Funny Drag: Newsletter Challenge (05/04/10)

The drag acts tangentially, decreasing that component of the velocity; but the downward diversion is accompanied by gravitational acceleration, increasing the radial component. Another way to look at it is that in equilibrium, orbital velocity increases as altitude decreases.

Well, I do know that a lower altitude in orbit requires a faster orbital velocity.

No orbit is perfectly circular, so there is always a perigee and a apogee to the orbit.

Atmospheric drag on a satellite at perigee lowers the altitude of the apogee because it actually lowers the orbital velocity at perigee. This causes the orbit to become more and more circular, until the entire orbit is at the perigee altitude, and the satellite soon falls from orbit, which is where the orbital speed increases as its orbit continues to decay.

"Well, I do know that a lower altitude in orbit requires a faster orbital velocity."

Doesn't this apply to cars on the road, whenever they fall into a pothole?

Yup, and the damage will go up with the cube of that velocity, or so it seems!

-J

"No orbit is perfectly circular, so there is always a perigee and a apogee to the orbit."

What about geostationary orbits? It would seem to me that perogee and apogee would be the same and we would see no change in altitude or position. I realize that would be in a perfect world (or space) but isn't that what the calculations describe?

This is really interesting. Is the speed determined from the distance covered by the orbit it self or from a fixed point on the ground? The reason I say that is if you take lets say, a one mile circumference and place an object going 60 miles an hour it takes one minute to complete. But if you place another object, going the same speed, with a one-half mile circumference it appears that its going twice the speed, but in fact is the same. Only the time to complete the trip gets shorter.

I know there are other things to consider because the rate of fall vs. the speed, and mass keeps them in orbit. I just find that question interesting....

"it appears that its going twice the speed,"

How funny indeed!!

Speed (velocity)= Distance/Time

It is going at twice the speed.

If not, please scrap the maths and let's all talk like as much "non-sense" as possible and make it all truly funny!!

Orbital speed is a function of altitude. The initial speed of the object is provided by booster rockets, but in orbit it is ballistic: it falls at approximately the same rate as it is moving forward. So as any object will increase speed as it descends from an initial altitude, so will a satellite. Its forward velocity is decreased by drag, but its vertical velocity will increase, until a new lower equilibrium orbital position is reached. The speed added is proportional to how far it fell from its original orbit, and its forward velocity does not need to be as high to maintain orbit since the radius is now smaller.

If we assume the orbit to be circular, then the circumference decreases at a rate of pi x the change in radius. So for a reduction of forward velocity of 3.14 feet per second, the object's orbital altitude will fall one foot and its overall speed will increase proportionally to the speed gained from that 1-ft fall, which is about 8 feet per second. Net increase in speed, just under 5 feet per second.

Because it just does.

Thanks for mentioning velocity and not speed...this question needs qualification.

The speed of a needle on a record is better...faster or slower when beginning or starting in relation to the surface of the record.

what about the sign of the radial speed?

Isn't it necessary?

there's no effect to the kinetic energy - but to the potential?

"G/r₁² = (v₁²)/r₁ and G/r₂² = (v₂²)/r₂. From this it can be seen that V₂ > V₁ if r₂<r₁ "

The Above conclusion is wrong: G/r₂² > G/r₁² implies that ONLY (v₂²)/r₂ > (v₁²)/r₁ Now, even if v_{2 =} v₁ we stil have (v₂²)/r₂ > (v₁²)/r₁ by virtue of your same conclusion about the r₂<r₁ !!

The Angular speed of the satelite is the entity that increases in this situation (you might call it Orbital speed but I call it Angular speed round the earth).

Hi LAA_Lucke, you wrote: " Now, even if v_{2 =} v₁ we still have (v₂²)/r₂ > (v₁²)/r₁ by virtue of your same conclusion about the r₂<r₁ !!"

Bob did qualify that he is working with circular orbits, or at least slowly inspiraling orbits, where what he wrote is essentially correct. I think it is only in fast inspirals where v₂=v₁ while r₂ < r_1, can perhaps happen.

What you are using is Kepler's second law of planetary motion, which on its own does not quite solve the dynamic orbit problem. One also needs the equation of motion: d²r/dt² = r dθ²/dt² - GM/r².

-J

Hi Jorrie,

Thank you for the clarification.

1- I wanted to first draw attention on the conclusion from the mathematic logics point. The conclusion would not be logical on its own without some Physics to explain how the V will change to support the conclusion.

2- Kepplers law will help focus on the fact that normally, in simple terms, the speed that will usually increase will be the angular rotation speed.

