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$r_H = 2.7 \times 10^9km$
$d=5.4 \times 10^9 km$

Giancoli 6th Edition, Chapter 5, Problem 59

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Chapter 5, Problem 59 is solved.

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Transcript for this Giancoli solution

We’re going to relate Halley’s Comet to the earth with Kepler’s Laws and then we’re going to talk about what this 'rh' actually means in this formula. So we have the period of Halley ’s Comet orbit squared divided by the period of earth’s orbit squared equals the radius of Halley’s Comet’s orbit cubed divided by the radius of earth’s orbit cubed. So this 'rh' isn’t really a radius of a circle so much instead it’s actually the semi major axis of the ellipse, the orbit being an ellipse instead of a circle. But I’ll draw a picture and explain that later, for now let’s just do the algebra and figure out what ‘rh’ is, so we’ll multiply both sides by the radius of the earth’s orbit cubed, in order to isolate ‘rh’ we’ll take he cube root of both sides as well and we’re left with: ‘rh’ equals the cube root of ‘rE’ cubed times ‘Th’ squared divided by one year squared, plugging in numbers and we get an answer: the cube root of one hundred and forty nine point six times ten raised to power six kilometers cubed times seventy six years squared divided by one year squared which gives us two point six eight four time ten raised to power nine kilometers, which is the semi-major axis of the ellipse of Halley’s Comet’s orbit. So let’s draw that. We’re told that the comet gets very close to the sun, which is one of the foci of the ellipse, and that’s a clue that the distance from the sun to the comet is negligible in comparison to the rest of the distance. Our job is to figure out the distance this comet gets away from the sun and we’ve figured out what 'rh' is, it’s the semi major axis and what our job is to find the distance from the sun to the outermost point, let’s call it ‘d’ and we’re going to say ‘d’ is essentially two times ‘rh’ although you can tell from the picture that’s not quite true. So we have: ‘d’ is approximately equal to five point four times ten raised to power nine kilometers. This gets really close to Pluto’s orbit which is five point nine times ten raised to power kilometers.