# Angular momentum, Spin and satelites

1. May 11, 2009

### Mahbod|Druid

Hello

we know that satelites turn around a star in ellipse path (or parabola/hyperbola)
I wanted to know where does the energy go when a satelite comes closer to a Star

we have angular momentum for the satelite itself and another angular momentum for spin of satelite

and we know that angular momentum = Constant
because f(r) = k , f($$\theta$$)=0

but if we calculate the velocity of satelite at diffrent distances we realize that angular momentum will give one answer which says that some Energy goes somewhere but where?

changes to spin ? ( Iw2/2 )
but angular momentum of spin must be constant too ?

Thanks

2. May 11, 2009

### Bob S

Review Kepler's second law: A line drawn from the Sun to the Earth describes equal areas in equal times.

3. May 11, 2009

### Nabeshin

Potential energy.

4. May 12, 2009

### Mahbod|Druid

but Kepler's law is equal to Angular momentum

and when a satelite comes closer to star potential energy turns into kinetic energy

satelites comes closer to star very tiny distance in circular orbits :

from angular momentum we have :
Iw1 = Iw2 => mR21w1 = mR22w2
R2 = R1 - dR
so:
R21w1 = R21w2 - 2R1dRw2

and : w2 = w1 $$\frac{R1}{R1 - 2dR}$$

and because of circular orbits : W*R = Velocity

w2 R2 = V2 = w1(R1 - dR) $$\frac{R1}{R1 - 2dR}$$
there we calculated Velocity by momentum

*******************************************************************
now lets calculate it with Energy : (dR is very small so we can get GM/R2 constant=Q)
m = mass of satelite
mQdR = $$\Delta$$ Kinetic energy
V2 = $$\sqrt{V^2 + 2QdR}$$ = V1 + QdR/V1

and here we calculated velocity from work done by Gravity field
and we see 2nd calculation depens on the mass of Star "Q = GM/R2)
calculation is not 100% correct but it will depend on mass of central star anyway ...

so where does the energy go ? into Spin ? then why is not angular momentum of satelite constant

Last edited: May 12, 2009
5. May 12, 2009

### D H

Staff Emeritus
You made several mistakes in the above. Some of them are
1. Assuming $v=\omega r$
2. Ignoring the change in potential energy
3. Using the wrong formula for potential energy.

You were told in post 3 that the change in kinetic energy is balanced by a change in potential energy. You cannot assume that the potential energy remains constant.

One way to express conservation of energy with orbits is with the vis-viva equation,

$$v^2 = GM_{\text{sun}}\left(\frac 2 r - \frac 1 a\right)[/itex] 6. May 12, 2009 ### Mahbod|Druid 1. Assuming $v=\omega r$ when we imagine orbits are Circular this can be right 2. Ignoring the change in potential energy mQdR = value of changed ptential when we throw a ball from height H1 and we want to calculate the speed at Height H2 while H2 is not far from H1 like 1000 meter we calculate like : mg(H2 - H1) = -(mV2^2 - mV1^2)/2 where g is constant = 9.8 i named Q as g where dR --> 0 (i know that we assume R --> 1000000000 , potential = 0 and as we come closer to star potential gets more minus but for easier caculation we can use mg*delta(H) ) 3. Using the wrong formula for potential energy. still we can say GMsun/R^2 = GMsun/(R + dR)^2 "You were told in post 3 that the change in kinetic energy is balanced by a change in potential energy" as i have calculated above all the change in potential energy doesnt turn in to Velocity Angular momentum does NOT depend on Mass of sun but potential energy does ... [tex]v^2 = GM_{\text{sun}}\left(\frac 2 r - \frac 1 a\right)$$

i have no idea what "a" is :-S

however gravity potential is not like a invisible spring it just depens the distance from Sun
so all changes in potential turns in to Work
so where does the energy go ? since Angular momentum = constant ?

7. May 12, 2009

### D H

Staff Emeritus
You are investigating a problem where the radial distance changes. You cannot assume circular orbit conditions for a non-circular orbit.

The correct equation for potential energy is $P=-\,GM/R$. You are using the equation for acceleration.

This is very telling. "a" is the semi-major axis of the orbit.

Rather than coming in, guns a blazing, claiming "energy is not conserved" you would have better served yourself by posting your analysis and asking "what have I done wrong?"

You have been told several times now that energy is conserved. One last time: Changes in kinetic energy are balanced by opposite changes in potential energy.

8. May 12, 2009

### Mahbod|Druid

You are investigating a problem where the radial distance changes. You cannot assume circular orbit conditions for a non-circular orbit.

Aha i think i got it ... !
this caused the problem assuming satelite changes its orbit without a close path caused the problem

if we assume ellipsal orbit all things will be right ...

thanks :D