# Rotation of satellite in geosynchronous orbit

1. Dec 2, 2007

### ~christina~

1. The problem statement, all variables and given/known data

One scheduled activities during a shuttle misison was the launching of a communications satellite. This 1220kg satellite Geo II is a uniform cylinder of diameter d= 1.18m and length L= 1.72m. It is identical in mass, density, and shape to Geo I, which is already in geosynchronous orbit around the earth. Prior to launching, a motor inside the shuttle bay takes one minute to set the satellite spinning from rest to 1.46 rev/s about the cylinder's axis. At this instance the spinning satellite is released from the bay compartment and placed in the same orbit as Geo I. GeoI has zero moment of inertia.

a) what force must the motor exert on the satellite to obtain this angular speed?

b) Before it's release, what is the magnitude of the linear velocity of a point on the curved surface of Geo II ?

c) what is the anglular displacement of Geo II before it's release?

d) Do the 2 satellites have the same total energy once Geo II is in orbit?
If yes, use physics principles to explain why. If not calculate the difference in total energies.

e) what are the linear and angular momenta of Geo I as ikt orbits about the Earth's center?

f) What is the gravitational potential energy of Geo I in this orgit?

[NOTE: treat the earth and the satellites as isolated bodies. Disregard frictional or drag forces]

http://img145.imageshack.us/img145/8650/38926826zx6.th.jpg [Broken]

2. Relevant equations
$$\sum W = 1/2 I \omega_f^2 - 1/2 I\omega_i^2$$

$$s= r \theta$$

$$v= r\omega$$

$$a_t = r \alpha$$

3. The attempt at a solution

I know:

$$m_2= 1220kg$$

diameter= 1.18m

Length= 1.72m

$$\omega_i = 0 rad/s$$

$$\omega_f= 1.46 rev/s$$

I guess I'd have to convert the angular acceleration first to rad/s

$$\omega_f= 1.46 rev/s$$

$$1.46 rev/s (2\pi rad/ 1 rev ) = 9.17 rad/s$$

$$\omega_f= 9.17 rad/s$$

a) what force must the motor exert on satellite to obtain this angular speed?
state assumptions...

well I was thinking of using this equation but I don't have the distance traveled unless I guess I consider that it is rad/s

$$\sum W = 1/2 I \omega_f^2 - 1/2 I\omega_i^2$$

and I don't have I either...

I need help on this.

Thank you

Last edited by a moderator: May 3, 2017
2. Dec 2, 2007

### ~christina~

Can anyone help me with this problem???
(pretty lost here)

3. Dec 3, 2007

### ~christina~

does anyone know how to do this question??? or is it supposed to go in advance physics??? (I don't think so but...)

desperate for help now...I actually started to do the question so It's not like didn't do any work.

4. Dec 3, 2007

### andrevdh

(a) To give the satellite an angular acceleration one need to apply a torque to it

$$\tau = I\ \alpha$$

where the angular acceleration assumed to be constant due to applying a constant torque (force). The angular acceleration can be evaluated along the lines you started:

$$\alpha \ = \ \frac{\Delta \omega}{\Delta t}$$

giving

$$\frac{(2 \pi 1.46) - 0}{60}$$

radians per second per second. Next you need to determine the moment of inertia of the satellite:

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html" [Broken]

Last edited by a moderator: May 3, 2017
5. Dec 5, 2007

### ~christina~

Hm..I got the angular acceleration

and after that I need to get the moment of inertia.

However it's different for different objects.

I'm not sure whether this is a solid cylinder or a hollow one.

I looked it up on hyper physics and It showed that for a solid cylinder, about it's axis the moment of inertia (I), was I= 1/2Mr^2 ..

looking at this page it showed that I would need the volume of the cylinder if it was hollow to find the density and use that but I don't think I'm given that information so should I assume it is solid? (it would make it much simpler)
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

Thanks andrevdh

Last edited by a moderator: May 3, 2017
6. Dec 5, 2007

### Staff: Mentor

I would assume by "uniform cylinder" that it's a solid cylinder, not a cylindrical shell.

7. Dec 5, 2007

### ~christina~

okay, then I assumed correctly Doc Al.

however I'm having a issue with whether both the satellites have the same energy when the Geo II is released.

I would think that linear velocity of satellites to be different. However I do not understand quite what they mean by 0 moment of inertia at that exact moment that Geo II is released.

I know that moment of inertia is $$\tau = I \alpha$$
but that would only apply I would think to the angular acceleration and torque on Geo I as it is already in orbit.

I would think Geo I has no torque because it has nothing to make it move but in space there are no forces to make it stop rotating either...

basically confused here.

