Calculating Halley's Comet Speed at Maximum Distance

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Homework Help Overview

The discussion revolves around calculating the speed of Halley's Comet at its farthest distance from the sun, given its speed at closest approach and the principle of conservation of angular momentum. The problem involves understanding the relationship between distance, speed, and angular momentum in the context of orbital mechanics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the conservation of angular momentum and how to set up equations relating the comet's speed and distance at different points in its orbit. Questions arise about the necessary variables and how to express angular momentum in terms of known quantities.

Discussion Status

Participants are actively engaging with the problem, attempting to clarify the steps needed to calculate angular momentum. Some have provided guidance on how to set up the equations, while others express confusion about the process and the role of mass in the calculations. There is a recognition of different learning styles, with some participants seeking more visual or step-by-step explanations.

Contextual Notes

There is an emphasis on the importance of attempting the problem independently, as per forum rules. Participants are encouraged to explore the concepts of angular momentum without direct solutions being provided.

physicsgurl12
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Homework Statement



As Halley’s comet orbits the sun, its distance from the sun changes dramatically, from 88700000000 m to 5614000000000 m. If the comet’s speed at closest approach is 53430 m/s, what is its speed when it is farthest from the sun if angular momentum is conserved?


Homework Equations



v=d/t ? i have no clue where to begin.

The Attempt at a Solution



my attempt was usless. i don't even know a formula to use.
 
Last edited by a moderator:
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welcome to pf!

hi physicsgurl12! welcome to pf! :smile:

what is its angular momentum at closest approach (call its mass "m", and assume it's a point mass)? :wink:
 
physicsgurl12 said:

Homework Statement



As Halley’s comet orbits the sun, its distance from the sun changes dramatically, from 88700000000 m to 5614000000000 m. If the comet’s speed at closest approach is 53430 m/s, what is its speed when it is farthest from the sun if angular momentum is conserved?

Homework Equations



v=d/t ? i have no clue where to begin.

The Attempt at a Solution



my attempt was usless. i don't even know a formula to use.

Hi Physicsgurl12, welcome to PF! :smile:

The relevant equation you need, is conservation of angular momentum L.
That is:

L = mvr

where v is the speed perpendicular to a distance r from the sun, and m is the mass of Halley's comet.

Can you set up an equation that equates the angular momentum at the closest point to the sun to the angular momentum at the farthest point to the sun?
 
Last edited:
Note that at closest approach and furthest from the Sun that the comet is at either end of its elliptical orbit. At those times, what can be said about the relative orientations of the velocity and radius vectors? Can you write formulae for the magnitude of the angular momentum (or specific angular momentum) at those points?
 

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this helps but i don't know how to fill in the equation using the numbers i have.
 
physicsgurl12 said:
this helps but i don't know how to fill in the equation using the numbers i have.

At the closest point you have a distance r = 88700000000 m
And you have a speed v = 53430 m/s
Since the comet is at its closest point to the sun, its speed will be perpendicular to its distance, which is what we need.
Furthermore we have an unknown mass m, which we will just leave as "m" for now.

Can you calculate the angular momentum L in this case?
 
physicsgurl12 said:
this helps but i don't know how to fill in the equation using the numbers i have.

Do you have both the distance and speed at a point in the orbit where the velocity is perpendicular to the radius vector?
 
I like Serena said:
At the closest point you have a distance r = 88700000000 m
And you have a speed v = 53430 m/s
Since the comet is at its closest point to the sun, its speed will be perpendicular to its distance, which is what we need.
Furthermore we have an unknown mass m, which we will just leave as "m" for now.

Can you calculate the angular momentum L in this case?

uh no, because we don't know M??
 
call the mass "m" (or "M")
 
  • #10
could you just post how you would solve it. i think that would help me understand better.?
 
  • #11
physicsgurl12 said:
uh no, because we don't know M??

That's ok, so L will be some factor that you can calculate times "m".

Can you say what this factor is?
 
  • #12
physicsgurl12 said:
could you just post how you would solve it. i think that would help me understand better.?

please read the forum rules

please attempt the question yourself

what is the angular momentum at closest approach (call the mass "m", and assume it's a point mass)?
 
  • #13
tiny-tim said:
please read the forum rules

please attempt the question yourself

what is the angular momentum at closest approach (call the mass "m", and assume it's a point mass)?

i can't attempt the problem if you can't tell me how to start it. We just started this chapter and its an honors class so there's not a lot of help. so mabey if you explain how to do like the first 2 steps i could go from there.
 
  • #14
I like Serena said:
That's ok, so L will be some factor that you can calculate times "m".

Can you say what this factor is?

honestly i don't know what that "factor" is. also I am super confused on how you CAN calculate times m.
 
  • #15
physicsgurl12 said:
i can't attempt the problem if you can't tell me how to start it. We just started this chapter and its an honors class so there's not a lot of help. so mabey if you explain how to do like the first 2 steps i could go from there.

We are really trying to help you here. Please believe me.
We are trying to explain how to do the first step.

The first step is taking the formula

L = m \times v \times r

and replacing as many symbols as you can by the numbers you know.
In my previous post I've already written what these numbers are.

[edit] "factor" is the word for a number with which you multiply. In this formula for instance "v" is a factor. [/edit]
 
  • #16
physicsgurl12 said:
could you just post how you would solve it. i think that would help me understand better.?

