Jump to content
Sign in to follow this  
Baby Astronaut

Energy needed to reach 99% c?

Recommended Posts

That depends on the mass of the object you we are talking about... but:

 

E2=(mc2)2 + p2c2

 

(I might have made a mistake in that don't use it very often)

 

But it's the total energy equation, if you remove the rest energy (mc2) you get the kinetic energy...

Share this post


Link to post
Share on other sites

Or KE = [math](\gamma -1)mc^2[/math]

 

Where [math]\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/math]

Share this post


Link to post
Share on other sites

My level of math.

 

+

-

X

/

2x + 5x = 7x.

 

My level of science.

 

E=mc2 (never actually calculated it)

The rest is things in basic English, with a few terms here and there.

 

Is it possible for me to get an answer like, "To move a kilogram up to 99% light's speed, it would take (number with digits only) joules of energy. And two kilograms would take double the number of joules"?

 

Or is that beyond the abilities of science?

Share this post


Link to post
Share on other sites

Since life is about letting everyone else do everything for you... Heh, well, I'll guide you through it.

 

Or KE = [math](\gamma -1)mc^2[/math]

 

Where [math]\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/math]

 

[math]\gamma = \frac{1}{\sqrt{1-\frac{0.99c^2}{c^2}}} = \frac{1}{\sqrt{1-0.99}} = \frac{1}{0.1} = 10[/math]

 

[math]KE = (\gamma -1)mc^2 = 9mc^2[/math]

 

There, this very simple equation describes the energy needed for a velocity of 0.99c. Now you just plug in the mass. For example 1kg to 0.99c requires... uh, about 808.88 PJ (petajoules). That's quite a bit, for example Tzar Bomba, the most powerful nuclear weapon ever detonated, released 50 megatons or 209.2 petajoules of energy. This is why "relativistic kill vehicles" are awesome.

 

I'm suffering a mild 37.5ish C fever so I might be completely botching up even these kinds of calculations, so some verification might be nice. Actually my result (808.88 PJ) is quite a bit higher than the convenient 547 PJ given in the Wikipedia RKV article but at least it's around the same orders of magnitude. Oh well.

Edited by Gilded

Share this post


Link to post
Share on other sites
Since life is about letting everyone else do everything for you... Heh, well actually I'm just bored.

 

 

 

[math]\gamma = \frac{1}{\sqrt{1-\frac{0.99c^2}{c^2}}} = \frac{1}{\sqrt{1-0.99}} = \frac{1}{0.1} = 10[/math]

 

[math]KE for v of 0.99c = (\gamma -1)mc^2 = 9mc^2[/math]

 

There. Now you just plug in the mass. For example 1kg to 0.99c requires... uh, about 808.88 PJ (petajoules). That's quite a bit, for example Tzar Bomba, the most powerful nuclear weapon ever detonated, released 50 megatons or 209.2 petajoules of energy.

 

I'm suffering a mild 37.5ish C fever so I might have completely botched the calculation up though, so some verification might be nice.

 

In your delirium you forgot to square the .99

 

Gamma is about 7 for this speed

Share this post


Link to post
Share on other sites
In your delirium you forgot to square the .99

 

Bleh, that's something I might've done even while in full health though. :P Thanks for the correction.

 

Gamma is about 7 for this speed

 

Yep, I got it now. [math]\gamma = \frac{1}{\sqrt{0.0199}} \approx{7.089}[/math]

 

Thus

[math]KE = (\gamma -1)mc^2 = (\frac{1}{\sqrt{0.0199}}-1)mc^2[/math]

Edited by Gilded
I can't seem to get anything right today :D

Share this post


Link to post
Share on other sites
I figured 6.65e9 Volts for a proton mass. Is this right?
Well all you have to do is plug the data in this equation:

 

[math]E=\frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}[/math]

Share this post


Link to post
Share on other sites
I figured 6.65e9 Volts for a proton mass. Is this right?

That depends on what you mean:

- 6.7e9 Volt definitely is not the proton mass.

- It is not even a mass.

- It is not an energy, either.

- The voltage needed to accelerate a proton in an electric field to 99% of c is not the value stated, either.

So it's probably not correct.

 

Well all you have to do is plug the data in this equation:

 

[math]E=\frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}[/math]

That does neither give the proton mass (that he said he was looking for but which simply is ~1000 MeV/c²), nor something in units of voltage (what his result was), nor the kinetic energy of a proton at 99% of c (what the thread was about). What it gives is the relativistic mass of an object.