Last edited: Dec 5, 2007
8. Dec 5, 2007

### ~christina~

d) Do the two satellites hav ethe same total energy once Geo II is in orbit? if yes use physics principle sot explain why. if no calculate difference in total E

I have 2 things that I would think that's how I'd find the energy

a)thinking I would find the energy change for Geo II from work done by external forces since...my text states..
"work done by external forces equals the change in the rotational E of the object"

so.. $$W= I\omega_f^2 - 1/2 \omega_i^2$$

b)or would it would be kinetic rotational E?

$$K_r = 1/2 I\omega^2$$

my text states this "an object rotating about a fixed axis remains stationary in space so there is no kinetic energy associated with translational motion. The individual particles making up the rotating object however, are moving through space; they follow circular paths. Consequently there is kinetic energy associated ith rotational motion"

and thus
What energy are they refering to??

Thanks

Last edited: Dec 5, 2007
9. Dec 5, 2007

### Staff: Mentor

If you are referring to the linear speed asked for in part b, that's just its surface speed with respect to its center.

That makes no sense to me either.

Did you solve part a yet?

10. Dec 5, 2007

### Staff: Mentor

The energy referred to in part d would be the sum of gravitational PE, linear KE, and rotational KE.

11. Dec 5, 2007

### ~christina~

part a? yes actually, I just didn't post it but I'll show what I did then.

a) force motor exerts on a satellite to obtain angular speed

$$\alpha = \Delta \omega / \Delta t$$

$$\alpha = (9.17rad/s - 0 rad/s) / 60s = .1528s^-2$$

[tex\I= 1/2 Mr^2[/tex]

$$1/2(1220kg)(.59m)^2 = 212 kg*m^2$$

b) linear v on point on curved surface of GeoII before release

$$v= r\omega$$

c) Angular displacement before release

$$\Delta\theta= 1/2(\omega_i + \omega_f)t$$

$$\Delta\theta = 1/2(0 + 9.17 rad/s)(60s)$$

$$274.1rad(57.3^o/ rad)= 15,763.23^o$$

This is what I did so far untill I got stuck at d)

I looked up information on a satellite orbiting earth and it seems that the velocity changes as the satellite comes closer to the earth and is slower when it is far from the earth b/c of it's elliptical orbit. However it also seems that the
rule KEi + PEi + Wext = KEf + PEf applies here but the Wext = 0 b/c gravity is internalized.

This I guess would apply to Geo I in orbit. For Geo II I'm not so sure since Wext in the equation wouldn't equal 0 since it had a force to spin it first before putting it into orbit but the question states " once Geo II is in orbit". I'm not sure about what would happen to the satellite when it is set into orbit.

Would it have the same energy or not? (I'm not sure how long it takes for the satellite just launched to have the conditions of a satellite that has been orbiting the earth for awhile)

Thanks

12. Dec 5, 2007

### Staff: Mentor

Keep in mind that the two satellites are the same size and have the same orbit (so same speed and distance from earth). What's the only thing different about Geo II?

13. Dec 5, 2007

### ~christina~

I'm not sure..I would say the work that has been done on it by the motor?

Oh and I was also curious to know whether what I did for a)- c) was incorrect since you didn't say whether my method was correct or not.

Thank You

14. Dec 6, 2007

### andrevdh

The 0.153 rad/s^-2 is the angular acceleration, $$\alpha$$, of the satellite due to the applied force to get it spinning before release into its orbit by the shuttle. I am not sure as to why the satellite is spun, maybe to provide stability in its orientation - like a bullet (a much lower rotational velocity is required to keep the same side pointing to earth while it is in orbit)? The 212 kg.m^2 is its rotational inertia, $$I$$, it is a measure of how much the satellite resists angular acceleration when a torque is applied to it, very much like the mass of an object appears in Newton's second law. The torque that the applied force needs to apply, $$\tau$$, can now be calculated from

$$\tau = I \alpha$$

to get the force that is needed for such a torque (to effect the spinning of the satellite) one need to decide at what distance from the axis the force need to be applied, that is the torque is provide by a force acting at a distance r from the axis. We therefore assume that the force is applied along the rim of the satellite and the force is acting tangentially to the rim:

$$\tau = Fr$$

see

http://www.saburchill.com/physics/chapters/0021.html" [Broken]

Once the satellite is released (and the torque is not applied any more) the satellite will keep on spinning about its axis at a constant rotational velocity since there is very little out there in outer space to rob it from its rotational kinetic energy.

Last edited by a moderator: May 3, 2017
15. Dec 6, 2007

### Staff: Mentor

Right. I guess that you're supposed to assume that Geo I is not spinning, so that's the only difference between them.

For a) you found the acceleration, but not the force. As andrevdh explains, you must find the torque and then the force.

