Did Isaac Newton, NASA and possibly Hillary Clinton deceive the world about...

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FalseProphet

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Did Isaac Newton, NASA and possibly Hillary Clinton deceive the world about kinetic energy?

I have a problem again.

When I want to accelerate an object from 0 to 2 km/s I need 2 Megajoule of energy per kg, because



Now I want to accelerate the same object by another 2 km/s, so I have to use another 2 Megajoule per kg.

Now the object has a velocity of 4 km/s.

I have invested 4 Megajoule per kg.

The object, having a velocity of 4 km/s, has a kinetic energy of 8 Megajoule per kg, according to the formula.

So I have used up 4 Megajoule per kg, but I got 8 Megajoule out of it.

How is that possible?

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rabinoz

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Did Isaac Newton, NASA and possibly Hillary Clinton deceive the world about kinetic energy?
What could NASA and Hillary Clinton have to do with it?

Isaac Newton's and James Prescott Joule's shoulders are strong enough to take the blame.

Quote from: FalseProphet
I have a problem again.
When I want to accelerate an object from 0 to 2 km/s I need 2 Megajoule of energy per kg, because



Now I want to accelerate the same object by another 2 km/s, so I have to use another 2 Megajoule per kg.

Now the object has a velocity of 4 km/s.

I have invested 4 Megajoule per kg.

The object, having a velocity of 4 km/s, has a kinetic energy of 8 Megajoule per kg, according to the formula.

So I have used up 4 Megajoule per kg, but I got 8 Megajoule out of it.

How is that possible?
It is possible because Work = Force times Distance = Energy. Check the link.

« Last Edit: December 10, 2018, 10:21:21 PM by rabinoz »

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FalseProphet

  • 3696
  • Life is just a tale
Did Isaac Newton, NASA and possibly Hillary Clinton deceive the world about kinetic energy?
What could NASA and Hillary Clinton have to do with it?

Isaac Newton's and James Prescott Joule's shoulders are strong enough to take the blame.

Quote from: FalseProphet
I have a problem again.
When I want to accelerate an object from 0 to 2 km/s I need 2 Megajoule of energy per kg, because



Now I want to accelerate the same object by another 2 km/s, so I have to use another 2 Megajoule per kg.

Now the object has a velocity of 4 km/s.

I have invested 4 Megajoule per kg.

The object, having a velocity of 4 km/s, has a kinetic energy of 8 Megajoule per kg, according to the formula.

So I have used up 4 Megajoule per kg, but I got 8 Megajoule out of it.

How is that possible?
It is possible because Work = Force times Distance = Energy. Check the link.

That's a dead link.

I get more energy out than I put into it, because energy is force x distance? How so? Isn't there something like a law of energy conservation?

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rabinoz

  • 26528
  • Real Earth Believer
Did Isaac Newton, NASA and possibly Hillary Clinton deceive the world about kinetic energy?
What could NASA and Hillary Clinton have to do with it?

Isaac Newton's and James Prescott Joule's shoulders are strong enough to take the blame.

Quote from: FalseProphet
I have a problem again.
When I want to accelerate an object from 0 to 2 km/s I need 2 Megajoule of energy per kg, because


Now I want to accelerate the same object by another 2 km/s, so I have to use another 2 Megajoule per kg.
Now the object has a velocity of 4 km/s.

I have invested 4 Megajoule per kg.

The object, having a velocity of 4 km/s, has a kinetic energy of 8 Megajoule per kg, according to the formula.

So I have used up 4 Megajoule per kg, but I got 8 Megajoule out of it.

How is that possible?
It is possible because Work = Force times Distance = Energy. Check the link.

That's a dead link. << not any more >>

I get more energy out than I put into it, because energy is force x distance? How so? Isn't there something like a law of energy conservation?
No the work you do to accelerate an object is the kinetic energy stored in it.

Suppose you apply a constant force of  20 N to accelerate a 1 kg object from 0 to 2 km/s
        The change in velocity is 2 km/s so this will take 100 secs and the average velocity is 1 km/s so it will travel 100 km (100 000 m).
         Hence the work done on it in is 20 x 100 000 J or  your 2 MJ.

Now to accelerate the 1 kg object from 2 to 4 km/s" will still take 100 secs but
         the average velocity is now 3 km/s so it will travel 300 km (300 000 m).
         Hence the work done on it this time is 20 x 300 000 J or  6 MJ adding up to the total of 8 MJ of stored kinetic energy.

