Mass of fuel would ALSO increase as spaceship comes near the speed of light

In summary, the conversation discusses the concept of relativistic mass and its relationship to the acceleration and energy of a spacecraft moving at high speeds. The conversation also addresses misunderstandings about the increase in relativistic mass equating to a real change in the quantity of fuel, and explains how the perspective of the observer affects their perception of the ship's mass and acceleration. Ultimately, the nature of the fuel is irrelevant in this scenario, as the key factor is the spaceship's ability to accelerate at a constant rate.
  • #36
@janus

Also, in your in an earlier post you mention that a ship could cross a vast distance in 20 yrs by its perspective and thus burn as much fuel as needed to travel at 1g for 20 years. You seemed to think that this would result in it burning less fuel form its perspective than as seen from an outside observer. You are wrong. If you actually calculated out how much fuel the ship figures it needs to travel at 1g for 20 yrs, it would equal the amount of fuel the outside observer determines that he needs. Because according to the outside observer, his acceleration drops off as he nears c.

JK- No the outsider would actually see less and less fuel being consumed in the spaceship as it approaches c. Just as he/shed sees all processes in the ship slow down, fuel is no exception. What i thought was that the inside observer would not suddenly experience more fuel consumption as he/she gets closer to c and wants to experience the same accelration g on his/her body.
 
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  • #37
I will say it again you should actually read the "How much fuel is needed?" part as it explains this.

Something is changing inside the ship, fuel is being burned so the ship is lossing the mass from the fuel as it's being burned off. The example he gave was calculated so that he started out each trip with the exact amount of fuel needed to make the trip. so for the short trip he only brought 10kg of fuel. But for the next trip he brought 57kg of fuel, he needed that much more fuel since for the first part of the flight he needs to accelerate the fuel for the entire flight. If you were too look at the last 3 years of the flights he would only be using that 10kg as the rest of the fuel will have burned off and he will no longer need to accelerate the mass of fuel he started with.
 
  • #38
@darkhorror

darkhorror said:
I will say it again you should actually read the "How much fuel is needed?" part as it explains this.

Something is changing inside the ship, fuel is being burned so the ship is lossing the mass from the fuel as it's being burned off. The example he gave was calculated so that he started out each trip with the exact amount of fuel needed to make the trip. so for the short trip he only brought 10kg of fuel. But for the next trip he brought 57kg of fuel, he needed that much more fuel since for the first part of the flight he needs to accelerate the fuel for the entire flight. If you were too look at the last 3 years of the flights he would only be using that 10kg as the rest of the fuel will have burned off and he will no longer need to accelerate the mass of fuel he started with.

JK- But if the mass of the fuel is burnt off then he should actually require even less fuel in the next 3 yrs of travel as compared to the first 3.6 yrs for his experiencing the same g on his body or no? And isn't he accelerating the entire flight anyways?
 
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  • #39
Please read the entire post before posting.

The fuel consumption does NOT remain constant to keep at a 1g acceleration. As the trip continues less and less fuel is consumed, since the mass of the fuel is diminishing as the trip continues. That is exactly what it is shown in the link that was given. If you have to go twice as far you are going to need much more than twice the mass of fuel because for the first part of the trip you have to accelerate the fuel for the rest of the trip.

Lets look at what they gave.
3.6 years Nearest star 10 kg
6.6 years Vega 57 kg

The fuel that is given is the fuel that they start the trip with. with the second ship going to Vega, the first 3.6 years does NOT consume 10kg of fuel but much more. It should have consumed 47kg of fuel in the first 3 years, then 10kg the last 3.6 years. As for the first 3 years it has to accelerate that extra 10kg of fuel, along with the rest of the unconsumed fuel. But when it only has 3.6 years to go it should only have 10kg of fuel left.
 
  • #40
@darkhorrow

darkhorror said:
Please read the entire post before posting.

The fuel consumption does NOT remain constant to keep at a 1g acceleration. As the trip continues less and less fuel is consumed, since the mass of the fuel is diminishing as the trip continues. That is exactly what it is shown in the link that was given. If you have to go twice as far you are going to need much more than twice the mass of fuel because for the
first part of the trip you have to accelerate the fuel for the rest of the trip.

JK- But i thght the ship is accelerating throughout the whole trip.

Lets look at what they gave.
3.6 years Nearest star 10 kg
6.6 years Vega 57 kg

The fuel that is given is the fuel that they start the trip with. with the second ship going to Vega, the first 3.6 years does NOT consume 10kg of fuel but much more. It should have consumed 47kg of fuel in the first 3 years, then 10kg the last 3.6 years. As for the first 3 years it has to accelerate that extra 10kg of fuel, along with the rest of the unconsumed fuel. But when it only has 3.6 years to go it should only have 10kg of fuel left.
[/quote]

JK- Oh i thought that this chart is about the same ship passing by these pts and at each of these they just mentioned the amount of fuel consumed whilst it passes by them. I also thought this one ship is continously accelerating. So none of these ships are accelrating for almost their whole journey till very close to their destination?
 
