In 1986, the distance of closest approach of Halley’s comet to the Sun is 0.57AU. (1 AU is the mean Earth-Sun distance = 150 million kilometers.) The comet’s speed at closest approach is

54.97 km/s.

(a) – By 1989, Halley’s comet had traveled to a distance corresponding to the orbit of Saturn at 9.56AU. What was the speed of Halley’s Comet at that time? Give your answer in units of km/s.

(b) – In 2024, Halley’s comet will be at the furthest distance from the sun corresponding to about 35AU. What will the speed of Halley’s Comet be at that time?

To find the speed of Halley's Comet at different distances from the Sun, we can use the conservation of angular momentum. The angular momentum of an object moving in a circular orbit is constant as long as no external torques act on it. Mathematically, we can write:

L = mvR

where L is the angular momentum, m is the mass of the comet, v is its velocity, and R is its distance from the Sun.

We can rewrite this equation as:

v = L / (mR)

where v is the velocity.

(a) To find the speed of Halley's Comet at a distance corresponding to the orbit of Saturn (9.56 AU), we can use the equation above. Given that the distance of closest approach is 0.57 AU and the speed at closest approach is 54.97 km/s, we can calculate the angular momentum L at closest approach:

L_closest = mv_closest * R_closest

L_closest = m * v_closest * 0.57 AU

Next, we can use the fact that angular momentum is conserved to calculate the velocity v at a distance of 9.56 AU:

v = L_closest / (m * R)

v = (m * v_closest * 0.57 AU) / (m * 9.56 AU)

Simplifying the equation, we get:

v = v_closest * (0.57 / 9.56)

Substituting the values, we find:

v = 54.97 km/s * (0.57 / 9.56) ≈ 3.29 km/s

So, the speed of Halley's Comet at a distance corresponding to the orbit of Saturn is approximately 3.29 km/s.

(b) To find the speed of Halley's Comet at a distance of 35 AU, we can again use the same equation. Given that the distance of closest approach is 0.57 AU and the speed at closest approach is 54.97 km/s, we can calculate the angular momentum L at closest approach:

L_closest = mv_closest * R_closest

L_closest = m * v_closest * 0.57 AU

Next, we can use the fact that angular momentum is conserved to calculate the velocity v at a distance of 35 AU:

v = L_closest / (m * R)

v = (m * v_closest * 0.57 AU) / (m * 35 AU)

Simplifying the equation, we get:

v = v_closest * (0.57 / 35)

Substituting the values, we find:

v = 54.97 km/s * (0.57 / 35) ≈ 0.89 km/s

So, the speed of Halley's Comet at a distance corresponding to about 35 AU will be approximately 0.89 km/s.

how do you solve this?