When applied to rockets this means that the rockets "efficiency" keeps increasing as the rocket accelerates but don't worry, you never get "something for nothing".

Hope it helps.

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FalseProphet

  • 3696
  • Life is just a tale
Did Isaac Newton, NASA and possibly Hillary Clinton deceive the world about kinetic energy?
What could NASA and Hillary Clinton have to do with it?

Isaac Newton's and James Prescott Joule's shoulders are strong enough to take the blame.

Quote from: FalseProphet
I have a problem again.
When I want to accelerate an object from 0 to 2 km/s I need 2 Megajoule of energy per kg, because


Now I want to accelerate the same object by another 2 km/s, so I have to use another 2 Megajoule per kg.
Now the object has a velocity of 4 km/s.

I have invested 4 Megajoule per kg.

The object, having a velocity of 4 km/s, has a kinetic energy of 8 Megajoule per kg, according to the formula.

So I have used up 4 Megajoule per kg, but I got 8 Megajoule out of it.

How is that possible?
It is possible because Work = Force times Distance = Energy. Check the link.

That's a dead link. << not any more >>

I get more energy out than I put into it, because energy is force x distance? How so? Isn't there something like a law of energy conservation?
No the work you do to accelerate an object is the kinetic energy stored in it.

Suppose you apply a constant force of  20 N to accelerate a 1 kg object from 0 to 2 km/s
        The change in velocity is 2 km/s so this will take 100 secs and the average velocity is 1 km/s so it will travel 100 km (100 000 m).
         Hence the work done on it in is 20 x 100 000 J or  your 2 MJ.

Now to accelerate the 1 kg object from 2 to 4 km/s" will still take 100 secs but
         the average velocity is now 3 km/s so it will travel 300 km (300 000 m).
         Hence the work done on it this time is 20 x 300 000 J or  6 MJ adding up to the total of 8 MJ of stored kinetic energy.

When applied to rockets this means that the rockets "efficiency" keeps increasing as the rocket accelerates but don't worry, you never get "something for nothing".

Hope it helps.

Actually it does help. Before I knew what I did not understand. Now I don't.

Cause on the one hand I can follow the calculation.

But on the other hand, that feels like a paradox to me. I always thought, for non-relativistic speed, when I add a constant amount of energy I get a constant acceleration. But now it seems I need 3 times the amount of fuel to go from 2 km/s to 4 km/s than I need to go from 0 km/s to 2 km/s. That is perplexing.

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rabinoz

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  • Real Earth Believer
Actually it does help. Before I knew what I did not understand. Now I don't.

Cause on the one hand I can follow the calculation.

But on the other hand, that feels like a paradox to me. I always thought, for non-relativistic speed, when I add a constant amount of energy I get a constant acceleration.
That might be the problem. Isaac Newton's Second Law of Motion boils down to force = mass x acceleration so a constant force gives a constant acceleration.

Quote from: FalseProphet
But now it seems I need 3 times the amount of fuel to go from 2 km/s to 4 km/s than I need to go from 0 km/s to 2 km/s. That is perplexing.
The rate of adding energy is power and power = force x velocity so the power you need (roughly rate of burning fuel) goes up with your velocity.

There are a number of physics tutorial sites and the Physics Classrooms isn't a bad one.
You can start at the beginning or dive into particular topics, though it's a good idea to "start at the very beginning, A very good place to start".

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FalseProphet

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Actually it does help. Before I knew what I did not understand. Now I don't.

Cause on the one hand I can follow the calculation.

But on the other hand, that feels like a paradox to me. I always thought, for non-relativistic speed, when I add a constant amount of energy I get a constant acceleration.
That might be the problem. Isaac Newton's Second Law of Motion boils down to force = mass x acceleration so a constant force gives a constant acceleration.

Quote from: FalseProphet
But now it seems I need 3 times the amount of fuel to go from 2 km/s to 4 km/s than I need to go from 0 km/s to 2 km/s. That is perplexing.
The rate of adding energy is power and power = force x velocity so the power you need (roughly rate of burning fuel) goes up with your velocity.


But wait! When you calculate the energy you need to accelerate a rocket, you only ask for delta-v. You do not ask how fast the rocket was at the first place. So it seems for rockets the initial velocity does not matter.

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rabinoz

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Actually it does help. Before I knew what I did not understand. Now I don't.

Cause on the one hand I can follow the calculation.

