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Originally posted by sonhousePower adds energy at a steady rate.
So how much energy does it take to make say, 1 kg, accelerate at one G?
It's not rocket science for god's sake🙂
No, wait....
Added energy adds to the kinetic energy.
K = mv^2/2
If K is 1 and m is 2,
1 = v^2
v = 1
If K is 4, v = 2
If K is 9, v = 3
If K is 16, v = 4
Now suppose K increases by 1 every second (similar to the OP problem). The rate of acceleration will slow because the velocity only increases with the square root of the added energy.
In short, constant power on constant mass apparently causes an acceleration which decreases over time. You demand a constant acceleration.
It sounds kind of like someone wanting a constant force to cause a constant velocity I guess.
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Originally posted by sonhouseThink about what your saying, this statment implies that you could with this finite amount of energy accelerate a mass indefinately........NOT POSSIBLE.
So how much energy does it take to make say, 1 kg, accelerate at one G?
It's not rocket science for god's sake🙂
No, wait....
Energy of a body in motion
E = 1/2*m*v^2 (eq 1)
v =a*t (eq 2)
substitute (eq 2) into (eq1)
E = 1/2*m*(a*t)^2
as it stands this equation has 4 variables of which you have only defined 2 of them. Either t or E must also be defined for there to be a solution
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Originally posted by AThousandYoungThat is 100 percent not true. If you look at the ion rockets flying in space right now, they shoot out of the rocket enough to give a constant acceleration with a constant input of energy, albeit a low acceleration and a lot of energy for the acceleration you actually achieve but it is a constant I can gaaaruuntee!
Power adds energy at a steady rate.
Added energy adds to the kinetic energy.
K = mv^2/2
If K is 1 and m is 2,
1 = v^2
v = 1
If K is 4, v = 2
If K is 9, v = 3
If K is 16, v = 4
Now suppose K increases by 1 every second (similar to the OP problem). The rate of acceleration will slow because the velocity only increases with th ...[text shortened]... rt, constant power on constant mass apparently causes an acceleration which decreases over time.
In a chemical rocket it's actually the other way round, an engine produces X amount of thrust and uses an enormous amount of burn per second shooting out the rear end, thousands of tons of it. In that case, if the rocket is causing the craft to accelerate at 3 G's (like the shuttle at launch or shortly thereafter) because the total mass of the rocket is going down second by second, the actual acceleration can actually increase because the mass is getting lower second by second. So to counter that they throttle back the engines if they want a constant acceleration, which of course can only last 20 odd minutes before it runs out of our sucky level of technology fuel. If they kept the engines wide open, the whole craft would mass maybe 1/5 of its starting mass and the resultant acceleration would go from 3 G's to 15 G's before the fuel ran out. They cannot allow that so they constantly control the thrust, reducing it second by second to achieve a constant acceleration.
In the Ion rocket, the mass of the propellant is so low there isn't much change of mass of the total system second by second or day by day so the acceleration stays pretty much equal till it also runs out of fuel but because the specific impulse rating of ion rockets is so high, something like 10 or more times that of chemical rockets, it can run for months on a stingy supply of propellant, which in this case is not being converted chemically, simply accelerated electrically to a very high exit velocity which is the name of the game in rockets.
Originally posted by joe shmoI already specified we are not talking about relativistic acceleration, just what you need to get around our solar system, say not going over 10 thousand miles per second or so.
Think about what your saying, this statment implies that you could with this finite amount of energy accelerate a mass indefinately........NOT POSSIBLE.
Didn't you see the piece about the VASIMIR rocket? It will use a 200 megawatt nuclear power supply and get to mars in 39 days, one fifth the time of our present round of sucky chemical rockets? What I am pointing out is even that pales in comparison to what could be done if you knew enough about physics, which the human race decidedly doesn't at this point in time. The Vasimir we already calculated in the science forum to be giving about 0.05 or so G's 50 MILLI G's of constant acceleration and it gets to Mars in 5 weeks! If it could do one whole G it would get there in a few days.
So playing forward, the same system would need 20 times the thrust for the same mass of the total system and therefore 20 times the power or 4 Gigawatts to squeeze out one G of accel, which aint' gonna happen any time soon. 200 megawatts we can maybe do in a few years or maybe even now if we want to bad enough but 4 gigawatts in a small size? We have to wait for anti-matter power.
