How much energy is required to continualy orbit the Earth in less than 84 minute

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Offline syhprum

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Shakespear sends Puck in orbit around the Earth in 40 minutes whereas a normal orbital time at ground level would be 84 (Puck being a supernatural being is not concerned with air resistance), how much energy per Kg would be required to maintain this orbit ?
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Offline Bored chemist

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None.

I think his reduced orbital time would mean he is trying to fly off into space.
But he doesn't; so the force needed to keep him here doesn't move through any distance and the work done is zero.
You could imagine him tied to a rail that runs round the earth. Provided that he's able to ignore that sort of friction too then there's no extra energy involved.

Incidentally if my original supposition about the direction of the force is wrong the same argument still holds but he needs roller skates.
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Offline syhprum

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Although the construction of a frictionless inverted railtrack under which he could orbit would be a convenient way to keep him orbiting once he had got up to speed in the absence of such a track he would need to provide a continuous thrust directing him towards the Earth to avoid flying out of orbit.
This would of course would require the expenditure of energy, how much ?.
 
« Last Edit: 19/05/2010 13:49:39 by syhprum »
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Offline Geezer

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Is the required thrust 10.8N ?
There ain'ta no sanity clause, and there ain'ta no centrifugal force ćther.

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Offline syhprum

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It is probably an over simplification but how I would go about calculating it would be to use mv^2/2 for the added energy for each orbit, the orbital speed for a normal orbit would be 40000000/(84*60) m/s = 7936, the speed for the 40 minute orbit would be 16666.7 m/s.now if we compute the stored kinetic energy at this speed we get 1.3889*10^8 joules per Kg from which I deduct the kinetic energy for normal orbital speed which is 0.3149*10^8 joules hence the additional energy that must be supplied each orbit is 1.074*10^8 joules per Kg.
Hence the power required is 44.75 MW per Kg.
I look forward to correction by more skilled computers. 
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Offline JP

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Is the required thrust 10.8N ?

I get something a little different...  Let me know what you think of my reasoning.

Centripetal force is F=mv2/r, so it depends on the square of your velocity.  My calculation goes like this:  Gravitational force is enough to hold you in orbit at the slow speed.  At high speed, your speed increases by a factor of f=84/40.  Since the force needed to hold you in orbit is proportional to v2, it increases by a factor of f2.  The total centripetal force you need to hold this faster orbit is f2Fgrav, where Fgrav is the regular orbital speed when that force is entirely provided by gravity.

Therefore, the extra force you need is: (f2-1)Fgrav.

Since the normal force of gravity at the earth's surface is 9.81 N for a 1 kg mass, the extra force required will be roughly 33.45 Newtons.

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Offline Geezer

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Is the required thrust 10.8N ?

I get something a little different...  Let me know what you think of my reasoning.

Centripetal force is F=mv2/r, so it depends on the square of your velocity.  My calculation goes like this:  Gravitational force is enough to hold you in orbit at the slow speed.  At high speed, your speed increases by a factor of f=84/40.  Since the force needed to hold you in orbit is proportional to v2, it increases by a factor of f2.  The total centripetal force you need to hold this faster orbit is f2Fgrav, where Fgrav is the regular orbital speed when that force is entirely provided by gravity.

Therefore, the extra force you need is: (f2-1)Fgrav.

Since the normal force of gravity at the earth's surface is 9.81 N for a 1 kg mass, the extra force required will be roughly 33.45 Newtons.

I was using the "easy math" version. I (incorrectly ) assumed that the total centripetal force would be 9.81*84/40N.

My mistake was assuming that Syhprum had rigged the numbers so that the answer would come out to some nice integer. I didn't think he was actually going to set a stinking exam question.  [::)]
« Last Edit: 14/05/2010 06:21:03 by Geezer »
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Offline JP

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I think you'll also need one more piece of information to answer this question: what's supplying that extra force?  If it's a rocket engine, you would need to specify the speed of the ejected mass or the rate at which it's ejected.  Then you can use conservation of momentum to figure out the energy of the ejected mass and you could use that to figure out the rate of energy expenditure. 

You can't just compare the kinetic energies of the two orbits because the energy being expended to keep Puck in orbit doesn't contribute to the orbital energy as BC pointed out.  All that extra energy is doing is keeping Puck from flying off into space. 

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Offline syhprum

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The Question I asked was how much energy Joules per sec MW etc was required to maintain this orbit please complete the calculation as it looks as though you are on the way to producing a correct answer.
Also another more complex question arises how much energy is required to take off, make a complete orbit and land as Puck was required to do but this is beyond me!

Geever

Now that calculators and computers are readily available I do not think there is any need to cook the questions to make the arithmetic easy.

