What Is the Speed of the Comet at a Different Distance?

AI Thread Summary
Comets move in elliptical orbits, and their speed varies with distance from the sun. The discussion centers on calculating a comet's speed at a different distance using conservation laws. Initially, angular momentum was considered, but it was noted that the positions given were not necessarily at aphelion or perihelion, complicating the use of this approach. Instead, the conservation of mechanical energy was suggested, leading to the specific mechanical energy equation, which remains constant throughout the orbit. Ultimately, the gravitational parameter is crucial for solving the problem, and the participant successfully found the correct speed.
Vanessa Avila
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Homework Statement



Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 2.2×10^4 m/s when at a distance of 2.6×10^11 m from the center of the sun, what is its speed when at a distance of 4.2×10^10 m

Homework Equations


L = rp = r(mv)

The Attempt at a Solution


I thought angular momentum is what I had to use for this one and thought of the two distances as perigee and apogee. I used conservation of momentum to attempt to solve the speed:
rmv = rmv
the m's cancel out
so I am left with r1v1 = r2v2
> 2.6 * 10^11m(2.210^4m/s)=4.2*10^10m(v2)
and I solved for the speed like that and got 136190. But that is wrong.
 
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Angular momentum would be a good approach if the two positions in question were aphelion and perihelion. Then the velocities would be perpendicular to the radii and you could apply conservation of angular momentum in a simple scalar form. But this problem does not say that the given positions are aphelion or perihelion, so there's no knowing what the angular relationship will be between the position and velocity vectors.

What other conservation law might you appeal to instead?
 
gneill said:
Angular momentum would be a good approach if the two positions in question were aphelion and perihelion. Then the velocities would be perpendicular to the radii and you could apply conservation of angular momentum in a simple scalar form. But this problem does not say that the given positions are aphelion or perihelion, so there's no knowing what the angular relationship will be between the position and velocity vectors.

What other conservation law might you appeal to instead?
Could conservation of mechanical energy work?
1/2mv2=GmM/r
 
Vanessa Avila said:
Could conservation of mechanical energy work?
Yes...
1/2mv2=GmM/r
...But not that way. The KE won't be equal to the PE (except for very specific circumstances that don't arise here).

What's the expression for the total mechanical energy of a body in orbit? Note that you can ignore the mass of the object itself and use what's known as the Specific Mechanical Energy. That's the energy per kg for the body in orbit.
 
gneill said:
Yes...

...But not that way. The KE won't be equal to the PE (except for very specific circumstances that don't arise here).

What's the expression for the total mechanical energy of a body in orbit? Note that you can ignore the mass of the object itself and use what's known as the Specific Mechanical Energy. That's the energy per kg for the body in orbit.
Would it be E=-1/2(GmM/r)
 
Vanessa Avila said:
Would it be E=-1/2(GmM/r)
That only accounts for the gravitational potential energy (actually half of the GPE). What's missing?
 
gneill said:
That only accounts for the gravitational potential energy (actually half of the GPE). What's missing?
woops.
So is it
E= 1/2mv^2 - GmM/r ?
 
Vanessa Avila said:
woops.
So is it
E= 1/2mv^2 - GmM/r ?
Yes, that's the total mechanical energy. As I mentioned previously, you can drop the "m" from the terms and it becomes the Specific Mechanical Energy which works just as well for these problems. You'll find that the "m" would eventually cancel out anyways if keep it.
 
Okay so then all we have to worry about is gnna be
1/2v^2 - GM/r ?
gneill said:
Yes, that's the total mechanical energy. As I mentioned previously, you can drop the "m" from the terms and it becomes the Specific Mechanical Energy which works just as well for these problems. You'll find that the "m" would eventually cancel out anyways if keep it.
 
  • #10
Vanessa Avila said:
Okay so then all we have to worry about is gnna be
1/2v^2 - GM/r ?
Yup. That will be the specific mechanical energy, which is a constant for the entire orbit.
 
  • #11
So how do i manipulate this to find the missing velocity given two radii and one velocity?
gneill said:
Yup. That will be the specific mechanical energy, which is a constant for the entire orbit.
 
  • #12
Vanessa Avila said:
So how do i manipulate this to find the missing velocity given two radii and one velocity?
The value that the equation delivers is a constant of the orbit. It applies everywhere along the trajectory.

Think in terms of conservation of energy problems that you've done for objects near the Earth's surface, where you equate initial and final energy sums.
 
  • #13
gneill said:
The value that the equation delivers is a constant of the orbit. It applies everywhere along the trajectory.

Think in terms of conservation of energy problems that you've done for objects near the Earth's surface, where you equate initial and final energy sums.
Oh okay so it's going to be like 1/2v^2 - GM/r = 1/2v^2 - GM/r?
 
  • #14
Yes, that's the idea.
 
  • #15
gneill said:
Yes, that's the idea.
But I do not have the M, so do i have to solve for that first by using another equation?
 
  • #16
Vanessa Avila said:
But I do not have the M, so do i have to solve for that first by using another equation?
Often the gravitational parameter μ = GM is a given value. Your course materials or textbook likely provide a value for it. If not μ then the mass of the Sun should be given.
 
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  • #17
Thank you so much for the help. I got the right answer finally! :)
gneill said:
Often the gravitational parameter μ = GM is a given value. Your course materials or textbook likely provide a value for it. If not μ then the mass of the Sun should be given.
 
  • #18
Vanessa Avila said:
Thank you so much for the help. I got the right answer finally! :)
I'm glad I could help!
 

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