Guess at slowing of Earth spin

(Post prior to edit:
Bear with me. This will be tough without LaTeX support. I will edit this reply once I have the math worked out with the primitive tools available here.)


We have two objects to consider. I will denote the properties of the combined objects with a suffix c and the individual properties with suffixes s and e for spacecraft and earth

• w_c is the angular velocity of the combined Earth+spacecraft system prior to launch
• I_c is the inertia tensor of the combined Earth+spacecraft prior to launch
• L_c is the angular momentum of the combined Earth+spacecraft prior to launch: L_c = I_c w_c
• m_s and m_e are the masses of the spacecraft and Earth
• I_s and I_e are the inertia tensors of the spacecraft and Earth about their respective centers of mass
• L_s and L_e are the angular momenta of the spacecraft and Earth about their respective centers of mass
• x_s and x_e are the position vectors of the spacecraft and Earth centers of mass in the combined center of mass frame
• r_s and r_e are the magnitudes of the position vectors x_s and x_e
• v_s and v_e are the linear velocities of the spacecraft and Earth centers of mass in the combined center of mass frame
• L_s,v and L_e,v are the angular momenta of the spacecraft and Earth centers of mass moving in the combined center of mass frame
• w_s and w_e are the angular velocites of the spacecraft and Earth about their respective centers of mass after launch.

I will do my work in the nonrotating combined Earth+spacecraft center of mass frame, which I am assuming is an inertial frame.

State prior to launch
The position vectors and masses are related by
m_cx_c + m_ex_e = 0

The spacecraft and Earth center of mass velocities are
v_s = cross(w_c,x_s)
v_e = cross(w_c,x_e)

The combined inertia tensor is, by Steiner's parallel axis theorem,
I_c = I_s + m_s(Ir_s - x_s^Tx_s) + I_e + m_e(Ir_e - x_e^Tx_e)
where I is the 3x3 identity matrix.

The combined angular momentum is
L_c = I_c w_c
= I_sw_c + m_s(Ir_sw_c - x_s^Tx_sw_c) + I_ew_c + m_e(Ir_ew_c - x_e^Tx_ew_c)

The vector triple product helps simplify this ugly mess:
A x (B x C) = (A . C) B - (A . B) C
Denoting A=C=x_s, B=w_c, then
Ir_sw_c - x_s^Tx_sw_c = dot(x_s,x_s)w_c - x_sdot(x_s,w_c) = cross(x_s,cross(w_c,x_s))

Thus
L_c = I_sw_c + I_ew_c + m_scross(x_s,cross(w_c,x_s)) + m_ecross(x_e,cross(w_c,x_e))

The kinetic energy prior to launch is purely rotational:
E_c = 1/2*dot(L_c,w_c)

Conservation of linear momentum
Prior to launch, the combined Earth+spacecraft system has zero linear momentum in the combined CoM frame. Linear momentum will be conserved across the very brief (instantaneous) interval between pre-launch and post-launch, and thus the velocities remain unchanged:
v_s = cross(w_c,x_s)
v_e = cross(w_c,x_e)

Conservation of angular momentum
The angular momentum of the spacecraft and Earth after launch is the sum of the individual angular momenta about their centers of mass plus the angular momenta of the spacecraft and Earth masses about the combined center of mass:
L_post-launch = I_sw_s + I_ew_e + m_scross(x_s,v_s) + m_ecross(x_e,v_e)
= I_sw_s + I_ew_e + m_scross(x_s,cross(w_c,x_s)) + m_ecross(x_e,cross(w_c,x_e))

Equating the pre-launch and post-launch total angular momentum,
I_sw_s + I_ew_e + m_scross(x_s,cross(w_c,x_s)) + m_ecross(x_e,cross(w_c,x_e)) = I_sw_c + I_ew_c + m_scross(x_s,cross(w_c,x_s)) + m_ecross(x_e,cross(w_c,x_e))

or

I_sw_s + I_ew_e = I_sw_c + I_ew_c

Conservation of energy
The above equation indicates that any change in the rotation rate of the Earth must be compensated with a corresponding change in the rotation rate of the spacecraft. The final step, which is just too ugly to do without LaTeX, is to realize that the kinetic energy of the combined system does not change between the instant before and instant after launch. With this,
w_s = w_e = w_c

i.e., there is no change in the Earth's rotation rate due to launch.

 

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What about the rocket exhaust?

 

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Okey dokey.

 

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There are two effects. When the rocket first takes off, the plume from the exhaust impacts the Earth. The spacecraft launches vertically and the plume is thus directed through the Earth's center of masy. The pluming therefore changes the Earth's translational velocity by some amount, but does not change the rotation rate by much at all (torque = cross(r,F), and r is nearly normal to F).

After the spacecraft has taken off and turned to the East, the spacecraft exhaust does have some component normal to its position vector. This exhaust is moving faster than the atmosphere (in inertial space). The exhaust soon transfers this energy to the atmosphere, increasing the Earth's angular velocity (by a tiny, tiny amount).

