# My orbitz problem

1. Jul 18, 2008

### atom888

Assume we have a sun-plannet system in a perfect, stable orbit.

The gravitation force cancel out with the centrifugal force which equates into:

Gm1m2/r^2 = m2v^2/r where m1= mass of the sun, m2 = mass of planet

simplify the equation above I get: Gm1/r = v^2

This implies that if we move the plannet faster in tangent direction, the distance between 2 bodies must reduce to achieve a stable orbit. The opposite is true. Slowing down the planet will have to move further away from the sun to attain stable orbit.

Let's consider kepler's second law and conservation of momentum. Basically, the law said that when the planet is half radius of a reference position, the velocity increase with the factor of 2. But according the the equation above, the velocity should increase with the factor of 4. Who can shed some light into this?

2. Jul 18, 2008

### tiny-tim

Hi atom888!
Not following you …
ah … Kepler's second law and conservation of momentum apply to the same orbit.

They don't apply if you start shoving extra energy in!

If a planet is in circular orbit with radius R, and you shove it backward a bit, it will go into an elliptical orbit, and its speed will indeed get more as it comes further in. Its speed at perihelion, at radius r say, will be proportional to R/r (and not to the square).

But the speed of a planet in circular orbit with radius r will be (R/r)2 times the speed of a planet in circular orbit with radius R.

3. Jul 18, 2008

### atom888

Hi tiny-tim,

What you're saying is circular orbits obey my equation and an elliptical orbit got nothing to do with this equation, in another world, obey Kepler's 2nd law and conserve momentum?

The part make you confused was that suppose we want to change the planet into another stable circular orbit, say futher out. We have to 2 ways to do this:
1) push it outward and decrease its velocity.
2) increase its velocity and decrease its velocity much more when it reach outter orbit.

In reality, it's almost impossible to move rotating planets due to it's gyro effect. I believe that's why planets rotates (to protect itself from external force thus changing orbits). Of course that somehow implies our moon is an artificial satellite.

Oh anothe thing. Which plannet in our solar system have elliptical orbit? As I record... none.

4. Jul 19, 2008

### Astronuc

Staff Emeritus
tiny-tim is correct.

The gravitation force cancel out with the centrifugal force which equates into: Gm1m2/r^2 = m2v^2/r is a force balance, which maintains that the centripetal force is equal and opposite to the gravitational force. This is a static situation.

If one wishes to change an orbit, one must apply a force to work such that the gravitational potential energy changes. Applying a force means the system is not in equilibrium, but rather it is perturbed. Then we perturb the system again to obtain a circular orbit.

Planetary orbits are almost circular, but not quite. A circular orbit has an eccentricity, e = 0. Planetary orbits have nonzero eccentricity. Please refer to this reference:

http://www.windows.ucar.edu/tour/li...cs/mechanics/orbit/eccentricity.html&edu=high

As for extrasolar planets - See Masses and Orbital Characteristics of Extrasolar Planets
http://exoplanets.org/ecc.html
http://exoplanets.org/planet_table.shtml

Last edited: Jul 19, 2008
5. Jul 19, 2008

### Janus

Staff Emeritus
Yes, your equation only works for the special case of a circular orbit. The full form that works for both circular and elliptical orbits is:
$$v= \sqrt{GM \left ( \frac{2}{r}- \frac{1}{a} \right )}$$

Where "a" is the semi-major axis of the orbit, or the average orbital distance.
Note for a circular orbit a=r and it reduces to the simple form you used.
For number 2, you increase its velocity to place it into the elliptical orbit which has its highest point at a distance equal to the radius of the new orbit you want. Then you increase its velocity to place it into a circular orbit. You need to increase the velocity because the object will lose velocity as it climbs away from the body that it is orbiting, and will being moving slower than the velocity needed to hold a circular orbit at the new altitude when it reaches it.
The gyroscopic effect of a rotating planet causes no resistance to changes in the planet's orbit. It causes resistance to changes in the direction of the axis of rotation.
As pointed out by Astronuc, all of them, to varying degrees.

For instance, due to the Earth's elliptical orbit, its orbital velocity varies from 29.5 km/sec to 30.4 km/sec.

6. Jul 19, 2008

### atom888

Now let's consider an over-exaggerated elliptical orbit so that from a far away point of view, it looks almost like linear. The motion one will see then is a bouncing object. How then one descibe the forces in this picture?

7. Jul 19, 2008

### tiny-tim

Hi Janus!
ooh … I've not seen that formula before …

I like that!
Hi atom888!

It'll be one-dimensional:

r'' = -GM/r2

so v2 = r'2 = 2GM(1/r- 1/rmax ).

8. Jul 19, 2008

### D H

Staff Emeritus
I like that equation, too. It is called the vis-viva equation.

9. Jul 19, 2008

### atom888

welcome back tiny-tim,

The cool part is we can interprete this picture as a pure elastic collision except the point of collision is not a physical collision. I'm trying to make this a model in microscopic scale. In theory sub atomic particle is so small that this model might suits well. I'll get back more on this.