  • #41
jonnyk said:
@janus



What i thought was that the inside observer would not suddenly experience more fuel consumption as he/she gets closer to c and wants to experience the same accelration g on his/her body.

No one is saying that he would. What does happen is that the acceleration of the ship decreases as seen by the external observer.
 
  • #42
jonnyk said:
@darkhorrow



JK- But i thght the ship is accelerating throughout the whole trip.

they are and are accelerating at 1g through the entire trip.

JK- Oh i thought that this chart is about the same ship passing by these pts and at each of these they just mentioned the amount of fuel consumed whilst it passes by them. I also thought this one ship is continously accelerating. So none of these ships are accelrating for almost their whole journey till very close to their destination?

They are different ships, each of these different ships do continuously accelerate at 1g. I am not sure where you are getting that they don't accelerate through there entire trip.
 
  • #43
jonnyk said:
@darkhorrow



JK- But i thght the ship is accelerating throughout the whole trip.

It is. Think of it this way. It takes more fuel to accelerate 2 kg up to a certain speed than it does to accelerate 1 kg. Say it takes 1kg of fuel to accelerate 1kg. After the fuel is gone, you can not accelerate any more.

Now let's say that you want to keep accelerating after that to double your final speed. That means that after the first leg of acceleration above you want to have 1kg of fuel left. But that means that at the start of your trip you have to accelerate 2kg (your initial 1kg +the fuel you want to have remaining.) So you have to burn 2kg of fuel during the first leg, so you have to start with 3 kg of fuel not 2 kg double your final speed.
JK- Oh i thought that this chart is about the same ship passing by these pts and at each of these they just mentioned the amount of fuel consumed whilst it passes by them. I also thought this one ship is continously accelerating. So none of these ships are accelrating for almost their whole journey till very close to their destination?

Yes the are accelerating the whole trip. And the amounts given are the starting fuel needed to reach those destinations with zero fuel left when you get to them.
 
  • #44
@darkhorror

darkhorror said:
they are and are accelerating at 1g through the entire trip.

JK- Ok

They are different ships, each of these different ships do continuously accelerate at 1g. I am not sure where you are getting that they don't accelerate through there entire trip.

JK- Ok so the 2nd shiip has 57kg fuel on board correct? So yes it would require more energy to get it to the same speed as ship 1. Now you say the 2nd ship would require abt 47kg fuel for the same distance traveled as the 1st ship which used 10kg because it has more fuel. HOWEVER isn't the mass of both ships equal? So if say the spaceships weigh 10000kg how come there's so much difference in accelerating a total initial mass of 10010kg as compared to 10057kg? Still doesn't make sense to me.
 
  • #45
"How much fuel is this? The next chart shows the amount of fuel needed (M) for every kilogramme of payload (m=1 kg)."
 
  • #46
@darkhorror

darkhorror said:
"How much fuel is this? The next chart shows the amount of fuel needed (M) for every kilogramme of payload (m=1 kg)."

JK- Is the small "m" the rest mass of the spaceship?
 
  • #47
Yes, you need 10kg of fuel for 1kg of spaceship.
 
  • #48
@darkhorror

darkhorror said:
Yes, you need 10kg of fuel for 1kg of spaceship.

JK- OH BOY! so in that article they are assuming a rocket of rest mass 1kg. Any decent sized spaceship would weigh around 50 tons at least so thatd make 500T fuel just to get to nearest star then. And that is assuming 100% efficiency too and converting all fuel matter into energy.
They talk about fusion at the bottom. Do you think that way it could work?
 
  • #49
jonnyk said:
@darkhorror



JK- OH BOY! so in that article they are assuming a rocket of rest mass 1kg.

No, they were giving a "mass ratio". They were telling you how much fuel would be needed per kilogram of rocket mass.
 
  • #50
something else you might want to think about is that the ship and fuel mass increasing close to the speed of light is going to put you closer and closer to collapsing into a black hole so the extra fuel mass will get you nowhere past a certain point. i do not remember where or how much but somebody somewhere has calculated how much energy can be in a local area before collapsing and it's not infinity. if you rearrange e=mc^2 you get m=e/c^2. the mass of a particular amount of energy is equal to it's energy in joules divided by the speed of light in meters per second squared. once this reaches 3.4 times the mass of sol it will collapse into a black hole.
 