Edited to add:
The exhaust is moving slower than the atmosphere until the rocket crosses the sound barrier (the ideal exhaust speed is the speed of sound). Once the rocket crosses the sound barrier, the exhaust is moving faster than the atmosphere inertially.

 

DH,

Thanks for the analysis. My understanding is this. Please correct me if I am wrong.

The fact that mass is moving away from the earth is exactly compensated for by the effect of the energy of the rocket exhaust being transferred to the atmosphere (indirectly to the earth). Yes? So, if we were to launch from the moon, with no atmosphere, simple conservation of angular momentum would dictate that the angular speed of the system must slow. Yes?

 

No. Once the masses are separated they do not interact rotationally (not quite true, but close enough). There is no change in angular velocity because of launch. The atmosphere is a secondary effect. Gravity gradient torque (something I did not discuss) is a secondary effect.

Remember that to conserve of angular momentum you must also consider that just after launch, the spacecraft and planet centers of mass have linear velocities that are normal to the spacecraft and planet position vectors relative to the combined center of mass. This adds to the angular momentum of the combined planet+spacecraft system. The math is a bit hairy, but the result is very simple. There is no change in angular velocity.

Think of the planet as a very large spacecraft. NASA knows quite well how docking and undocking change vehicle states; they have been doing this since the 1960s. While docking changes the angular velocity, undocking does not.

 

Mainly to D.H.:

I have only skimmed very briefly first part of your long post, and did not even read most of it, because I think you are concerned with the angular momentum of the Earth + satellite system about the sun. If this is the case, I agree with you that is will not change "one iota" by placing a satellite in orbit about the Earth.

We are concerned with the spin rate change of Earth alone, about its axis, when some angular momentum (originally part of the Earth's total) is removed from Earth by being placed into orbit.

I too have worked a little with satellites. I assume you know well the de-spin system often used, which has two small symetric weights tied at ends of cords, which are initially wrapped around the satellite (and held so until one wants to "de spin" satellite).

For benefit of those not so familiar with this cheap, fuel-less highly reliable de-spin system: When the weights are released, they move away from the body of the spinning satellite, but much more slowly because they are partially restrained by the force in the cords tying them to the circumference of the satellite. This force, times the radius of the satellite, provides a torque to de-spin the satellite (unless your technician wrapped them the wrong way around.

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Then as my space group would say: "you are in deep yogurt" with a more rapidly spinning satellite)

Now imagine that the iron core of the Earth is all that will remain of "Earth" and we throw all the rest of current Earth off into space in two symetric halves, (like the de-spin weights just discussed, but without any cords. For example, as if some "dark energy" condensed on the iron core and cancelled the gravity of the iron core on the higher layers so they just went flying off into space.) Most of the original angular momentum of current Earth would be found in the throw off crust parts and the iron core would be "de spun" just as a satellite can be.

Our case is not this extreme de-spining of "remaining Earth" case - just the opposite. Only a very small part of the Earth's angular momentum is removed from what is a reduced Earth Mass. Answer is made harder to guess as it is expresed in altitude change of the geostationaly orbit with a slower spinning "remainder Earth"

Answering you, has made me realize, consciously for the first time, that the moment of intertia of "Remainder Earth" has decreased as the 1000Kg of satellite was originally at full Earth radius - so inertia "I" of "remainder earth" will be very slightly less. This reduction of I tends to speed up Earth if it were the only thing happening.

This stimulates, in me, an possible new pole for guessing:

How does the current tidal slowing of Earth by moon compare to the slowing man has made by digging mines?

(All mined matterial assume to be now on the surface of a perfectly spherical Earth.) I will let someone else set that one up, if the want to do so.

D. H. - If you were concerned about our problem (and not the angular momentum about the Sun problem, where "not one iota" is correct) please say so, and I will fully read your long post.

 

DH and Billy,

Thanks very much. I am an engineer, but I deal largely with things like volts, ohms, bytes and code loops.

 

Billy T,

First, I did not mention the sun at all. In fact, I ignored that the sun even exists. I am concerned with the rotational and translation states of the Earth and satellite as observed in the non-rotating reference frame with origin at the Earth+satellite center of mass. To answer your final question, I am concerned with the problem posed in this thread.

Your de-spin example works because the weights are tethered to the satellite and because no external forces and torques are involved. The weights are still coupled to the spacecraft. Strings are attached!

This example does not apply to a spacecraft launch. The spacecraft, once it launches, is no longer connected to the Earth. The spacecraft exerts no torque on the Earth (well, almost no torque; I ignored gravity gradient torque exerted on the Earth by the spacecraft).

No strings are attached -- how could the spacecraft exert any torque? Because the spacecraft exerts negligible torque on the Earth, it doesn't matter where the spacecraft goes -- LEO, GEO, or Mars. The question then becomes how does spacecraft launch change the angular momentum of the Earth?

