# Angular momentum, Spin and satelites

• Mahbod|Druid
In summary, Kepler's law states that the area of a circle described by a line from the sun to the earth is equal in every period of time. Some energy is lost during the orbit, but the total amount of energy remains the same.
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

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

Mahbod|Druid said:
some Energy goes somewhere but where?

Potential energy.

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

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 let's 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:
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,

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

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 into Work
so where does the energy go ? since Angular momentum = constant ?

Mahbod|Druid said:
1. Assuming $v=\omega r$
when we imagine orbits are Circular this can be right
You are investigating a problem where the radial distance changes. You cannot assume circular orbit conditions for a non-circular orbit.

3. Using the wrong formula for potential energy.
still we can say GMsun/R^2 = GMsun/(R + dR)^2
The correct equation for potential energy is $P=-\,GM/R$. You are using the equation for acceleration.

$$v^2 = GM_{\text{sun}}\left(\frac 2 r - \frac 1 a\right)$$

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

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

## 1. What is angular momentum?

Angular momentum is a physical quantity that describes the rotational motion of an object or system. It is the product of an object's moment of inertia and its angular velocity, and is conserved in the absence of external torque.

## 2. How is angular momentum related to spin?

In quantum mechanics, spin is a property of particles that describes their intrinsic angular momentum. It is related to the angular momentum of an object in classical mechanics, but it is a fundamental property of particles and cannot be fully explained by classical concepts.

## 3. How do satellites use angular momentum?

Satellites in orbit around the Earth use angular momentum to maintain their position and velocity. This is known as orbital angular momentum, and it is used to keep the satellite in a stable orbit without the need for constant propulsion.

## 4. Can angular momentum be changed?

Yes, angular momentum can be changed by applying an external torque to an object or system. This can cause a change in its rotational motion and can also affect the spin of particles.

## 5. How is angular momentum measured?

Angular momentum is measured in units of kilogram meters squared per second (kg m^2/s) in the International System of Units (SI). It can be calculated by multiplying an object's moment of inertia by its angular velocity, or by measuring the rotational motion of an object using specialized equipment.

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