Halley comet which passes around the sun every 76 years has an elliptical orbit. when closest to the sun it is at a distance of 8.823 x 10^10 m and moves with a speed of 54.6 km/s. The greatest distance between hally's comet and the sun ( aphelion ) is 6.152 x 10^12 m. calculate its speed at aphelion. answer should be in m/s

The easiest way to answer this question is to use Kepler's second law. At perihelion and aphelion, the rate of sweeping out area of the ellipse is the same. It is related to the law of conservation of angular momentum.

Thus,
(R^2*V)aphelion = (R^2*V)perihelion

Vaphelion = 54600 m/s*[(8.823*10^10)/6.152*10^12]^2
= __?

11.23 m/s

To calculate the speed of Halley's Comet at aphelion, we need to use Kepler's second law, which states that the area swept out by a planet or comet in equal time intervals is equal. We can use this law to determine the speed at different points in its elliptical orbit.

Given:

Perihelion distance (closest to the Sun) = 8.823 x 10^10 m
Speed at perihelion = 54.6 km/s = 54.6 x 10^3 m/s

Aphelion distance (greatest distance from the Sun) = 6.152 x 10^12 m
Speed at aphelion = ?

We know that the area swept out by the comet at perihelion is equal to the area swept out at aphelion.

Let's assume that the time taken to sweep out these areas is the same. Therefore, we can equate the two areas:

(1/2) * r1 * v1 = (1/2) * r2 * v2

Where:
r1 = Perihelion distance = 8.823 x 10^10 m
v1 = Speed at perihelion = 54.6 x 10^3 m/s
r2 = Aphelion distance = 6.152 x 10^12 m
v2 = Speed at aphelion (which we need to find)

Rearranging the equation and substituting the known values:

r1 * v1 = r2 * v2

(8.823 x 10^10 m) * (54.6 x 10^3 m/s) = (6.152 x 10^12 m) * v2

Simplifying the equation:

v2 = (8.823 x 10^10 m * 54.6 x 10^3 m/s) / (6.152 x 10^12 m)

Calculating further:

v2 = (480750 x 10^13 m^2/s) / (6.152 x 10^12 m)

Simplifying the units:

v2 = 78.211 m/s (approximately)

Therefore, the speed of Halley's Comet at aphelion is approximately 78.211 m/s.

To find the speed of Halley's comet at aphelion, we can use the conservation of angular momentum. Angular momentum is conserved for an object in orbit, which means that it stays constant throughout its entire orbit.

Angular momentum can be expressed as the product of the moment of inertia and the angular velocity. In this case, the moment of inertia (I) can be considered constant since the mass of the comet remains the same. So, we can write the equation as:

I1 * ω1 = I2 * ω2

Where I1 and ω1 represent the moment of inertia and angular velocity, respectively, at the perihelion (closest to the Sun), and I2 and ω2 represent the moment of inertia and angular velocity, respectively, at the aphelion (greatest distance from the Sun).

The moment of inertia (I) is given by the mass (m) times the square of the distance (r) from the center of mass of the comet to its axis of rotation:

I = m * r^2

Since we are comparing the perihelion and aphelion distances, we can write the equation as:

m * r1^2 * ω1 = m * r2^2 * ω2

The mass (m) cancels out, giving us:

r1^2 * ω1 = r2^2 * ω2

Now let's substitute the given values into this equation. Given that at perihelion, the distance (r1) is 8.823 x 10^10 m and the speed (v1) is 54.6 km/s, and at aphelion, the distance (r2) is 6.152 x 10^12 m (aphelion distance) and we want to calculate the speed (v2):

Convert the speed at perihelion to meters per second:
v1 = 54.6 km/s = 54.6 * 10^3 m/s

Now let's rearrange the equation to solve for ω2:

ω2 = (r1^2 * ω1) / r2^2

Substitute the given values:

ω2 = (8.823 x 10^10 m)^2 * (54.6 * 10^3 m/s) / (6.152 x 10^12 m)^2

Now, calculate ω2:

ω2 = (8.823^2 * 10^(10 * 2) * 54.6 * 10^3) / (6.152^2 * 10^(12 * 2))

Simplifying:

ω2 = (77.77 * 10^20 * 54.6 * 10^3) / (37.82 * 10^24)

ω2 = (4237.162 * 10^23) / (3782 * 10^24)

ω2 = 1.1208 * 10^(-1) radians/second

Finally, to find the speed at aphelion (v2), we can relate angular velocity (ω) and speed (v) by the formula:

v = r * ω

Substituting the given values:

v2 = r2 * ω2

v2 = 6.152 x 10^12 m * 1.1208 * 10^(-1) radians/second

Now, let's calculate v2:

v2 = 6.883 * 10^(12 - 1) m/s

v2 = 6.883 * 10^11 m/s

Therefore, the speed of Halley's comet at aphelion is approximately 6.883 x 10^11 m/s.