Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: syhprum on 26/01/2016 06:43:17
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The normal minimum orbit time for the Earth is about 84 minutes so to orbit it in 40 would require the continuous expenditure of energy , how much kilowatts per Kg of pucks mass.
possibly Puck being an ethereal being he might have zero mass which would of course reduce the energy required !
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Who is Puck?
Anyway an orbit of 40 minutes gives an orbital height of 11,451.5Km above sea level and an orbital speed of 167,973.4 Km per hour.
Now puck can't have zero mass as that would have Puck traveling at the speed of light and as such Puck is going far too fast to be in any orbit around the Earth. Certainly far too fast for a 40 min orbit.
Also that orbital height is well within the atmosphere and as you surmised would require a continuous energy input to overcome atmospheric drag. Speaking of atmospheric drag, at that speed Puck would be a very hot massive etherial being.
How much energy?
That is much too complex for us to work out here. It depends so much on aerodynamics and I'm not too sure where you would find a wind tunnel to test at those speeds so you can come up with energy figures.
Puck would also present one hell of an aeroplane hazard.
EDIT: That orbital height should read 11.5 Km. Didn't convert from SI units.
Sorry about that.
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The atmospheric drag at 11,000 km could be ignored for a single orbit, which is all Puck promised to do.
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The atmospheric drag at 11,000 km could be ignored for a single orbit, which is all Puck promised to do.
In that case, if the speed is brought up to just over 170,000 km/hr, with a really streamlined shape, Puck could orbit in 40 minutes without thrust. So no energy input during that one orbit. Totally Geodesic.
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Oops just noticed That is meant to be 11.5 kms not 11,451.5
My apologies for the confusion. I wrote the last comment and then thought, ("Hang on what does he mean atmospheric effects can be ignored). Calculate in SI units and convert you fool.
Again apologies..
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The normal minimum orbit time for the Earth is about 84 minutes so to orbit it in 40 would require the continuous expenditure of energy
Perhaps the scenario considered here can be imagined like this:
- The ISS has an altitude of 300-400 km above sea level (ie above most of the atmosphere), and orbits in free-fall in around 90 minutes, with an orbital velocity of about 42,000km/90 minutes = 7.6 km/second.
- However, it is possible to have a satellite at the same altitude (ie still above most of the atmosphere) that orbits in 40 minutes, with an orbital velocity of 42,000km/40 minutes = 17.5 km/second.
- Now, 17.5 km/s exceeds Earth's escape velocity (which is around 11.5 km/s). So you need to be firing a rocket away from the Earth to keep this space probe in a circular orbit at the same altitude as the ISS.
- This space probe will not be in free fall, but will be continuously accelerated towards the Earth by the rocket.
- Firing this rocket continuously takes power (which you measure in kW).
how much kilowatts per Kg of pucks mass?
This is where I am struggling with the physics, but I'll have a go...
- The acceleration due to gravity is (approximately) 1G at the altitude of the ISS.
- This acceleration is enough to bend the ISS orbit into a circular path.
- For Puck, the orbital velocity has increased by a factor of 2.3.
- The centripetal(?) force required to keep it in orbit increases as the square of the velocity, so the acceleration required is 5.2G.
- But the acceleration due to gravity at this altitude is only 1G, so the rocket must supply 4.2G of acceleration.
- For a Puck weighing 1kg, the rocket must apply approximately 42 Newtons of thrust, continually.
However, at this point I don't know how to turn a rocket's thrust into kW.
Who is Puck?
Could Puck be that "merry wanderer of the night" imagined by Shakespeare?
I haven't seen "A Midsummer Night's Dream (http://en.wikipedia.org/wiki/Puck_(A_Midsummer_Night%27s_Dream))", so I wouldn't know...
Maybe we could ask a Canadian hockey player?
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However, it is possible to have a satellite at the same altitude (ie still above most of the atmosphere) that orbits in 40 minutes, with an orbital velocity of 42,000km/40 minutes = 17.5 km/second.
Now, 17.5 km/s exceeds Earth's escape velocity (which is around 11.5 km/s). So you need to be firing a rocket away from the Earth to keep this space probe in a circular orbit at the same altitude as the ISS.
I bet you were a power PC user back in the day. Yes that is the perfect solution. Love it.