Share this post


Link to post
Share on other sites

Proton mass is about 939 MeV, or rounded up, 1 GeV. So you'd want a KE of ~6 GeV

Share this post


Link to post
Share on other sites
That does neither give the proton mass (that he said he was looking for but which simply is ~1000 MeV/c²), nor something in units of voltage (what his result was), nor the kinetic energy of a proton at 99% of c (what the thread was about). What it gives is the relativistic mass of an object.
Sorry! My Wrong! I wasn't thinking:-(

Share this post


Link to post
Share on other sites
That depends on what you mean:

- 6.7e9 Volt definitely is not the proton mass.

- It is not even a mass.

- It is not an energy, either.

- The voltage needed to accelerate a proton in an electric field to 99% of c is not the value stated, either.

So it's probably not correct.

 

 

That does neither give the proton mass (that he said he was looking for but which simply is ~1000 MeV/c²), nor something in units of voltage (what his result was), nor the kinetic energy of a proton at 99% of c (what the thread was about). What it gives is the relativistic mass of an object.

 

 

What is the voltage (potential) required to accelerate a single proton to c?

Share this post


Link to post
Share on other sites

Since a photon by definition is always travelling at c, can it legitimately be said to accelerate? It's not like it starts at zero mph and works it's way up. It's always travelling at c, so I posit that it never "accelerates." Can anyone confirm or reject this conjecture based on something more substantial than my WAG?

Share this post


Link to post
Share on other sites
Since a photon by definition is always travelling at c, can it legitimately be said to accelerate? It's not like it starts at zero mph and works it's way up. It's always travelling at c, so I posit that it never "accelerates." Can anyone confirm or reject this conjecture based on something more substantial than my WAG?

Proton, not photon :)

I meant .99c

Best of luck on getting an answer.

 

To whom it may concern: I browsed the web seeking tips on how to ask questions, and while that site has great advice for how to ask a question, there's one glaring flaw in the article. Sometimes there's a good reason for a person to write a brief question: to save the reader time. To write out that I did do research and found nothing uses your precious time. And having you read my explanation here does likewise. Another glaring flaw: everyone has their area of expertise. Sometimes you get help solving a problem and other times you help someone else.

 

Honestly, how would I learn calculus in a timely manner to do the equation? If someone can point me to an online calculator and a good tutorial on how to use the calculator with no knowledge, then perhaps, but otherwise, I need a starting point. From that info, I'll be able to calculate other things.

 

For example, if you say it takes 4,000 Joules to bring a 30-ton object to 99.99% light-speed, and 3,000 Joules to bring a 22.5 ton object to 99.99% light-speed, then I can do the math from there onwards and calculate that it'll take 200,000 Joules to bring an 1,500 ton object to 99.99% light-speed.

 

But if you give me equations that I don't understand, it's a waste of time for both of us.

Edited by Baby Astronaut

Share this post


Link to post
Share on other sites

To whom it may concern: I browsed the web seeking tips on how to ask questions, and while that site has great advice for how to ask a question, there's one glaring flaw in the article. Sometimes there's a good reason for a person to write a brief question: to save the reader time. To write out that I did do research and found nothing uses your precious time. And having you read my explanation here does likewise. Another glaring flaw: everyone has their area of expertise. Sometimes you get help solving a problem and other times you help someone else.

 

Honestly, how would I learn calculus in a timely manner to do the equation? If someone can point me to an online calculator and a good tutorial on how to use the calculator with no knowledge, then perhaps, but otherwise, I need a starting point. From that info, I'll be able to calculate other things.

 

For example, if you say it takes 4,000 Joules to bring a 30-ton object to 99.99% light-speed, and 3,000 Joules to bring a 22.5 ton object to 99.99% light-speed, then I can do the math from there onwards and calculate that it'll take 200,000 Joules to bring an 1,500 ton object to 99.99% light-speed.

 

But if you give me equations that I don't understand, it's a waste of time for both of us.

 

I appreciate your position. However …

 

How is someone answering a question to know your level of math skills? I don't find it unreasonable to expect algebra (not calculus) skills to be a given.

 

Also, one has to appreciate that some questions do not have simple answers, and that for the answer to make some sense, you need to have learned some background information. At some point, the asker is responsible for obtaining that education. The link you provided had some good insight into this.