But on the other hand, that feels like a paradox to me. I always thought, for non-relativistic speed, when I add a constant amount of energy I get a constant acceleration.
That might be the problem. Isaac Newton's Second Law of Motion boils down to force = mass x acceleration so a constant force gives a constant acceleration.

Quote from: FalseProphet
But now it seems I need 3 times the amount of fuel to go from 2 km/s to 4 km/s than I need to go from 0 km/s to 2 km/s. That is perplexing.
The rate of adding energy is power and power = force x velocity so the power you need (roughly rate of burning fuel) goes up with your velocity.


But wait! When you calculate the energy you need to accelerate a rocket, you only ask for delta-v. You do not ask how fast the rocket was at the first place. So it seems for rockets the initial velocity does not matter.
Yes, you are right there.

The initial velocity matters very much in calculating the delta-V required for a manoeuvre.
It is true, however, that the initial velocity does not matter in the burn (fuel used) to achieve that delta-V.
There is no conservation of energy paradox because the fuel the rocket needs for that delta-V has been accelerated (has added kinetic energy) by previous burns.

There are some interest follow-ons of this.
One is that a rocket can be accelerated in the gravitational field of a planet and then the delta-V is applied at this higher velocity.
This is called an Oberth maneuver and is described in Wikipedia, Oberth effect. This is a "powered fly-by" as opposed to the "passive fly-by" of a simple "gravity assist" manoeuvre.

Careful, you'll soon be pushing my limited knowledge of Rocket Science ;).

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FalseProphet

  • 3696
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Actually it does help. Before I knew what I did not understand. Now I don't.

Cause on the one hand I can follow the calculation.

But on the other hand, that feels like a paradox to me. I always thought, for non-relativistic speed, when I add a constant amount of energy I get a constant acceleration.
That might be the problem. Isaac Newton's Second Law of Motion boils down to force = mass x acceleration so a constant force gives a constant acceleration.

Quote from: FalseProphet
But now it seems I need 3 times the amount of fuel to go from 2 km/s to 4 km/s than I need to go from 0 km/s to 2 km/s. That is perplexing.
The rate of adding energy is power and power = force x velocity so the power you need (roughly rate of burning fuel) goes up with your velocity.


But wait! When you calculate the energy you need to accelerate a rocket, you only ask for delta-v. You do not ask how fast the rocket was at the first place. So it seems for rockets the initial velocity does not matter.
Yes, you are right there.

The initial velocity matters very much in calculating the delta-V required for a manoeuvre.
It is true, however, that the initial velocity does not matter in the burn (fuel used) to achieve that delta-V.
There is no conservation of energy paradox because the fuel the rocket needs for that delta-V has been accelerated (has added kinetic energy) by previous burns.

There are some interest follow-ons of this.
One is that a rocket can be accelerated in the gravitational field of a planet and then the delta-V is applied at this higher velocity.
This is called an Oberth maneuver and is described in Wikipedia, Oberth effect. This is a "powered fly-by" as opposed to the "passive fly-by" of a simple "gravity assist" manoeuvre.

Careful, you'll soon be pushing my limited knowledge of Rocket Science ;).

I am not sure if I understand you. What do you mean by "the fuel has been accelerated". How can that matter?

Would it make a difference if I carry the fuel or if I would use some other propulsion methods, like for example momentum exchange tethers or anything that does not require a fuel? Wouldn't different propellants that have different specific energies have a different impact on delta-v?

Sorry, now I'm really confused.

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wise

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is a shit because then two cars run opposite direction and crush then stops; energy goes to zero. But according to Newton it turns to energy of "crash, voice, heat, ets". It does not wash!
1+2+3+...+∞= 1

Come on bro, just admit that the the earth isn't a sphere, you won't even be wrong

is a shit because then two cars run opposite direction and crush then stops; energy goes to zero. But according to Newton it turns to energy of "crash, voice, heat, ets". It does not wash!
Wash?

Energy goes to zero and thats why a car crash equals explosions and crushed metal
Be gentle

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sokarul

  • 19303
  • Extra Racist
He seems to be using wash as in equal.

https://www.merriam-webster.com/dictionary/it's%20a%20wash

I wonder what translater translates equal to wash.
ANNIHILATOR OF  SHIFTER

It's no slur if it's fact.

Ah thanks for that
Be gentle

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wise

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He seems to be using wash as in equal.

https://www.merriam-webster.com/dictionary/it's%20a%20wash

I wonder what translater translates equal to wash.

