# Comet Elliptical Orbits Question

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1. Apr 21, 2013

### Chaso

1. The problem statement, all variables and given/known data
Comets move around the sun in very elliptical orbits. At its closet approach, in 1986, Comet Halley was 8.79 x 10^7 km from the sun and moving with a speed of 54.6 km/s.

What was the comet's speed when it crossed Neptune's orbit in 2006?

2. Relevant equations

Mv1r1=Mv2r2

3. The attempt at a solution

What I did was use the equation above, and solve for v2. But it says I'm wrong. Any Suggestions???

Last edited: Apr 21, 2013
2. Apr 21, 2013

### Simon Bridge

What did you use for r2?
What did you get for v2?

3. Apr 21, 2013

### Chaso

For r2, I use the radius from Neptune to the sun.

4. Apr 21, 2013

### Simon Bridge

Oh I see ... excuse me, didn;t read it properly.
That formula should be:

$\vec{r}_1\times m\vec{v}_1=\vec{r}_2\times m\vec{v}_2$

At Neptune's orbit, the velocity won't be tangential to the radius.

5. Apr 21, 2013

### Staff: Mentor

The formula you've chosen would apply when the velocities are both perpendicular to the radii, say at perihelion and at aphelion. Here this holds true for only one of the given points (the closest approach).

Instead, consider a conservation of energy approach.

6. Apr 21, 2013

### Chaso

So use K2 + U2 = K1 + U1

expansion to:
(1/2)(Mc)(v2^2) + -(G)(Me)(Mc)/(R) = (1/2)(Mc)(v1^2) + -(G)(Me)(Mc)/(R2)

R = Distance of neptune from sun
R2 = 8.79 x 10^7

Is this what I do?

Last edited: Apr 21, 2013
7. Apr 21, 2013

### Simon Bridge

Mc is the mass of the comet and Me is the mass of the body the comet is orbiting?

Note: another way thinking about it is "potential energy gained equals kinetic energy lost".