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?

(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).

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.

To find the maximum distance from the Sun, we need to consider the eccentricity of Halley's comet's orbit.

The eccentricity of an orbit is a measure of how elongated or flattened the orbit is. It ranges from 0 to 1, with 0 being a perfect circle and 1 being a highly elongated ellipse.

Given that Halley's comet is 0.56 AU from the Sun at perihelion (the point in its orbit closest to the Sun), we can use this information to find the maximum distance from the Sun, called aphelion (the point in the orbit farthest from the Sun).

To do this, we need to use the equation of an ellipse:

r = a(1 + e)

Where r is the distance from the Sun, a is the semi-major axis, and e is the eccentricity.

We already know that the semi-major axis of Halley's comet's orbit is 1 AU, since 1 AU is defined as the semi-major axis of the Earth's orbit.

Substituting the given values into the equation, we have:

r = 1(1 + e)

At perihelion (closest to the Sun), r = 0.56 AU.

0.56 = 1(1 + e)

Solving for e, we get:

0.56 = 1 + e

e = 0.56 - 1

e = -0.44

Since we are looking for a positive value of eccentricity, we take the absolute value:

e = 0.44

Now that we have the eccentricity, we can find the maximum distance from the Sun (aphelion).

Using the equation r = a(1 + e), we can now substitute the known values to calculate the maximum distance from the Sun.

r = 1(1 + 0.44)

r = 1.44 AU

Therefore, the maximum distance from the Sun is 1.44 AU, and the eccentricity of the orbit is 0.44.

To find the maximum distance from the Sun and the eccentricity of Halley's comet's orbit, we can use the information given and some of Kepler's laws.

1) Maximum distance from the Sun:
The maximum distance from the Sun is called the aphelion. We are given that the comet's perihelion distance (closest distance to the Sun) is 0.56 AU.

Since the orbit is elliptical, we know that the distance from the Sun at aphelion is the sum of the perihelion distance and the length of the major axis:

Aphelion = Perihelion + 2 * Length of Semimajor Axis

Since 1 AU is equal to the semimajor axis of Earth's orbit, we can convert 0.56 AU to an actual distance using the fact that 1 AU is about 93 million miles.

0.56 AU * 93 million miles/AU ≈ 52.08 million miles

So, the aphelion distance is approximately 52.08 million miles.

2) Eccentricity of the orbit:
The eccentricity of an elliptical orbit can be determined using the following formula:

eccentricity = (Distance between the foci) / (Length of Major Axis)

For Halley's comet, we can assume that the distance between the foci is negligible compared to the length of the major axis. Hence, the eccentricity can be approximated as:

eccentricity ≈ (Perihelion Distance) / (Length of Major Axis)

Using the given perihelion distance of 0.56 AU and the fact that 1 AU is equal to the semimajor axis of Earth's orbit, we can convert to actual distance:

0.56 AU * 93 million miles/AU ≈ 52.08 million miles

Since the length of the major axis is twice the semimajor axis, we get:

Length of Major Axis = 2 * 1 AU = 2 * 93 million miles ≈ 186 million miles

Now we can calculate the eccentricity:

eccentricity ≈ (52.08 million miles) / (186 million miles) ≈ 0.28

Therefore, the maximum distance from the Sun (aphelion) is approximately 52.08 million miles, and the eccentricity of Halley's comet's orbit is approximately 0.28.