Calculating Comet Halley's Speed Across Neptune's Orbit

AI Thread Summary
To calculate Comet Halley's speed when it crossed Neptune's orbit in 2006, the conservation of energy principle is applied, using initial and final distances from the Sun. At perihelion in February 1986, the comet was 8.79x10^7 km from the Sun, moving at 54.6 km/s. The mass of the comet cancels out in the calculations, simplifying the process. The mean distance of Neptune from the Sun is essential for determining the final position. With this information, the necessary calculations can be completed to find the comet's speed at that point.
musicmar
Messages
95
Reaction score
0

Homework Statement


At its perihelion in February 1986, Comet Halley was 8.79x10^7 km form the Sun and was moving at 54.6 km/s. What was its speed when it crossed Neptune's orbit in 2006?


Homework Equations


K0+U0=K+U


The Attempt at a Solution


I know I need to do this using conservation of energy. I have v-initial and r-initial, and presumably I can look up r-final. The mass of the comet will cancel, and big M will be the mass of the sun(also can look up). I am just confused about the r's, though. I was wondering why it is necessary to know the dates or that it was at its perihelion. How do I find the distance it was from the sun when it crossed Neptune's orbit?
 
Physics news on Phys.org
I figured it out. You actually just have to use the mean distance of Neptune from the Sun, and then you have all of the necessary information to use conservation of energy.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...

Similar threads

Replies
6
Views
8K
Replies
2
Views
2K
Replies
1
Views
2K
Replies
5
Views
3K
Replies
3
Views
2K
Replies
8
Views
305
Back
Top