It is not "it's wash". It is "It does not wash". Try again.



https://lmgtfy.com/?q=%22it%20does%20not%20wash%22what%20means
1+2+3+...+∞= 1

Come on bro, just admit that the the earth isn't a sphere, you won't even be wrong

Calculus it.
Youre using the proportional formula.
You need to use the full formula.
Your velocityprime is not zero the 2nd time.

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rabinoz

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is a shit because then two cars run opposite direction and crush then stops; energy goes to zero. But according to Newton it turns to energy of "crash, voice, heat, ets". It does not wash!
Either show where Newton's theories claim any of that or admit that you are totally ignorant on these matters.

You are undoubtedly referring to the "Law of Conservation of Energy" and that did not come from Isaac Newton, who lived from 1643 to 1727.

But "In 1842, Julius Robert Mayer discovered the Law of Conservation of Energy."

And whether you like it or not, or accept it or not, "noise" and "heat" are forms of energy, so you are wrong on all counts!

Wise, I do wish that you would stop cluttering up the threads simply to prove that you are totally ignorant about anything to do with Science!

is a shit because then two cars run opposite direction and crush then stops; energy goes to zero. But according to Newton it turns to energy of "crash, voice, heat, ets". It does not wash!
Either show where Newton's theories claim any of that or admit that you are totally ignorant on these matters.

You are undoubtedly referring to the "Law of Conservation of Energy" and that did not come from Isaac Newton, who lived from 1643 to 1727.

But "In 1842, Julius Robert Mayer discovered the Law of Conservation of Energy."

And whether you like it or not, or accept it or not, "noise" and "heat" are forms of energy, so you are wrong on all counts!

Wise, I do wish that you would stop cluttering up the threads simply to prove that you are totally ignorant about anything to do with Science!

No wise is correct.
Kinetic energy does go to zero, if assuming all the parts stop and none fly off.
However, it is transferred to thermal energy, sound energy, light energy, breaking of bonds energy all sorts of not kinetic energy because by definition...

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rabinoz

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No wise is correct.
Kinetic energy does go to zero, if assuming all the parts stop and none fly off.
Yes, "Kinetic energy does go to zero".

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sokarul

  • 19303
  • Extra Racist
He seems to be using wash as in equal.

https://www.merriam-webster.com/dictionary/it's%20a%20wash

I wonder what translater translates equal to wash.

It is not "it's wash". It is "It does not wash". Try again.



https://lmgtfy.com/?q=%22it%20does%20not%20wash%22what%20means

The English you quoted does not agree with you.
ANNIHILATOR OF  SHIFTER

It's no slur if it's fact.

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wise

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No, it does.
1+2+3+...+∞= 1

Come on bro, just admit that the the earth isn't a sphere, you won't even be wrong

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rabinoz

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  • Real Earth Believer
No, it does.
But you were still wrong when you wrote
is a shit because then two cars run opposite direction and crush then stops; energy goes to zero. But according to Newton, it turns to energy of "crash, voice, heat, etc". It does not wash!
Because:
  • The "Law of Conservation of Energy" was not from Isaac Newton, who died in 1727 but from Julius Robert Mayer discovered the Law of Conservation of Energy 1842.

  • And whether you like it or not, or accept it or not, the kinetic energy of a vehicle before a crash is turned into "noise" and "heat" and these are forms of energy.

So whether you think it washes or not you are wrong on all counts![/quote]

Heres your issue in reverse, as another example for you, wise.

Your car is off.
You get in and turn it on.
The gas is lit on fire converting chemical potential energy into heat and kinetic.
Your car makes sound, gets warm, goes forward.
Eventually your tank is empty because youve converted all that chemical gasoline into sound, heat, motion.
You give the gas attendant money and repeat process.

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sokarul

  • 19303
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No, it does.

What comes after "it does not wash" is important.
ANNIHILATOR OF  SHIFTER

It's no slur if it's fact.

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Heiwa

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Did Isaac Newton, NASA and possibly Hillary Clinton deceive the world about kinetic energy?

I have a problem again.

When I want to accelerate an object from 0 to 2 km/s I need 2 Megajoule of energy per kg, because



Now I want to accelerate the same object by another 2 km/s, so I have to use another 2 Megajoule per kg.

Now the object has a velocity of 4 km/s.

I have invested 4 Megajoule per kg.

The object, having a velocity of 4 km/s, has a kinetic energy of 8 Megajoule per kg, according to the formula.