As I have shown, the Earth's rotation rate at the instant after launch is exactly the same as it was just before launch. Spacecraft launch does not change the Earth's rotation rate by one iota.

For your final example, suppose the Earth is quartered like an apple, with the core remaining intact, and the four quarters go flying off with the linear velocity w x r. The inner core would not be "de-spun" unless strings are attached.

 

I voted for the first option, because the rocket exhaust accelerates a section of air, which doesn't affect earth's rotation too much.

 

I guess my answer is for a satellite with a long string attached then.

 

Through the exhaust. The rocket pushes on the exhaust, the exhaust pushes on the atmosphere, the atmosphere pushes on the Earth.

Before the rocket reaches the exhaust speed, the exhaust is acting to slow the Earth through the atmosphere.

After reaching the exhaust speed, subsequent exhaust will act to speed up the Earth, but the total speed-up must be smaller than the total slow-down. Do you see why?

Think about the momentum in the Earth-atmosphere frame of: 1) slow exhaust 2) fast exhaust 3) the rocket. They must sum to zero, right?

 

I suspect that undocking would change vehicle states if the undocking ship fired it's boosters in the direction of the larger ship. However in the case of the rocket and the planet - it seems like gravity would counteract much of the force of the blast. Gravity is pulling the two objects together (but it should be said that the energy of attraction was greater at launch because of the closer proximity), chemestry is seperating the two. I think the net effect would be close to neutral depending how the smaller vehicle accelerated to excape velocity and if it accelerated to excape velocity. The planet may not excape a net effect if the rocket left very quickly in a straight line away from the planet. In other words, the effect of a rocket varries depending on how it leaves the planet - (it could rotate the planet picking up speed for example but that would change the rotational velocity of the planet).

But the point I wanted to make is that the pull of gravity is a substantial element and is pulling the planet toward the rocket ever so slightly as the rocket tries to excape.

 

Wait. Hold on. Is this true or not:

Conservation of angular momentum is defined by:

I_initW_init = I_finalW_final

Right?

So, the final angular velocity of anything is:

W_final = I_initW_init / I_final

Yes? So, since I is dependent on mass, if we pitch some mass away from the earth, normal to it's surface, it must spin faster. Right?

I think DH started out with this. I also think DH said that once the rocket leaves the surface it no longer has any effect whatsoever on the L of the earth.

Now, if a rocket starts out normal to the surface and stayed that way, the earth would spin up by the equation above. If the rocket starts normal, then slowly tips over high up in the atmosphere, the reaction of the exhaust (opposite to the earths rotation of course) will tend to slow the rotation of the earth, but the transfer of linear momentum must be incredibly small on the thin atmosphere.

 

No. Three things are wrong here. First, the pre-launch inertia tensor is a combination of the Earth inertia tensor, the satellite inertia tensor, and the separation of Earth and satellite centers of mass from the combined center of mass (Steiner parallel axis theorem).

Secondly, the system after launch comprises two objects. You have too account for the angular momentum from each. Both the Earth and spacecraft are rotating after launch.

Finally, you also have to account for the post-launch angular momentum that results from the Earth and satellite velocity components that are normal to their position vector with respect to the combined Earth+satellite center of mass.

 

Agreed. I edited my post on the effect of the exhaust to add that qualification.

I don't see that. So many factors come into play:

• Atmospheric pressure degrades rocket performance. The exhaust speed (relative to the rocket) is lower at high pressure that at vacuum. The exhaust velocity increases as the rocket gains altitude and gains speed. This argues for increasing the Earth's rotation rate.
• Escape velocity versus exhaust velocity. The exhaust velocity is ideally the speed of sound at the temperature and pressure at the bell of the rocket. This is on the order of 1000 mph, much lower than escape velocity. The rocket spends more time getting from the exhaust velocity to the escape velocity than it does in getting to the exhaust velocity in the first place. This argues for increasing the Earth's rotation rate.
• Atmospheric friction. The spacecraft is drilling a hole through the atmosphere. This argues for increasing the Earth's rotation rate.
• The rocket equation. It takes a lot of energy just to get the rocket off the launch pad. The acceleration increases as the rocket depletes mass. This argues for decreasing the Earth's rotation rate.

In any case, the total effect is tiny.

 

Well, we can certainly agree on that much.

 

Just one question:

Isn't this force canceling out some of the force of the launch? I mean if you fired a projectile from the earth the resistence from the atmosphere would keep that energy in a closed system?

But I guess that the atmosphere is a barrior and if the rocket was pushing against it, that could affect rotation velocities. But the gas that was accelerated out of the rocket has an equal and oppisite effect doesn't it?

 

This isn't a closed system. The rocket engines are firing.

 

Can we agree that if a rocket was up in the atmosphere at a 70 degree angle or so, and it encountered a barrior, and was pushing and pushing against it - the net effect would be 0. The particles from the engine effect the earth when they or particles they have influenced hit the earth. Can we agree on that?