 

Your original question was ill-formed, and eventually an answer was given. You need a kinetic energy of 6 times the mass energy

 

What is the voltage (potential) required to accelerate a single proton to c?

 

I meant .99c

 

Not a simple situation, because once you accelerate a charge, it starts radiating, so it loses energy. The basic equation is that E = qV, so it's a Volt of potential per electron-Volt of energy. Ignoring that energy loss, the answer is ~6GV.

Share this post


Link to post
Share on other sites
I figured 6.65e9 Volts [...]. Is this right?

 

What is the voltage (potential) required to accelerate a single proton to [99% of] c?

 

Not a simple situation, because once you accelerate a charge, it starts radiating, so it loses energy. The basic equation is that E = qV, so it's a Volt of potential per electron-Volt of energy. Ignoring that energy loss, the answer is ~6GV.

Even though the correct answer to the original and some additional questions are already flying around in this thread, there is so much confusion about so basic things that I find it appropriate to post a little summary of equations and solutions.

  1. I'll assume that above [math]\gamma = 7[/math] is correct. See above for the definition of [math]\gamma[/math]. There was no confusion, as far as I see.
  2. The rest mass of a proton is [math]m = 938[/math] MeV/c² (off my head, the value might be off by ~5 but that will not matter).
  3. The total energy of a (free) proton at rest is [math]E_0 = 938[/math] MeV. This value comes from the mass plugged in into the (in-)famous E=mc².
  4. The total energy of a (free) moving proton is [math] E_\gamma = \gamma E_0 [/math]. In this case [math] E_\gamma = 7 \cdot 0.938 \cdot 10^9 \ eV \ = 6.57 \cdot 10^9 [/math] eV. That is pretty much the value you (=gre) had, except that you called it "mass of the proton" and called the unit "Volt" rather than electronvolt. The relativistic mass (perhaps that is what you meant by mass?) of this proton would be [math]6.57 \cdot 10^9 [/math] eV/c². Note the factor c² that make the units match.
  5. The total energy is the sum of the kinetic energy and the energy at rest. To get the kinetic energy (what the thread originally was about) you have to substract the rest energy from the total energy, i.e. [math]E_{kin} = E_\gamma - E_0 = 5.63 \cdot 10^9 [/math] eV. This is the step that a lot of people in this thread might have had a problem with.
  6. To accelerate a proton to such an energy in an electric field since the proton carries one elementary charge (apart from the sign, depending on definition) the potential difference the proton has to go through -ignoring any effects of losing energy along the way- would then be [math]5.63 \cdot 10^9 [/math] V.

Share this post


Link to post
Share on other sites
How is someone answering a question to know your level of math skills? I don't find it unreasonable to expect algebra (not calculus) skills to be a given.

Post #3 I highlighted my math skills.

 

Also, one has to appreciate that some questions do not have simple answers, and that for the answer to make some sense, you need to have learned some background information. At some point, the asker is responsible for obtaining that education. The link you provided had some good insight into this.

That link had flaws I pointed out, but yes, all questions do in fact have simpler answers than the usual ones. I only asked how many joules of energy is needed to go 99% the speed of light. I did neglect to supply the amount of mass, but I'm sure a person could have said "well, you didn't specify the mass, but if it were a grain of sand, it would take ___ joules to accelerate it". A simple answer is possible.

 

Your original question was ill-formed,

Very true. Not purpusely, though.

 

and eventually an answer was given.

I'm probably the only person who missed it and still does.

 

You need a kinetic energy of 6 times the mass energy

Thanks. It's the first instance I actually learned something in this thread. A useful piece of information that can be used to arrive at other conclusions.

 

For example, if the pure energy converted from a bucket of sand would power New York City for several days, or if the pure energy converted from a grain of sand is enough to boil 10 million kettles, then I can make the following deductions.

 

The amount of energy needed to accelerate a bucket of sand to 99.99% light-speed would be the nearly equal to the amount of energy needed to power New York City for approximately 20 days to a month.

 

The amount of energy needed to accelerate a grain of sand to 99.99% light-speed would be the nearly equal to the amount of energy needed to boil 60 million kettles.

 

I can visualize the energy requirements to approach 99.99% c much better, and I'm certain others reading this can now too.