  • #51
schonovic said:
something else you might want to think about is that the ship and fuel mass increasing close to the speed of light is going to put you closer and closer to collapsing into a black hole so the extra fuel mass will get you nowhere past a certain point. i do not remember where or how much but somebody somewhere has calculated how much energy can be in a local area before collapsing and it's not infinity. if you rearrange e=mc^2 you get m=e/c^2. the mass of a particular amount of energy is equal to it's energy in joules divided by the speed of light in meters per second squared. once this reaches 3.4 times the mass of sol it will collapse into a black hole.

Er.. please be careful with your response. The "m" in that equation is the invariant mass. Relativistic mass is not an appropriate term anymore. Please read the FAQ in the General Physics forum.

Your application of Special Relativity is faulty here. You are also responding to a thread that is more than a year old.

Zz.
 
  • #52
sorry about responding to an older thread. i didn't think i used the term relativistic mass in that response but o.k., what does invariant mass mean? do you mean that a heavy mass close to the speed of light will not collapse into a black hole even if it is massive as a black hole?
 
  • #53
Did you even attempted to read the FAQ that was suggested?

You did refer implicitly to relativistic mass when you said:

...the ship and fuel mass increasing close to the speed of light...

If that isn't a reference to relativistic mass, then you've just violated conservation of energy.

Zz.
 
  • #54
i seem to be missing the point, even if the fuel mass increases with the ship mass: so what, nothing can exceed the speed of light anyways so why do we care if the fuel mass increases...does it make a difference?
 
  • #55
schonovic said:
i seem to be missing the point, even if the fuel mass increases with the ship mass: so what, nothing can exceed the speed of light anyways so why do we care if the fuel mass increases...does it make a difference?

Yes, because you are using that faulty "mass increase" to cause a "... collapse into a black hole.."

I'm assuming that you now know what "invariant mass".

Zz.
 
  • #56
o.k. Zz i stand corrected. it is invarient mass. i simply use m=e/c^2 to calculate how much recoil the weapon systems on my science fiction starships experience. if i fire a 10^17 joule laser on a 450,000 M.T. starship how much kick will the ship experience? however it seems i can't do it that way. maybe i'll have to learn a new way, thanks.
 
  • #57
schonovic said:
o.k. Zz i stand corrected. it is invarient mass. i simply use m=e/c^2 to calculate how much recoil the weapon systems on my science fiction starships experience. if i fire a 10^17 joule laser on a 450,000 M.T. starship how much kick will the ship experience? however it seems i can't do it that way. maybe i'll have to learn a new way, thanks.
If m is the rest mass rather than the relativistic mass, then E=mc^2 is the energy of an object at rest (so 0 kinetic energy), for a moving object the total energy is given by E^2 = m^2 c^4 + p^2 c^2, where p is the relativistic momentum given by p = mv/sqrt(1 - v^2/c^2) for massive objects (for photons this formula for relativistic momentum doesn't work, but since a photon has m=0 you can reduce the previous formula to E=pc for photons, so p=E/c for photons). If you let M equal the relativistic mass, so M=m/sqrt(1 - v^2/c^2), it turns out that the formula for the total energy of a moving object reduces to E=Mc^2.

So if all the energy of the laser goes into increasing the linear kinetic energy of the starship (not a totally realistic assumption but never mind), and if the starship was initially at rest in the frame where the oncoming laser had an energy of 10^17 joules, then before being accelerated the ship must have had a rest energy of E = (450,000,000 kg)*(299792458 m/s)^2 = 4.04439830431568 * 10^25 kg m^2 / s^2 = 4.04439830431568 * 10^25 joules. Then after it's accelerated, the new total energy of the ship must be 4.04439830431568 * 10^25 + 10^17 joules = 4.04439831431568 * 10^25 joules. So this energy must be the one that comes out of E=Mc^2=mc^2/sqrt(1 - v^2/c^2), which means we have:

4.04439831431568 * 10^25 joules = 4.04439830431568 * 10^25 joules / sqrt(1 - v^2/c^2)

This indicates that the gamma-factor 1/sqrt(1 - v^2/c^2) must be given by:

(4.04439831431568 * 10^25 joules) / (4.04439830431568 * 10^25 joules) = 1.00000000247256

Solving 1/sqrt(1 - v^2/c^2) = 1.00000000247256 for v gives a velocity of about 7*10^-5 * c, or about 21,000 meters/second, in about the same range as the Voyager 1 spacecraft (around 17,000 meter/second).

edit: sorry, I thought you wanted to know how much kick the ship would gain if it absorbed the energy of a laser, but rereading I see you were asking about the recoil when it fires the laser. This is a bit more complicated, since in order to shoot out a laser the ship must convert some internal potential energy into photons, and thus its rest mass will actually change slightly (the rest mass of a multiparticle system includes internal potential energy and heat along with the rest mass of all the particles individually), along with a change in its kinetic energy. I'm not sure how to solve this problem exactly, although if m is its rest mass before firing the laser and m' is the rest mass after firing, then since energy is conserved the following should be true in the frame where the ship was initially at rest:

mc^2 = m'c^2/(1 - v^2/c^2) + 10^17 joules

...since m'c^2/(1 - v^2/c^2) incorporates both the ship's rest energy and its kinetic energy after firing the laser, and the total energy after firing the laser is (ship's rest energy) + (ship's kinetic energy) + (laser's energy).
 