Parts b & c are fine. (I do get a slightly different number for part c--I think you have a typo in the third digit of your radians.)

16. Dec 6, 2007

### ~christina~

Oh...yes I did do that but forgot to go and in my rush to leave for school I went and forgot to type that but I got :

$$\sum \tau= I\alpha$$

$$\sum\tau= (212kg*m^2)(.1528s^-2)= 32.3936 N*m$$

Hm..I didn't do this though.

doing that

$$\tau = Fr$$

$$F= \tau/r$$

$$F= 32.3936 N*m/ .59m = 54.904 N$$

Oh.. that was what I was curious about. So do I assume as well that Geo II eventually stops spinning?

so assuming it is not spining then from what you said before (quoted below)

then since they have the same orbit then the speed that you're refering to is the speed at which they are orbiting the earth right?

then GeoII has angular velocity but Geo I has none.

This may be wrong but I know that.
The change in the kinetic energy of an object is equal to the net work done on the object.
And if there is no velocity then => KEi + PEi + Wext = KEf + PEf only depend on the
work done on the object like I had deduced before.

and based on the work energy principle for rotational motion the change in rotational kinetic energy is equal to the net work:

$$W_{net}= 1/2 I\omega_f^2 - 1/2 I\omega_i^2$$

since for Geo I it would = 0 since I= 0 and $$\omega= 0$$ since it isn't rotating then totally the $$W_{net}= 0$$ as well.

For Geo II if I'm right..

$$W_{net}= 1/2 I\omega_f^2 - 1/2 I\omega_i^2$$

$$W_{net}= 1/2 I\omega_f^2 - 0$$

it's 0 since it was at rest.

$$W_{net}= 1/2(212kg*m^2)(9.17rad/s^2)- 0$$

$$W_{net}= 8,913.423J$$ ==> and since $$W_{net}= 0J$$ for Geo I

the difference would just give me...E= 8,913.423J as the energy difference.

Is this right?

Yes I did that above. And after checking part c), the numbers are funny and corrected it is:

c) Angular displacement before release

$$\Delta\theta= 1/2(\omega_i + \omega_f)t$$

$$\Delta\theta = 1/2(0 + 9.17 rad/s)(60s)$$

$$274.1rad(57.3^o/ rad)= 15,705.93^o$$

Thank You

Last edited by a moderator: May 3, 2017
17. Dec 7, 2007

### ~christina~

since nobody answered I shall move to the rest of the question.

e) what are the linear and angular momenta of Geo I as it orbits about the Earth's center?

well since the satellite of Geo I is not rotating. (and moment of inertia is 0)

$$L= I \omega$$ and since the Geo I is not rotating..it's angular acceleration is = 0
and since moment of inertia is 0 as well as stated by the problem then L= 0.

However I'm not sure about how I find the linear momentum of Geo I as it orbits around Earth's center.

f) What is the gravitational potential energy of Geo I in this orbit?
How would I find this if I do not know how high up the GeoI is in the orbit?
(it's not given)

$$PE_{gravitational} = mgh$$

Thank you.

18. Dec 8, 2007

### Staff: Mentor

Looks good.

No reason to assume that. Once you set the satellite in orbit, there'd be nothing exerting a torque on it to stop the spinning.

Right. If they're in the same orbit, they must have the same orbital speed.

Right.

Yes, but keep it simple. You know Geo II is rotating, so its rotational KE must be $1/2 I \omega^2$. Since Geo I is not rotating (presumably), that's the only difference between them.

Good.

19. Dec 8, 2007

### Staff: Mentor

parts e & f

In order to tackle parts e and f, you must figure out the orbital speed and orbital radius of the satellite. The key piece of information is that it is in a geosynchronous orbit: That means it appears to just sit there in the sky as the earth rotates. So what must be its period for one orbit?

Once you have the period, you can use Newton's 2nd law (for circular motion) and gravity to figure out the orbital radius and speed.

That bit about the moment of inertia being zero is nonsensical. Someone else brought it up elsewhere, see my comment here: https://www.physicsforums.com/showthread.php?t=203162

Careful! While it's true that the angular momentum about its axis is zero, it's still orbiting the earth and will have angular momentum with respect to the earth. (Hint: First find the linear momentum.)

This is an orbital mechanics/circular motion problem. See my comments above.

First you have to figure out the radius of its orbit around the earth. Then use the appropriate formula for gravitational PE. But not mgh--that's only good close to the earth's surface.

20. Dec 8, 2007

### ~christina~

I don't have the time to go through this now since I have to study for a final but I will get back to this on tuesday or wednesday after that test and I'll get back to this then.

Thanks Doc Al