So I have used up 4 Megajoule per kg, but I got 8 Megajoule out of it.

How is that possible?
First of all you must establish what Force (N) to apply to your object (kg) to reach  speed 2000 m/s neglecting influence of gravity forces of other objects in the vicinity. You need 1N to accelerate 1 kg 1 m/s assuming no other forces are involved and then it takes 2000 seconds to reach 2000 m/s speed.
If you apply 10 N to 1 kg, you will accelerate 10 m/s and it takes 200 seconds.
If you apply 100 N to 1 kg, you will accelerate 100 m/s and it takes only 20 seconds.
So how do you apply force 1 or 100 N to a 1 kg object during 2000 or 20 seconds ? Do you use a rocket engine? So just use a 1N rocket engine and fire it during 2000 seconds and it will increase the speed of any 1 kg object attached to it to 2000 m/s.
Use a rocket engine with low fuel consumption (kg/s)! I prefer a fantastic rocket engine with 0 fuel consumption. Then you will never run out of fuel and can fly anywhere and ignore the effect of gravity from celestial bodies. Your man to do it is Elon Musk of SpaceshitX company. He is today sending things into LEO on behalf of NASA. I describe it at http://heiwaco.com/moontravele.htm

« Last Edit: December 13, 2018, 07:12:09 AM by Heiwa »

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Pezevenk

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Lol it's not that confusing, it's just how kinetic energy is defined... Increasing the velocity two times increases the kinetic energy 4 times, it shouldn't be a surprise...
Member of the BOTD for Anti Fascism and Racism

It is not a scientific fact, it is a scientific fuck!
-Intikam

Read a bit psicology and stick your imo to where it comes from
-Intikam (again)

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FalseProphet

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Lol it's not that confusing, it's just how kinetic energy is defined... Increasing the velocity two times increases the kinetic energy 4 times, it shouldn't be a surprise...

For me the confusing thing is that how much energy you need to accelerate an object by a given value depends on the initial velocity of an object. After all velocity is relative.

My mistake seems to be that I think of kinetic energy as if it was force x time instead of force x length. But that is impetus, not kinetic energy. Historically that is also what physicists thought prior to the times of Newton and Bernoulli, when they asked for the "true measure of movement", so I am in good company, when you can call people who have been dead for 300 years "good company" that is.

But why is the formula for kinetic energy 1/2 mv^2 and not just mv^2?

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markjo

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Did Isaac Newton, NASA and possibly Hillary Clinton deceive the world about kinetic energy?

I have a problem again.

When I want to accelerate an object from 0 to 2 km/s I need 2 Megajoule of energy per kg, because



Now I want to accelerate the same object by another 2 km/s, so I have to use another 2 Megajoule per kg.

Now the object has a velocity of 4 km/s.

I have invested 4 Megajoule per kg.

The object, having a velocity of 4 km/s, has a kinetic energy of 8 Megajoule per kg, according to the formula.

So I have used up 4 Megajoule per kg, but I got 8 Megajoule out of it.

How is that possible?
It's possible because you aren't using the right formula to calculate the change in kinetic energy.  You have to take the initial conditions into account when calculating the change.
Science is what happens when preconception meets verification.
Quote from: Robosteve
Besides, perhaps FET is a conspiracy too.
Quote from: bullhorn
It is just the way it is, you understanding it doesn't concern me.

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FalseProphet

  • 3696
  • Life is just a tale
Did Isaac Newton, NASA and possibly Hillary Clinton deceive the world about kinetic energy?

I have a problem again.

When I want to accelerate an object from 0 to 2 km/s I need 2 Megajoule of energy per kg, because



Now I want to accelerate the same object by another 2 km/s, so I have to use another 2 Megajoule per kg.

Now the object has a velocity of 4 km/s.

I have invested 4 Megajoule per kg.

The object, having a velocity of 4 km/s, has a kinetic energy of 8 Megajoule per kg, according to the formula.

So I have used up 4 Megajoule per kg, but I got 8 Megajoule out of it.

How is that possible?
It's possible because you aren't using the right formula to calculate the change in kinetic energy.  You have to take the initial conditions into account when calculating the change.


Thanks, people have already pointed that out.

Can you explain me why, to accelerate a rocket, I only have to consider delta-v and not the initial velocity of the rocket, when calculating the energy I need?

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rabinoz

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Thanks, people have already pointed that out.

