x1012*1.9665
= 5.3x1012m
The ratio of maximum to minimum orbital speed =rmax/rmin = (1+e)/(1-e) =59
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.