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I don't know how to turn a rocket's thrust into kW
How about applying F=ma (the second most-popular equation in science, according to Brian Cox):
- Puck has a mass of 1kg
- The force is 42N
- The acceleration is 42m/s2
- In 1 second, the mass will reach a velocity of 42m/s, perpendicular to its orbital velocity of 17.5km/s
- After 1 second, the kinetic energy in this direction will be E=½mv2=882 Joules
- So the power delivered by the rocket is 0.9kW continuously, for a perfectly efficient rocket.
But real rockets are not that efficient. Even ignoring the mass of the rocket and the fuel tank:
- If Puck had a solid booster rocket with a specific impulse (http://en.wikipedia.org/wiki/Specific_impulse#Examples) of 250, Puck would disappear in a puff of smoke in just 1 minute - not enough to complete 1 orbit. The chemical reaction would release about 50kW.
- However, if Puck had an ion drive with a specific impulse of 10000, Puck could continue his shadowy trail for 40 minutes - just enough to complete 1 orbit. If my calculations are correct, the ion drive would consume a massive 600kW of power - that's a lot of solar cells!
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Puck could orbit in 40 minutes without thrust. So no energy input during that one orbit. Totally Geodesic.
We can work out what it would take to get a 40 minute free-fall orbit by applying Kepler's third law (http://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion) to the ISS:
- The ISS orbits in 90 minutes, at a radius of around 6700km from the center of the Earth
- According to Kepler, the square of the orbital period (P2) of an object is directly proportional to the cube of the semi-major axis of its orbit (a3).
- For a circular orbit, a is equal to the orbital radius.
- To get a free-fall orbit of 40 minutes, the ratio of the periods is 40/90=44%
- And the ratio of the orbital radii is 58%, or 3800km, which is well inside the Earth's 6400km radius.
- Don't worry about air resistance - Puck would need to be a very fast mole!
- However, Kepler's laws assume that planetary bodies are geometrical points, and his laws are invalid if you go subterranean.
- You would need to compress the entire Earth a lot denser than its current 5.5g/cm2 to achieve a 40 minute free-fall orbit - and that will take a lot of kW!
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We can work out what it would take to get a 40 minute free-fall orbit by applying Kepler's third law to the ISS:
The ISS orbits in 90 minutes, at a radius of around 6700km from the center of the Earth
According to Kepler, the square of the orbital period (P2) of an object is directly proportional to the cube of the semi-major axis of its orbit (a3).
For a circular orbit, a is equal to the orbital radius.
To get a free-fall orbit of 40 minutes, the ratio of the periods is 40/90=44%
And the ratio of the orbital radii is 58%, or 3800km, which is well inside the Earth's 6400km radius.
Don't worry about air resistance - Puck would need to be a very fast mole!
However, Kepler's laws assume that planetary bodies are geometrical points, and his laws are invalid if you go subterranean.
You would need to compress the entire Earth a lot denser than its current 5.5g/cm2 to achieve a 40 minute free-fall orbit - and that will take a lot of kW!
I stand totally corrected. Please ignore this fossil's poor excuse for very rusty back of envelope math.
I put pen to paper to solve for a 40 minute orbit and I wasn't even close as your calculation showed. Using Orbit r = The "3rd root of GMP²/4π²", Where G is the gravitational constant, M is the Mass of Earth, and P is the orbital period in seconds, I too got a radius of 3,873.441 Km.
With the Earth radius at 6,371 Km, I think you were being extra kind to me to suggest a very fast mole Puck could do it.
My math is obviously in need of some practice.
Thank you for pointing out my totally misleading first answer.
Now how much energy would it take to compress the Earth to inside that radius? I wouldn't even know where to start for that calculation.
Now your brute force suggestion, 4.2g for 40 minutes I believe is quite achievable for Puck.
Bugger the energy requirements. That's for the engineers to figure out.
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I did not intend that the method of generating the downward thrust should be gone into too deeply only power absorbed generating the downwards thrust.
0.9 kilowatts per Kg sounds very modest.
perhaps I should have explained who or what puck was but I thought Shakespears sprite was well known.
Perhaps the problem could looked at in a different way let us assume that Puck is travelling at a normal obital speed what power is required to accelerate him to 17.5 m/s in 40 minutes ?