Homework Help: Orbital Energy Conservation

1. Mar 10, 2012

Sekonda

Hey, here is the problem:

The minimum distance of a comet from the Sun is observed to be half the radius of the
Earth’s orbit (assumed circular) and its speed at that point is twice the orbital speed v(earth) of the Earth. The Earth’s and comet’s orbits are coplanar. Find the comet’s speed in terms of
v(earth) when it crosses the Earth’s orbit, and the angle at which the orbits cross.

I believe this is an energy conservation problem and I used the equation:

0.5mv^2 + (L^2)/(2mr^2) - GMm/r

as well as the fact the earths orbital velocity : v(earth) = vsin(θ)

Though I'm not sure if these equations are correct for this particular scenario,

Thanks for any help!
S

2. Mar 10, 2012

Staff: Mentor

Since angular momentum is conserved separately, you probably won't need to consider angular kinetic energy in the mechanical energy conservation relation. The specific mechanical energy ($\xi$) of the comet's orbit should come in handy. What does it tell you about the overall shape of the comet's orbit?

3. Mar 11, 2012

Sekonda

Thanks for the help, I think I have attained the velocity and the angle.

Well the shape of the orbit or trajectory depends on the energy of the system, I think the shape will be hyperbolic; though I think it depends on whether or not the kinetic energy is greater or equal to or less than the gravitational potential energy; though I'm not sure if there is anyway of working this out without the provided constants i.e. the velocity of the earth, mass of the sun, radius of earth's orbit and Gravitational constant.

Or is there another way of determining the shape of the orbit?

4. Mar 11, 2012

Staff: Mentor

You are given speed and radius at a location: v = Ve*2, r = AU/2, where Ve is the Earth's orbital speed (Assume a circular orbit. What's the formula for its speed?), and AU is the Earth's orbital radius. You can plug the various expressions into the total mechanical energy expression. You should be able to determine whether the result will be less than, equal to, or greater than zero.

5. Mar 14, 2012

Sekonda

And once again, I think I have solved this problem due to your help.

Thank you once again gneill!

6. Mar 14, 2012

Staff: Mentor

You're very welcome.