This post was meant to appear before JP,s  post
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Offline Bored chemist

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Although the construction of a frictionless inverted railtrack under which he could orbit would be a convenient way to keep him orbiting once he had got up to speed in the absence of such a track he would need to provide a continuous thrust directed away from the Earth to avoid flying out of orbit.
This would of course would require the expenditure of energy, how much ?.
 
Still none.
Work = force times distance.
No matter what force you end up calculating, since Puck remains at the same height that force acts through no distance so the work done is zero.
Since, as you have accepted, a rail could provide the force you have to ask how much work can a lump of metal do?
Can it provide any energy? If so then you have solved the world's energy crisis.

The only extra energy needed is because to go round the world quicker, Puck moves faster and has more kinetic energy. Calculating that is trivial as long as you know the size of the Earth.
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Offline Geezer

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Now that calculators and computers are readily available I do not think there is any need to cook the questions to make the arithmetic easy.


What? Do you mean my trusty slide rule is redundant? That's really quite good news because the numbers are pretty much worn off mine now, and nobody seems to make them anymore.
There ain'ta no sanity clause, and there ain'ta no centrifugal force ćther.

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Offline Geezer

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Still none.
Work = force times distance.
No matter what force you end up calculating, since Puck remains at the same height that force acts through no distance so the work done is zero.
Since, as you have accepted, a rail could provide the force you have to ask how much work can a lump of metal do?
Can it provide any energy? If so then you have solved the world's energy crisis.

The only extra energy needed is because to go round the world quicker, Puck moves faster and has more kinetic energy. Calculating that is trivial as long as you know the size of the Earth.

I think Syhprum is saying no rail is allowed. However, as he didn't specify the radius of the orbit, can we assume a reduced radius for the faster orbit so that the only centripetal force is provided by gravity so that no rail is required and no work is done?
There ain'ta no sanity clause, and there ain'ta no centrifugal force ćther.

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Offline JP

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Still none.
Work = force times distance.
No matter what force you end up calculating, since Puck remains at the same height that force acts through no distance so the work done is zero.
Since, as you have accepted, a rail could provide the force you have to ask how much work can a lump of metal do?
Can it provide any energy? If so then you have solved the world's energy crisis.

The only extra energy needed is because to go round the world quicker, Puck moves faster and has more kinetic energy. Calculating that is trivial as long as you know the size of the Earth.

I think Syhprum is saying no rail is allowed. However, as he didn't specify the radius of the orbit, can we assume a reduced radius for the faster orbit so that the only centripetal force is provided by gravity so that no rail is required and no work is done?

We'd be inside the earth!  Then we might be eaten by mole people!


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Offline syhprum

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I assumed a ground level orbit no lower orbit is possible.

I think we have a similar case here to electrons circulating around a Cyclotron once the magnetic field is established by the super conducting magnet no energy is required to maintain it (it plays the part of a railtrack) the electrons continually radiate energy that must be replaced by the RF generator.
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Offline syhprum

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Even if we dodged the mole people we would find that gravity is less underground.
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Offline Geezer

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Still none.
Work = force times distance.
No matter what force you end up calculating, since Puck remains at the same height that force acts through no distance so the work done is zero.
Since, as you have accepted, a rail could provide the force you have to ask how much work can a lump of metal do?
Can it provide any energy? If so then you have solved the world's energy crisis.

The only extra energy needed is because to go round the world quicker, Puck moves faster and has more kinetic energy. Calculating that is trivial as long as you know the size of the Earth.

I think Syhprum is saying no rail is allowed. However, as he didn't specify the radius of the orbit, can we assume a reduced radius for the faster orbit so that the only centripetal force is provided by gravity so that no rail is required and no work is done?

We'd be inside the earth!  Then we might be eaten by mole people!



Ah, but as there is no friction, who cares. The mole people would never catch up.
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Offline syhprum

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JP

You raised a very relevent point ie the efficiency of the propulsion system, I had not considered this and had unconsciously assumed 100%, maybe Puck could use a photon drive or something similar
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Offline syhprum

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I must apologise for the poor phrasing of the question, what I meant to ask was what power must Puck expend to maintain this orbit hence my answer of 44.75 MW per Kg.
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Offline JP

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JP

You raised a very relevent point ie the efficiency of the propulsion system, I had not considered this and had unconsciously assumed 100%, maybe Puck could use a photon drive or something similar

Even outside of the efficiency, you need to know something about the mass being ejected or its velocity.  Since this isn't an energy conservation problem (the energy added to the orbiting body is zero!) you can't answer it by only knowing details about the orbits.  Your answer could be zero (in the case BC talks about) or non-zero (in the case of a rocket engine providing the thrust).  Because you can't get the energy from only knowing the orbit, you need to specify details about the source of the force holding Puck in place instead.