Can you explain me why, to accelerate a rocket, I only have to consider delta-v and not the initial velocity of the rocket, when calculating the energy I need?
The energy needed is not usually calculated explicitly but is implicit in the Tsiolkovsky rocket equation as described in:
Quote from: Wikipedia
Tsiolkovsky rocket equation
The Tsiolkovsky rocket equation, classical rocket equation, or ideal rocket equation, describes the motion of vehicles that follow the basic principle of a rocket: a device that can apply acceleration to itself using thrust by expelling part of its mass with high velocity and thereby move due to the conservation of momentum.

The equation relates the delta-v (the maximum change of velocity of the rocket if no other external forces act) with the effective exhaust velocity and the initial and final mass of a rocket, or other reaction engine.

For any such maneuver (or journey involving a sequence of such maneuvers):
The equation relates the delta-v (the maximum change of velocity of the rocket if no other external forces act) with the effective exhaust velocity and the initial and final mass of a rocket, or other reaction engine.

For any such maneuver (or journey involving a sequence of such maneuvers):
         
where:
          Δv is delta-V the maximum change of velocity of the vehicle (with no external forces acting).
          m0 is the initial total mass, including propellant, also known as wet mass.
          mf is the final total mass without propellant, also known as dry mass.
          ve is the effective exhaust velocity relative to the rocket.
          ln is the natural logarithm function.
The simplest way to explain it is that the initial rocket mass, including fuel (often RP-1 and LOX), is m0 but only the final mass, mf, is accelerated to the final velocity, vf.

If the rocket is being accelerated from an initlial velocity, say v0, any fuel in the rocket is already travelling at v0.  Hence the energy is in this fuel is its own chemical energy plus its kinetic energy because it is moving at v0.

The SpaceX Falcon Heavy has a launch mass on 1,420,788 kg but puts "only" about 63,800 kg (a lot of that is usually more propellant) into Low Earth Orbit.

I hope this makes sense, because I'm finding it hard to put it simply - probably because i'm no Rocket Scientist so have an imperfect understanding myself.

<< Fixed a few v's thst were m's  :( >>
« Last Edit: December 13, 2018, 04:21:29 PM by rabinoz »

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FalseProphet

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Thanks, people have already pointed that out.

Can you explain me why, to accelerate a rocket, I only have to consider delta-v and not the initial velocity of the rocket, when calculating the energy I need?
The energy needed is not usually calculated explicitly but is implicit in the Tsiolkovsky rocket equation as described in:
Quote from: Wikipedia
Tsiolkovsky rocket equation
The Tsiolkovsky rocket equation, classical rocket equation, or ideal rocket equation, describes the motion of vehicles that follow the basic principle of a rocket: a device that can apply acceleration to itself using thrust by expelling part of its mass with high velocity and thereby move due to the conservation of momentum.

The equation relates the delta-v (the maximum change of velocity of the rocket if no other external forces act) with the effective exhaust velocity and the initial and final mass of a rocket, or other reaction engine.

For any such maneuver (or journey involving a sequence of such maneuvers):
The equation relates the delta-v (the maximum change of velocity of the rocket if no other external forces act) with the effective exhaust velocity and the initial and final mass of a rocket, or other reaction engine.

For any such maneuver (or journey involving a sequence of such maneuvers):
         
where:
          Δv is delta-v the maximum change of velocity of the vehicle (with no external forces acting).
          m0 is the initial total mass, including propellant, also known as wet mass.
          mf is the final total mass without propellant, also known as dry mass.
          me is the effective exhaust velocity.
          ln is the natural logarithm function.
The simplest way to explain it is that the initial rocket mass, including fuel (often RP-1 and LOX), is m0 but only the final mass, mf, is accelerated to the final velocity, mf.

If the rocket is being accelerated from an initlial velocity, say v0, any fuel in the rocket is already travelling at v0.  Hence the energy is in this fuel is its own chemical energy plus its kinetic energy because it is moving at v0.

The SpaceX Falcon Heavy has a launch mass on 1,420,788 kg but puts "only" about 63,800 kg (a lot of that is usually more propellant) into Low Earth Orbit.

I hope this makes sense, because I'm finding it hard to put it simply - probably because i'm no Rocket Scientist so have an imperfect understanding myself.

I am not sure if that is what I mean. When you use means of acceleration without propellant there is no difference between m0 and mf, but still you only consider delta-v, not initial velocity. But maybe that is because you always work in a gravitational field and the delta-v in this case corresponds to lower and higher orbit.