Does Halley's Comet travel past the outer bounds of the Oort Cloud?

Question

We know Halley's Comet returns every 75-76 years. We can reasonably compute its elliptical orbit.

We know that the Oort Cloud is a cloud of predominantly icy planetesimals proposed to surround the Sun at distances ranging from 2,000 to 200,000 au. We divide it into two regions: a disc-shaped inner Oort cloud (or Hills cloud) and a spherical outer Oort cloud.

We know that some comets are thought to have originated from the Oort cloud.

My question is: Does Halley's Comet travel past the outer bounds of the Oort Cloud?

Answer 1

No. Halley's Comet has an aphelion of 35 AU, which is far less than the believed boundaries of the Oort cloud.

Any object on an elliptical orbit with a perihelion of 0.6 AU and an aphelion of 2,000 AU has a semimajor axis of about 1,000 AU, and orbital period of about 31,600 years, much longer than Halley's Comet's current orbital period of 75-76 years. To get out to 200,000 AU and back on an elliptical orbit, you'd be looking at an orbital period of more than 31.6 million years.

As mentioned by @NilayGhosh and @planetmaker, 1P/Halley likely started as an Oort Clout object, was perturbed into becoming a long-period comet, and probably subsequently became the prototypical periodic comet through gravitational interactions with the solar system's gas giants at some point in the last couple hundred thousand years.

As such, though it is possible that Halley's comet may have been in an orbit whose aphelion went beyond the outer boundary of the Oort Cloud in its long-period-comet days, it certainly does not do so now.

Answer 2

Without knowing anything specific about this comet, you can use Kepler's third law to get an idea whether this could be the case. It says that the cube of the semi-major axis is proportional to the square of the orbital period (with the same factor for all bodies orbiting the same central body): $$a^3 \propto T^2 $$

Comparing this to earth (which orbits the same star): $$ \frac{a^3_H}{T^2_H} = \frac{a^3_E}{T^2_E}$$ and thus (because $a_E = 1\,AU$ and $T_E = 1 \,\text{year}$, and using the 76 years from the question): $$ a_H = \Bigl(\frac{T_H}{T_E}\Bigr)^{\frac23} ·a_E = 76^{\frac23} ·a_E \approx 18 \,AU $$ This is the semi-major axis, and the aphelion can be at most double this distance (it is actually 35 AU, see the answer by notovny), but a lot less than the 2000 AU asked in the question.