# Homework Help: Gravity, Kepler's Laws

1. Apr 19, 2005

### Felix83

Use the following data to calculate the speed vp of comet Halley at perihelion. Comet Halley orbits the Sun with a period of 76 years and in 1986 had a distance of closest approach of 8.90×1010 m and an aphelion distance of 5.30×1012 m.

I'm stumped so far on this one. I'm assuming the closest approach distance is useless info thrown in there. I know I could use keplers law of periods to find what the perihelion distance is, and I know that angular momentum is constant, so I could find the velocity at aphelion if I knew vp, but I can't figure out how to get vp.

2. Apr 19, 2005

### James R

Total energy (gravitational potential + kinetic) is also conserved, since only gravity is acting on the comet.

3. Apr 19, 2005

### Felix83

so how would you find vp?

4. Apr 19, 2005

### Janus

Staff Emeritus
I'll toss you a bone here. The closest approach distance is the perhelion.

5. Apr 20, 2005

### Felix83

oh ok, but still, even if i know both distances, i dont see how i would solve for vp

6. Apr 20, 2005

### OlderDan

7. Apr 20, 2005

### Janus

Staff Emeritus
There are a number of ways to get this answer, here's one:

Use the the two distances given to find the semi-major axis of the orbit (the average orbital distance.)

With this and the given period of the orbit you can get the orbital velocity at that distance.

This in turn gives you the angular momentum of the comet.

Find the velocity at the perhelion distance which gives you the same angular momentum.

8. Apr 20, 2005

### Felix83

Using conservation of angular momentum relative to the Sun, find the ratio of the comet's perihelion distance Rp to its aphelion distance Ra from the values of vp and va.

since distance of closest approach - perihelion distance and the aphelion distance are given, you would just divide the two to get the ratio. so it would be either 0.0168 or 59.5. however, the computer says both of these are wrong...?

9. Apr 20, 2005

### Felix83

ok using the formula v= (GM(2/ra - 1/a))^.5 i can find va if i calculate a, since i already know ra.

I used kepler's law of periods to calculate a=2.68*10^12m

I plugged that back into the formula along with ra to calculate
va=752m/s , which the computer says is wrong

do you guys get the same numbers when you calculate this?

10. Apr 20, 2005

### Felix83

ok, there are 3 parts to the problem:
1) vp=?
2) va=?
3) rp/ra=?

once you have vp, I could easily find the other 2. i was surfing around the internet and happened to find that vp=54200m/s, which i plugged in and it was correct. then i calculated that va=963m/s and that rp/ra=0.0178 which were both also correct.

then I decided to try v= (GM(2/r - 1/a))^.5 for vp, and sure enough i got 54200m/s. the reason va, and rp/ra were wrong before is that i tried to calculate them before vp, and the rounding changes the answers slightly, so you have to calculate the 3 parts in a specific order.

11. Apr 20, 2005

### Felix83

anyone have a proof of the forumula v= (GM(2/r - 1/a))^.5 ?