 

It's one of the principle reasons for having first started another thread where I learned squat.

 

People aren't obligated to supply answers, but the "do the math yourself" reasoning is a probably a fallacy of sorts. Many people here browse Wikipedia to supplant their knowledge. Yet instead of it being dotted with "figure it out yourself" statements, Wikipedia has lots of things spelled out by volunteers. After reading it, you're on better footing to research its contributed material and thus ensure its accuracy.

 

I do lots of inter-disciplinary research, and so I must focus attention where my talents lay, and depend on the courteous help of those who understand the more vague areas better than me.

 

Otherwise, learning becomes inefficient. I do my part in helping others understand. If someone had their college professor explain and re-explain the material after class, and they still cannot quite grasp it, I'll go read the textbook chapter, deduce what parts they likely don't understand (and reasoning out why not), and then I'll review the chapter with the person but substituting in my words and personalized descriptions and ignoring the book's terminology. Comprehension dawns in their eyes, and often a laugh of how simple it could be.

 

The person goes on to get an "A" on the test.

 

Had I just tossed the person a book and said "read and learn", then how would it advance their knowledge?

 

We're all great at some things, and poor at others. Sometimes one does need things spelled out. I for my part do attempt to find out on my own first, but when you've searched in the library, and your eyes gloss over the paragraphs of jargon, and it takes many readings to simply end up not having grasped even the basic concepts, then is it really a bad thing to just ask for it straight?

 

The information I seek is usually a precursor onto a further level of understanding. I'm better at visualizing than doing formulas. I'll see the universe's mechanics better in that light. So please, take into account how our minds perhaps function differently and with piles of other research on my plate, I don't focus on learning a new discipline whenever I hit a snag, but rather I'll follow a more efficient path and be helped when necessary, and in turn I'll help others (experts included) who lack my expertise.

 

It's about communication, the transportation for ideas.

Share this post


Link to post
Share on other sites

Can we start like stars do. Give huge energy at the start only in an instant. start with a very high velocity by giving huge enery at the start. In universe all the big changes take place in a very short while anyways.

Share this post


Link to post
Share on other sites
E=mc2 (never actually calculated it)

 

 

c=fy

and

c2 = F / md

Written as one equation

c2 = (fy)2 = F / md

 

Where;

c = velocity of light

f = frequency

y = Compton wavelength values given on CODATA proton http://physics.nist.gov/cgi-bin/cuu/Value?pcomwlelectron http://physics.nist.gov/cgi-bin/cuu/Value?ecomwl

F = Force = constant (ka) by amplitude (x) = kax

md = mass density = mass (m) / wavelength (y) = m / y

Let us assume that amplitude is proportional to wavelength

F = kax = kwy

Energy (E) = force (kwy) by distance (y) thus;

E = kwy2

Einstein's E=mc2

c2 = F / md

We can re-write this substituting F = kwy and md = m / y

c2 = kwy2 / m

Substitute E= kwy2

 

Thus E = mc2

To deduce mass

kwy2 = mc2

m = kwy2 / c2

m = kwy2 / (yf)2

Thus m = kw / f2

Planck's constant h and E = hf

Substitute c = fy into E = mc2

E = mc2 = mcfy = (mcy)f

If we substitute known values of mass(m), c(speed of light) and y(compton wavelength) for electron we get;

 

mcy = 9.1 × 10-31× 3 × 108 × 2.4 ×10-12 = 6.5 ×10-34 = Planck's constant h

Thus h = mcy

E = (mcy)f

Thus E = hf

 

It is important to note that the compton wavelength is still subject to some uncertainty and as such value slight smaller or greater than hf are to be expected

 

Given all of the above

 

e=mcfy

e=mcv

e=pc

 

where p is the momentum. where c is much less than the speed of light we use the classical kinetic energy + the rest energy

 

Kinetic Energy Ek = mv 2/ 2

Rest Energy Er = mc2

 

E = mv2/2 + mc2

 

which reduces to

 

E^2= (pc)2+ (mc2)2

Edited by fiveworlds

Share this post


Link to post
Share on other sites

Can we start like stars do. Give huge energy at the start only in an instant. start with a very high velocity by giving huge enery at the start. In universe all the big changes take place in a very short while anyways.

 

Of course. It doesn't make any difference if you accelerate really slowly or really quickly. It still takes the same amount of energy.

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
Sign in to follow this  

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.