Let's check the 100% efficient rocket engine case, and assuming that the rocket fuel doesn't actually carry any mass away from Puck (which obviously is just an approximation).  If he's shooting out particles with momentum, p, at a rate of N particles per second, then the force provided is pN.  This isn't enough to answer the question unless you know something else, since you basically know the ejected momentum now, which is mv, but you can't get energy from this, since it's mv2/2.  However, if you're assuming that you're ejecting photons, then you know that p=E/c for each photon,  So...

pN=EN/c=Fgrav

More interestingly, since c is a constant and Fgrav is constant for this particular orbit, that means that it's the product of E (energy per photon) and N (number of photons per second) that matters.  You can eject more photons at lower energy or fewer photons at higher energy.

At any rate, the total rate of energy you're releasing here is EN, so in that case

EN=Fgravc≈1010 Joules per second per kilogram. 

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Offline syhprum

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That seems rather a lot, my answer of 44.75 MW would seem to be out by a factor of 225 times at least !
could this be that a photon drive is a pretty inefficient way to propel a body moving at .00000518 c
« Last Edit: 14/05/2010 10:10:20 by syhprum »
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Offline JP

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Well, your initial calculation has an error in it.  The error you made is that you just compared the kinetic energy of the two orbits.  This just tells you the difference in the energy required to accelerate Puck to the two different speeds in the first place. 

Once he's at the lower speed, he requires zero energy to maintain his orbit (neglecting air resistance).  Once he's at the higher speed, he needs an additional force of roughly 4 earth gravities constantly pushing him towards the earth.  It's the energy needed to constantly provide that extra force that you're asking about in this question.  And that energy entirely depends on the process that's generating that force.

(By the way, I made that calculation very quickly, so I wouldn't be surprised if there's an error in it.)

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Offline syhprum

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WIKI quotes a figure of 300MW/N as 33.45N is required the power needed by this route is 10^10 watts which is the figure you gave, there must be a better way to do it!
Shakespeare talks of putting a girdle around the Earth in 40 minutes, I take it this includes take off and landing after one orbit I wonder how much energy this would require ?
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Offline Bored chemist

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OK, it's 40,000 Km round the earth.
To go round in 40 mins would mean a speed of 40,000KM /(60*40)=17000m/s
To do it in 84 mins means travelling at 8000/s
Each Kg of puck takes 1/2  V^2 of energy to get him going so it's 32MJ for an 84 minute lap and 145MJ for a 40 min lap time.
The difference is 112.5 MJ per kilo

Once you have got him up to speed you can stop using any more power.

Can I just ask those who think that they need to burn millions of Watts to keep him going please tell me against what is all that work done?
Puck maintains the same height, so it's not done against gravity. He keeps a constant speed, so it's not work done to increase kinetic energy.
There's nothing else involved so the energy doesn't seem to have anywhere else to go.

When it comes down to it, you can achieve this effect with a fixed rail.
A stationary rail cannot do work.
So achieving this does not need any work to be done.
Any way of doing it that dissipates power is inefficient so all the question seems to be asking is asking is "just how inefficient can we make this system?"

You could replace the rail with a rocket firing downwards to produce the force needed to keep Puck in low orbit.
In that case the rocket will need to dissipate power- but the power needed is not defined by the force it delivers- the exit velocity of the gases is also involved (not to mention other inefficiency) so the question isn't defined properly.

Where are you putting all this energy you are all talking about?
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Offline Geezer

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Where are you putting all this energy you are all talking about?


If there is no rail and there is only a rocket, the energy is being sent into space in the kinetic energy of the matter that is ejected by the rocket.
There ain'ta no sanity clause, and there ain'ta no centrifugal force ćther.

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Offline syhprum

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Puck does a constant speed but it is velocity we are concerned with, energy is constantly required to divert his natural hyperbolic path that would take him out Earth orbit to a circular one around the Earth.
I think the question of efficiency can be addressed by comparing in the case of the photon drive the speed of the ejected photons with the speed of Puck in the direction of the centre of the Earth which is a very high ratio.
Photon drive is only efficient to propel a high velocity vehicle 
« Last Edit: 14/05/2010 20:12:11 by syhprum »
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Offline Geezer

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Puck does a constant speed but it is velocity we are concerned with energy is constantly required to divert his natural hyperbolic path that would take him out Earth orbit to a circular one around the Earth

We can calculate the required force to maintain the path, but as JP points out, without some information about the nature of the thing (presumably it's some sort of rocket) that's producing the force, I don't think we can get at the energy consumed.

