Halley's comet moves about the sun in an elliptical orbit with its closest approach to the SUn being .59AU and its greatest distance being 35AU (1AU=Earth-sun distance) if the comet's speed at closest approach is 54km/s what is its speed when it is farthest from the Sun? you may neglect any change in the comet's mass and assume that its angular momentum about the sun is conserved.
btw the answer in the back of the book is 910 m/s
what does it mean when it says the momentum is conserved
To find the speed of Halley's comet when it is farthest from the Sun, we can make use of the conservation of angular momentum.
The angular momentum of an object revolving around a central point is given by the equation:
L = (mvr)
Where:
L = angular momentum
m = mass
v = velocity
r = distance from the central point
Since the mass of the comet remains constant, and it is given that the angular momentum of Halley's comet about the Sun is conserved, we can equate the angular momentum values at closest approach and farthest distance from the Sun.
At closest approach:
L1 = m * v1 * r1
At farthest distance:
L2 = m * v2 * r2
Since the mass of the comet is the same in both cases, we can ignore it.
Next, we can equate the two expressions for angular momentum:
v1 * r1 = v2 * r2
Given that the closest approach distance (r1) is 0.59 AU, the farthest distance (r2) is 35 AU, and the speed at closest approach (v1) is 54 km/s, we can rearrange the equation to solve for v2:
v2 = (v1 * r1) / r2
Substituting the values:
v2 = (54 km/s * 0.59 AU) / 35 AU
To calculate this, we need to convert the distance values to the same unit. 1 AU is equal to the Earth-Sun distance, which is approximately 149.6 million kilometers.
So, 0.59 AU is equivalent to 0.59 * 149.6 million km.
v2 = (54 km/s * 0.59 * 149.6 million km) / (35 * 149.6 million km)
Simplifying the expression:
v2 = (54 km/s * 0.59) / 35
Now, we can calculate the value of v2:
v2 = 910 m/s
Hence, the speed of Halley's comet when it is farthest from the Sun is 910 m/s.