If you had an anti-matter power source, you could accelerate for months at one G, which it turns out you get close to c after a year of such acceleration, of course you are talking relativistic effects and you may THINK you are going faster than light but we on earth see you as not getting to the speed of light even though your onboard clocks might say it looks like you are going faster than c. It is more like the ruler gets shorted as you approach the speed of light and your velocity has changed space around and your time flow internally so you think maybe you got to Andromeda Galaxy in say two years but a couple million years still passes by on Earth, but I am not talking about that kind of interstellar journey, just very fast solar system trips. One G gets you pretty much anywhere in the solar system in about a week.
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Originally posted by joe shmoWhat do you think I defined? E. One megawatt. Jeez.
Think about what your saying, this statment implies that you could with this finite amount of energy accelerate a mass indefinately........NOT POSSIBLE.
Energy of a body in motion
E = 1/2*m*v^2 (eq 1)
v =a*t (eq 2)
substitute (eq 2) into (eq1)
E = 1/2*m*(a*t)^2
as it stands this equation has 4 variables of which you have only defined 2 of them. Either t or E must also be defined for there to be a solution
Getting close to c is what the fitzgerald contraction formula is all about.
Did you read my PM's?
Of course you add kinetic energy to a body under acceleration, call it a battery, so what? Ion rockets do that by accelerating at a constant but low acceleration, the total kinetic energy always goes up. That's just the way the cookie crumbles!
Originally posted by sonhouseI will have look at them throughout the week(they look to be information dense)
What do you think I defined? E. One megawatt. Jeez.
Getting close to c is what the fitzgerald contraction formula is all about.
Did you read my PM's?
Of course you add kinetic energy to a body under acceleration, call it a battery, so what? Ion rockets do that by accelerating at a constant but low acceleration, the total kinetic energy always goes up. That's just the way the cookie crumbles!
First of all you asked for the "ENERGY", and Second a MegaWatt is NOT a unit of ENERGY, its a unit of power. Power= Energy/time
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Originally posted by joe shmook, Power, I thought I mentioned using a constant input of energy per second which is power. Is that what is tying this all up?
I will have look at them throughout the week(they look to be information dense)
First of all you asked for the "ENERGY", and Second a MegaWatt is NOT a unit of ENERGY, its a unit of power. Power= Energy/time
Acceleration is ALREADY a measure of power, you have to have the time in the equation or you get no acceleration.
Maybe I should have said one megawatt constantly applied.
When I see a power supply rated as one kilowatt, I automatically think per second. Just my techie bias I guess.
Still, even at that, one megawatt applied for one second can give you how much mass the thing can accelerate for one second which will be the same answer I gave in my PM.
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Originally posted by sonhouseacceleration is not a measure of power
ok, Power, I thought I mentioned using a constant input of energy per second which is power. Is that what is tying this all up?
Acceleration is ALREADY a measure of power, you have to have the time in the equation or you get no acceleration.
Maybe I should have said one megawatt constantly applied.
When I see a power supply rated as one kilowatt, I aut ...[text shortened]... much mass the thing can accelerate for one second which will be the same answer I gave in my PM.
in SI units
units of accelration: (m/s^2)
units of Power: [(kg*m/s^2)*m]/s or (kg *(m^2)/(s^3))
They are not the same.
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Originally posted by joe shmoYou still have to use power to accelerate and acceleration implies power by having time as one of the units so it sure looks related to me, not directly but indirectly. You have to use power for a given amount of time to show an acceleration. So you still don't agree with my analysis given I said applying 1 megawatt continuously?
acceleration is not a measure of power
in SI units
units of accelration: (m/s^2)
units of Power: [(kg*m/s^2)*m]/s or (kg *(m^2)/(s^3))
They are not the same.
BTW, I hope you don't think I am trying to put you down or something, it is interesting the way this post has gone, it causes me to think more carefully about my constructs.
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Originally posted by sonhouseThis is explicitly solvable as stated
You still have to use power to accelerate and acceleration implies power by having time as one of the units so it sure looks related to me, not directly but indirectly. You have to use power for a given amount of time to show an acceleration. So you still don't agree with my analysis given I said applying 1 megawatt continuously?
BTW, I hope you don't th ...[text shortened]... eresting the way this post has gone, it causes me to think more carefully about my constructs.
"one megawatt applied for one second can give you how much mass the thing can accelerate for one second(sonhouse)" (You also stated how they were the same thing, that however is incorrect)
none of the others (from you OP and including up to the above statment) had solutions as stated.
Here is the solution to your properly worded construct.
P*t = 1/2 *m*(a*t)^2
m= (2*P*t)/(a*t)^2
using your numbers I come up with 20,825 kg