Also, if it is a rocket, I have a suspicion that it will need to exhaust a good bit more than a kilogram of matter for every kilogram of Puck's mass during a circuit, so, as we have to account for the changing mass of the fuel, the force will be far from constant and the math is going to become rather tricky.

It's not rocket science, er, well, I suppose it really is.
« Last Edit: 14/05/2010 20:39:13 by Geezer »
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Offline syhprum

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The calculation has been done for a photon drive and comes out at 10^10 Watts but of course photon drives are only efficient driving high speed vehicles and the velocity increment applied to Puck in the direction of the centre of the Earth is very low compared with the the velocity of the emitted photons.
If my estimate of the required power as 44.75 MW is correct the efficiency is .045%.

I did only ask how much power was required not how it was generated. 
« Last Edit: 15/05/2010 08:56:56 by syhprum »
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Offline syhprum

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An interesting question arises would it be possible with available technology to actually do an orbit of the Earth including take of and landing in 40 minutes, I don't think it could be done.
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Offline Geezer

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I did only ask how much power was required not how it was generated.
 

In reality, would it not have to be far greater than that? I think the only way we could produce the force would be with a rocket, but the rocket would also have to transport its fuel, so the required thrust would also have to account for the mass of the fuel which I think will be much greater than the mass of Puck.

I think you're estimate might be off by a couple of orders of magnitude.
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Offline syhprum

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I was only concerned with how much power was required to maintan the orbit not to design a system to produce the power, as you say it would require a large rocket which in turn would have to generate a great deal of power to orbit its own mass and fuel.
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Offline Bored chemist

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In summary, if you use the most efficient means to achieve the stated task then the energy is zero and if you use a less efficient means then the answer is "it depends".
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Offline syhprum

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Can't argue with that !
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Offline tommya300

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The attempt to, of the intended event is in question. Not the fact that the attempt of the intended event was successful!
For the event to be successful the duration needed to be at least longer then 84 minutes.
Therefore the orbit was not accomplished and nothing is needed to sustain the failed event.
Like in baseball I through the ball to the outfield, in Yellowstone Park,
 (big ball park)! Can we assume I was successful in making it reach the outfield?

"An orbit is a regular, repeating path that one object in space takes around another one."
« Last Edit: 19/05/2010 10:42:37 by tommya300 »

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Offline imatfaal

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mmm - not sure about this one.  cannot see any holes in arguments, but still...
There’s no sense in being precise when you don’t even know what you’re talking about.  John Von Neumann

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Offline Bored chemist

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I think tommya means that this
"An interesting question arises would it be possible with available technology to actually do an orbit of the Earth including take of and landing in 40 minutes, I don't think it could be done."
wouldn't be an orbit because it's a one-off.

Incidentally, I gather that bored researchers at the North and South poles fill in time by having races round the world.
With a bit of messing about (like getting to the Arctic), you could walk round the world in 84 mins, if you jumped in the air at the start and finish of this circumnavigation would that count as take off and landing?
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Offline wolfekeeper

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FWIW 44 minutes is probably much too quick for humans- the g-forces get ridiculous, if you include acceleration and deceleration time, the g-force really mounts, IRC you're hanging upside down pulling 10g for most of it- not my idea of fun!

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Offline Bored chemist

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So, you didn't read the bit about being able to walk round the world in less than 84 minutes?
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Offline wolfekeeper

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That's not an orbit.

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Offline Bored chemist

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Nor's the original post's question.
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Offline daveshorts

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Just to make it more complex, if the energy or reaction mass supply for puck's rocket are carried on him, his mass will change as he orbits. Even with photon drives, his mass is going to decrease as it fires a bit. So you have to solve a fun differential equation.

Essentially
a = F/m

F= (g Isp /m)  dm/dt

m = mass of fuel
a = acceleration required = 33.45 m/s2
Isp = figure of merit for your rocket engine for shuttle = 450s
g = earth's gravitational field (from defintition of ISP) = 9.81

if you integrate those up you get
m = eat/gIsp

so if he has to orbit for 40mins = 2400s
m = e18.2
  = 80 000 000kg of rocket per kg of puck

As this is hydrogen oxygen fuel the energy is going to be silly at anout 285kJ/mol of water formed this will use of the order of 1 x 1015J /kg of puck

You would be better off using something with a higher Isp like an ion engine using an ion thruster with an Isp of around 3000 you would only need 15kg of fuel per kg of puck. The energy required will be the integral of the force times about 40kW/N

so I get about 340MJ/kg

which may or may not be right according to my ability to integrate.

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Offline syhprum

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I only really wanted to calculate how much power was required, I had not got as far as devising a system to generate it.
Many thanks for your heroic computation.

Gino (syhprum)
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