As the other posters have tried to explain, that's not how Homework Help works here at the PF. You have been given some great turorial help so far -- please try to use it and start writing out some equations to see where you get...

You should also consider doing a bit of reading at websites like wikipedia.org to get more background information. You could try search terms like orbital mechanics, for example :wink:
 
  • #17
I like Serena said:
We are really trying to help you here. Please believe me.
We are trying to explain how to do the first step.

The first step is taking the formula

L = m \times v \times r

and replacing as many symbols as you can by the numbers you know.
In my previous post I've already written what these numbers are.

[edit] "factor" is the word for a number with which you multiply. In this formula for instance "v" is a factor. [/edit]

okay well i understand that i can fill in r with 88700000000 and v with 53430 but i don't understand what you want me to do from here. arnt we looking for l? and i don't know the mass so i can't fill that in either?
 
  • #18
physicsgurl12 said:
okay well i understand that i can fill in r with 88700000000 and v with 53430 but i don't understand what you want me to do from here. arnt we looking for l? and i don't know the mass so i can't fill that in either?

Does looking over the wikipedia page that I suggested help your understanding?
 
  • #19
physicsgurl12 said:
okay well i understand that i can fill in r with 88700000000 and v with 53430 but i don't understand what you want me to do from here. arnt we looking for l? and i don't know the mass so i can't fill that in either?

Could you please write that down in the formula?
And could you multiply the numbers for v and for r?

The next step is to do the same for the farthest distance.
 
  • #20
berkeman said:
As the other posters have tried to explain, that's not how Homework Help works here at the PF. You have been given some great turorial help so far -- please try to use it and start writing out some equations to see where you get...

You should also consider doing a bit of reading at websites like wikipedia.org to get more background information. You could try search terms like orbital mechanics, for example :wink:

i understand why you would say that's not how it works, but mabey you should think about how others not just yourself. what i mean is that everyone learns in a different way. some people learn by seeing something done, some are more hands on and some can just listen to something being explained and get it. but i am not one of those people. i myself am a visual person. also i have tryed wikipedia and the advanced vocabulary it uses to explain things ,like many sites, are hard to comprehend.
 
  • #21
I like Serena said:
Could you please write that down in the formula?
And could you multiply the numbers for v and for r?

The next step is to do the same for the farthest distance.

L=(4.739241*10^15)*m
so for the next step i do the exact same thing but change the r to 5614000000000??
 
  • #22
physicsgurl12 said:
i understand why you would say that's not how it works, but mabey you should think about how others not just yourself. what i mean is that everyone learns in a different way. some people learn by seeing something done, some are more hands on and some can just listen to something being explained and get it. but i am not one of those people. i myself am a visual person. also i have tryed wikipedia and the advanced vocabulary it uses to explain things ,like many sites, are hard to comprehend.

[Hijack]

I understand your point, but this thread in the Feedback forum should help you understand why we have the rules the way that we do:

https://www.physicsforums.com/showthread.php?t=373889

[/Hijack]
 
  • #23
physicsgurl12 said:
L=(4.739241*10^15)*m

Now L and m are constant, so what is v at the furthest point?
 
  • #24
berkeman said:
[Hijack]

I understand your point, but this thread in the Feedback forum should help you understand why we have the rules the way that we do:

https://www.physicsforums.com/showthread.php?t=373889

[/Hijack]

i totally get the rules you have but this perticular assignment is online, everyones problems are different and we had a sub today so we couldn't ask questions on how to do any of the worksheet. mabey, just mabey you should trust people a bit more. i mean the kids who are taking physics especially when its honors or not a required class most of those kids arnt going to be the ones are going to cheat.
 
  • #25
physicsgurl12 said:
L=(4.739241*10^15)*m

That's fine! :smile:

physicsgurl12 said:
so for the next step i do the exact same thing but change the r to 5614000000000??

Yes and no.
Yes, you need to do the exact same with r is 5614000000000, but you do not know the speed v at the farthest distance yet, so you need to leave v as it is.

Could you write that down again please?

The next step is to say that the two results for L are equal to each other.
 
  • #26
tiny-tim said:
Now L and m are constant, so what is v at the furthest point?

isnt it 5614000000000?
 
  • #27
I like Serena said:
That's fine! :smile:



Yes and no.
Yes, you need to do the exact same with r is 5614000000000, but you do not know the speed v at the farthest distance yet, so you need to leave v as it is.

Could you write that down again please?

The next step is to say that the two results for L are equal to each other.

so L=m*(4.739241*10^15) and L=mv5614000000000
then m*94.7392441*10^15)=mv5614000000000 tthen m cancles out and then you get v=844.1825793 m/s?
 
  • #28
physicsgurl12 said:
so L=m*(4.739241*10^15) and L=mv5614000000000
then m*94.7392441*10^15)=mv5614000000000 tthen m cancles out and then you get v=844.1825793 m/s?

Yep! You've got it! :smile:
 
  • #29
I like Serena said:
Yep! You've got it! :smile:

i entered it and it said it was correct. thanks so much. now i understand what you were trying to say earlier.
 
  • #30
physicsgurl12 said:
i entered it and it said it was correct. thanks so much. now i understand what you were trying to say earlier.

You're welcome. :)
I'm glad we got it cleared up.
 

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