3- You are correct when you involve the drag and your dynamic equations, that the linear speed (tangential to orbit) might increase under specific conditions depending on the incidental angles: This is mainly relevant when an object travelling from non orbital trajectory enters an orbital zone and bounces or something like that.

Finally, I am not an expert in dynamic motions in orbits etc and may be not accurate in the math/physics of the phenomenon.

Thanks

Hi LAA, I did not show the full working out because I didn't think it was necessary. My equations can be simplified by multiplying both sides by the common radius terms to get the following G/r₁ = v₁² and G/r₂ = v₂². From this it can be seen that V₂ > V₁ if r₂<r_1.

thank you Bob.

MRQ wrote: "Thus, if r gets smaller, the orbital rate w in radians/sec will increase."

Right. This is essentially Kepler's 2nd law of planetary motion: an orbit sweeps out equal areas in equal time intervals. Smaller r, same area, larger angle. Can also be viewed as the law of conservation of orbital angular momentum L = m r² ω (until atmospheric drag messes it up).

-J

Brilliant!

This is the most clear answer I have read, easy to understand even for somebody not that familiar with that type of calculations, like my self.

Mochilo

Please clarify whether or not r > radius of earth (Height of orbit (above earth's surface)+radius of earth)?

Thanks in advance.

The symbol r refers to the radius of the orbiting satellite, that is, to the radius of the earth plus the height of the satellite above the ground. For simplicity of calculation, I have assumed a circular orbit though that would not necessarily be true.

Short of submitting all the equations etc... the simple answer is:

The Angular speed of the satelite increases when the orbit decays while the linear speed decreases due to the drag and makes the orbit decay.

Just as a rock skips across a pond and increases speed so do satellites.

It can be because this is not a complete question. Drag initially slows the tangential velocity, causing the satellite to drop, increasing the velocity so that it's not stable at that radius, causing it to climb back up to where it doesn't have enough velocity, ad infinitum. In other words, the drag pulls it into an elliptical orbit and it picks up speed when dropping and loses speed (to the drag) when climbing. If you compute speed as dA/dt where A is the area swept out, it loses to drag.

Hi TVP45, I think what you refer to does not quite happen to normal satellites in almost circular orbits. The drag is almost constant and will lower the orbit - actually it will spiral down, slowly at first and much later 'exponentially' and reenter.

If the orbit is very elliptical, dipping into the atmosphere at perigee will lower the apogee a little at a time, as AH pointed out before. So it will eventually circularize the orbit and then spiral in...

-J

I'd have to think about that a while. I haven't done satellite dynamics for over thirty years now and I'm pretty rusty.

Yes, you are correct. I was not doing the perigee correctly and was getting a time term that does not exist in nature. Ach, I miss my neuron; I should never have put it up for collateral for those %%$@# CDOs.

I will let others deal with this answer, but would like to further add to the question.

How does one calculate the effect of atmospheric drag on the earth itself?

There is a considerable amount and the answer is quite simple.

Is not the reason that the speed increases only due to the fact that the potential energy of the satellite is converted to kinetic energy and the reduction in energy at higher altitudes due to drag is less than the gain due to the potential energy change. Of course, at some point this is no longer the case, and the drag component will surpass the gain due to the potential energy change, thus the satellite will slow down (will never hit the surface at 22,000 mph).

The question "drag actually increases the speed. How can this be?" is kind of like, if you jump out of a helicopter, why does your speed increase, even though there is drag? Although of course, the satellite has a tangential velocity, but the premise is the same. Seems most of the answers get into the orbital mechanics of why orbital velocity is faster at lower altitudes.

This is really very simple the earth is spinning and the atmosphere drags the satellite along with it, thus increasing the speed of the satellite to try and match the spin of the earth.

I think not, unless the atmosphere is spinning in excess of orbital speed.

As a matter of fact;

The atmosphere DOES 'spin' faster than the earth at times, and therefor, (relatively)faster than orbital speed.

This is part and parcel to my addition to the question, previoulsy posted.

A satellite in low Earth orbit will orbit the Earth in about 90 minutes, traveling in excess of 17,000 miles per hour. This is roughly equal to mach 22 (22 times the speed of sound).

So yes, the atmosphere may travel relatively faster that the surface of the Earth itself (which is moving at about 1,000 miles per hour at the equator)-- but certainly not faster than a satellite near the edge of the Earth's atmosphere.

Guest at 7:54PM 05/04/2010

Your thinking is on track if we are talking about 'relative' being the distance the sat has to travel in comparison with the earths surface.

But I am saying that 'at any given distance', if it is 1 inch or 1000 miles, the atmosphere IS moving slower or faster than the earth itself.

Studied, proven, and recorded on a daily basis by several geo, astrological, and meteorological scientists.