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  • #58
schonovic said:
sorry about responding to an older thread. i didn't think i used the term relativistic mass in that response but o.k., what does invariant mass mean? do you mean that a heavy mass close to the speed of light will not collapse into a black hole even if it is massive as a black hole?
If it is not a black hole in the frame where it is at rest then it will not be a black hole in any other frame regardless of its total energy (which can be arbitrarily high) in that other frame.
 
  • #59
JesseM thanks for taking all the effort over a simple science fiction spaceship. I'm sorry i did not actually completely elaborate on how i calculate the kick my laser gives my ship. i first use m=e/c^2 to calculate the mass of the laser energy, then i multiply the lasers mass times its velocity to figure how many Newtons of force it exerts durring the fring. then i apply that force to the mass of my ship to figure the kick of the laser. if this is wrong i simply figure it is for science fiction as long as i always follow the same rules for all ships I'm o.k. I'm just a simple sci-fi creator and do not have a degree in physics.
 
  • #60
schonovic said:
JesseM thanks for taking all the effort over a simple science fiction spaceship. I'm sorry i did not actually completely elaborate on how i calculate the kick my laser gives my ship. i first use m=e/c^2 to calculate the mass of the laser energy, then i multiply the lasers mass times its velocity to figure how many Newtons of force it exerts durring the fring.
That won't work at all--mass times velocity gives momentum, not force! (force is mass * acceleration) But figuring out the momentum (p) of the laser should actually work, since conservation of momentum implies the ship should gain an equal and opposite momentum (in the frame where the ship was originally at rest before firing the laser). For photons with have zero rest mass, the equation E^2 = m^2 c^4 + p^2 c^2 reduces to E=pc, so the momentum of the laser (and the equal and opposite momentum of the ship after firing) would be given by p=E/c. For a laser with energy 10^17 joules = 10^17 kg meters^2 / second^2, that works out to a momentum of (10^17 kg meters^2 / second^2 ) / (299792458 meters/second) = 333564095 kg meters/second. So that should be the momentum change of the ship, and although relativistic momentum is given by m*v/sqrt(1 - v^2/c^2), for velocities that are small compared to the speed of light it's reasonably accurate to just use the classical momentum m*v. So:

m*v = 333564095 kg meters/second
v = (333564095 kg meters/second) / m

And with m = 450000000 kg, this gives the ship a velocity of 0.74 meters/second after firing the laser. But caution, this is a very different answer from the one I got using energy--since this calculation is a lot simpler, I think the previous one probably contained an error somewhere.

Anyway, to sum up, if a ship of mass m fires a laser with energy E, then the ship's change in velocity v (in the frame where it was previously at rest) can be calculated with this simple equation:

v = E/mc

(make sure your units are consistent though--for example, if you use joules for energy, then since 1 joule = 1 kilogram*meter^2/second^2, you must use kilograms for the mass m of the ship and c = 299792458 meters/second)
 
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  • #61
i know you say my way will not work but i'v tried it your way and my way and they get the exact same answer. it's my presentation that is faulty. I'm not describing my method in clear detail. i'll try again. first i use m=e/c^2. then i multiply m*light speed to get Newtons. then i divide Newtons by ships mass to get velocity after firing the laser. maybe it works because you could be accelerating @ 300000000m/sec^2 for one second. it's wrong i guess i just got lucky somehow it gets the same answer. anyways thanks for the work and the nice handy dandy equation. I'm glad we have guys like you to steer us fumbling laymen in the right direction again thanks and let me get off this year ond thread. bye.
 
  • #62
Hi jonnyk
If you look closer at the discussion of relativistic rockets at Baez's site you'll notice total energy is conserved. Read it a bit closer and you'll see what I mean.
 
  • #63
qraal said:
Hi jonnyk
If you look closer at the discussion of relativistic rockets at Baez's site you'll notice total energy is conserved. Read it a bit closer and you'll see what I mean.
Egads! I just noticed how old the post I am replying to was. Oops.
 
  • #64
mass times velocity gives momentum but that momentum and mass applied to an inert object has an effect.
 

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