They use GPS on a cosmic scale. Radio telescopes around the globe are monitoring specific quasar emissions as time stamps to record the slowing or speeding up of 1 earth day.

The reason for the time variation?

The variable drag, or lack of it so to speak, that the atmosphere exerts on the surface of the earth!

Yeah, your right! Sometimes the atmosphere is going slower or faster than the Earth!! Sometimes standing. still.

I think those astrological scientists were working for NASSAU, right?

Asfter all, everthing is relative. It is variable drag and sometimes the lack of the drag energy.

nice - a wave-tripper!

"This is really very simple the earth is spinning and the atmosphere drags the satellite along with it, thus increasing the speed of the satellite to try and match the spin of the earth."

For that to happen the satellite would have to be orbiting the earth at a rate less than the rotation of the earth, i.e. less than one orbit every 24 hours. For a satellite that is at the edge of the atmosphere this is much too low a speed to maintain orbit. Satellites in that altitude range have an orbital period of a few hours.

an Orbit is an equality of gravitational and centrifugal force.

the air decreases the speed of the orbital object - the centrifugal force decreases too and the gravitational force acclerates the orbital objcet.

This acceleration increases the speed of the orbital object.

this answer inspires the question:

Is there an altitude (and if at what height) where the decrease of the speed by air will increasae the orbital distance from earth (or any other object)?

This should be definied be the gradiation of the air brake effect and the change of the gravitation.

WHK, you asked: "Is there an altitude (and if at what height) where the decrease of the speed by air will increase the orbital distance from earth...?"

For nearly-circular orbits and non-aerodynamic bodies, this can surely not happen. They will spiral in, once the drag is significant.

For an object with wings, your scenario surely possible - dive in and then zoom up, but it won't reach the same altitude at the same speed that it started at; some orbital energy is lost as heat.

An object in an elliptical/parabolic/hyperbolic orbit that just dips into the atmosphere will almost always increase in orbital distance again, but also with the loss of some orbital energy and angular momentum. It is used in aero-braking to lower the apoapsis of probes visiting other planets.

-J

Atmospheric drag is so slight taht the actual slowing affects of drag aer relatively negligible. However, the drag does have the affect of reducing the orbiting object's altitude. Becasue the orbit's circumference is samller, the law of Conservation of Angular Momentum causes the object to speed up. The sattelite maintain the same rotational momentum, so as it draws closer to the gravitational focus of it orbit, it speed must increase to balance the lost diastance.

Guest wrote: "Becasue the orbit's circumference is samller, the law of Conservation of Angular Momentum causes the object to speed up. The sattelite maintain the same rotational momentum, so as it draws closer to the gravitational focus of it orbit, it speed must increase to balance the lost distance."

Not strictly true. The drag robs the satellite of angular orbital momentum, so it does not stay the same - that's why it spirals in. Angular orbital momentum is only constant in drag-free space. When drag gets severe during reentry, this loss is so great that it loses speed as well.

-J

".............this loss is so great that it loses speed as well."

In which case, the drag is NOT funny, after all those knowledgeable calculations!!

"air" speed? or ground speed?

The question doesn't state this effect occurs on any particular type of satellite, so I think it's flawed thinking that the atmospheric push will accelerate the object. I believe this effect would only be true for those satellites in geosynchronous orbit but only with very limited effect. However there are many types of orbits that are neutral or counter to the rotation of the earth's atmosphere where this would be a detriment. Furthermore, the question states that "the orbit is very close to the end of the atmosphere and are subjected to air drag". I take it to mean that the satellite ALWAYS encounters air drag. The question doesn't state that the object is without drag and then experiences it at a later time. So, the only time the centrifugal force and gravitational forces are equal is at the very beginning of it's life in orbit. Since it is always experiencing air braking it's angular velocity is always less than it was previously. In addition, the question doesn't clarify which of the two speeds is increasing. Everyone's assuming it's the angular velocity that increases, I believe the atmosphere is decreasing the angular velocity and as it does the satellite can no longer perfectly balance the pull of gravity. Therefore, as the angular velocity decreases the downward velocity toward the earth increases due to the pull of gravity.

Guest wrote: "Since it is always experiencing air braking it's angular velocity is always less than it was previously."

I don't think this is generally true. Angular velocity is ω = dθ/dt and for low drag, will generally increase with smaller r, due to Kepler's 2nd law of planetary motion: an orbit sweeps out equal areas in equal time intervals. It is only when the drag becomes severe, e.g.at or near reentry, that ω will decrease.

-J

While I'm sure many of you are correct with your equations (I have not taken the time to look over them all) I offer the average person the most simplistic of answers. As an object sees drag while in orbit it's tangential speed slows down. This decrease in tangential speed causes the object to fall faster to Earth as it can no longer keep its same orbit. Hence, as an object sees drag it actually falls to Earth faster.

Gravity.

On level ground, the car the two effects are wind and rolling resistance.

The satellite encounters wind resistance to an extent, but the effect of gravity pulling it towards Earth, or better put, down and over the horizon, makes it accelerate.

The same would be placing the car on the down slope of a hill where gravity now comes into play. Wind and rolling resistance will slow the car, but gravity will make it speed up.

Doug

EV TECH, Inc.

don't forget the drivers intelligence!!!

The satellite loses energy through drag and moves to a lower-energy orbit. When you allow for the change of gravitational force with radius, this turns out to be lower and faster (by which I mean higher linear velocity). The energy loss is accounted for by the fact that potential energy decreases more than kinetic energy increases.

Atmospheric drag on satellites increases their angular velocity (relative to the gravitational field/"pull" of the earth). This is especially true over the poles, which are somewhat flattened.

The drag of the air causes a satellite to lose energy and fall closer to earth, but the closer an object is to the earth, the faster it falls. A satellite is falling all the time; its forward speed just keeps it falling past the curvature of the earth.

hmmm, but drag affects _direction_ of moment, as much as it does forward or downward inertia,,, so there must be something else involved

Air firiction is very much less due to very close to the atmosphere but the horizontal component of centriptle force (gravitational force) is acted upon satillites and pulls it in the tengential direction of the orbit.

Anis A. siddiqui.

The direction of the drag is the same with the orbital movement of the satellites.

Maybe drag works with gravity to act as a tether, or downward force, resulting in centrifical acceleration?

Now that is a layman's answer if I ever offered one!

To look at it in the simplest terms, drag causes them to fall down (ie towards the earth). That which falls down gains speed. Of the two opposing effects, the gain wins.

There is low air pressure in front of satellite. As the insignificant amount of air travels around the satellite the air pressure increases at the rear side thus propelling the satellite forward. Note; the increased air pressure at the rear side of the satellite provide more energy than the energy produced by the drag. Example: a rocket has a tremendous amount of drag however the air pressure (energy) under the rocket is far greater than the energy the drag produces.

because a technician at control HQ detected it and turned up the juice

the earth spins and the satellites are not atached so the air speeds them up

I guess I am not clear - increases the speed relative to what? The ground, the object, or the fluid?

Drag is the difference in the momentum of an object comparied to the fluid surrounding it. So if the momentum of the fluid is low compared to the object then from the frame of the fluid the object slows down and from the fram of the object the fluid is locally speeding up. If the fluid has a higher speed then the drag causes the object to speed up from the frame of the fluid because it was impeding the flow. The problem I have is where am I (the observer)? Am I in the fluid stream (object speeding up), on the object (the fluid is slowing down) or on some other reference which could be slowing down or speeding up depending on what I am doing.

the earths rotation is faster than that of an object out of the atmosphere so the satellites that are at the edge will have the air drag moving faster than they are therefore they will be pushed along instead of slowed down. Like having a tail wind compared to having a head wind.

I must ask speed related to what ?

Maybe this is a question that is a trick or are you concerned about satellites being faster than your car but what type of satellite geo-stationary or what.

Speed is relative is it not ??

Let me try to get my point across this way since I cant do the math in my head. 1/4 of a earth rotation measured at the equator is X miles at Y miles per hour. Something in orbit staying directly above any given point along the equator at the edge of our atmosphere for that 1/4 rotation would be subject to a fast moving atmosphere on one side and a non moving environment on its other side thus being either towed by the movement or pushed by the movement whichever. If that makes any sense.

where X is 10000km and Y is 6 hours

The Y is not 6 hours. The Y represents the miles per hour of rotation. Not 1/4 of 24 hours ( 1 rotation / orbit / day ).

correct - speed is way by time, for known way the time is equivalent to speed!

Speed is an infinit number, time is only one digit!

So mean speed in related to the earths surface like the needle of a record.

So you are wrong...the closer the rotating object gets to the center of the arrangement the less distance covered in a given time...forgetting about the earths rotational movement.

Still I ask "Speed" in relation to what...you could sit on the satellite and you would not notice any "speed" increase....maybe you should have said velocity rather than speed.

you're sitting on a photon and the hole cosmos is drifting away!

This is lots of brilliant maths and "stuff" for a very bad question.

Speed is relative...end of story.

The satellite is just floating about in space it has no speed until you have a reference point to measure the speed.

This is what I learn when I was 11 years old and I believe it.

I mean the universe is expanding so fit that into the equation !!!!

WHAT IS SPEED ???? !!!

is speed electromagnetic?