ROCKET FORMULA

[loucca]

Hi, I was wondering if someone can help me with the rocket formula.

Use the rocket formula v = s ln(Mfuel/Mpayload - 1) to find the ratio Mfuel/Mpayload in the case that v = 100s. This could correspond to the case of a chemical fuel with exhaust velocity s = 3 km/s and a final velocity after fuel exhaustion of v = 300 km/s which is only 0.1% of light speed. If your rocket weight 100 tons, how much fuel would you need, in tons?

If the first answer is in Kpc.. is it 1,5 Kpc?

I got lost in the second question but I would like to know how to do it

Could anybody help me? Thanks.

[glowjet]

Hi Loucca and welcome to SGL, this is essentially an Astronomy forum, and questions so diverse as relating to rocket science could, quite well, not receive any response. Although it is not without possibility that some knowledgeable forum member may have the answer to your inquiry. If your interests also happen to be Astronomy related, please enjoy the forum

[Tiny Small]

Hi, I was wondering if someone can help me with the rocket formula.

Use the rocket formula v = s ln(Mfuel/Mpayload - 1) to find the ratio Mfuel/Mpayload in the case that v = 100s. This could correspond to the case of a chemical fuel with exhaust velocity s = 3 km/s and a final velocity after fuel exhaustion of v = 300 km/s which is only 0.1% of light speed. If your rocket weight 100 tons, how much fuel would you need, in tons?

If the first answer is in Kpc.. is it 1,5 Kpc?

I got lost in the second question but I would like to know how to do it

Could anybody help me? Thanks.

If you are talking about movement in space, your velocity and acceleration is calculated by how much mass you dump out the back and how much mass you have to move. This isn't quite as simple to calculate as it first appears as it is not just a force calculation. As the mass of your fuel source is changing constantly, the mass you have to move is also changing constantly and as a result you need to calculate the integral of the graph of fuel and original mass combined. So you need to be able to use integral calculus.

If you are talking about escaping earth, it's the same calculation but taking into consideration a negative acceleration due to gravity of almost 10ms^-2 and achieving and escape velocity of 7000 ms^-1.

This of course means that you need to phrase your question mathematically in order to form a quadratic equation of the form ax²+bx+c=0 so that you can calculate the integral whilst also calculating the efficiency of your fuel as the only fuel that is 100% efficient is total fusion... which doesn't show up anywhere in our observable universe.

The easiest way to do this is to use already known information (from previous launches and force calculations etc) to form a graph on a Cartesian plane with one axis representing acceleration and the other representing the declining mass. From there you can relatively easily form a quadratic equation and thus calculate the definite integral from 0 ms^-1 to final velocity in ms^-1 of the fuel requirement.

The fuel requirement would be equal to the total area under the graph right upto the vertical asymptote of the parabola, approaching infinitely close to the line if allowed to continue and approaching the line to a finite degree if velocity is capped, hence the requirement for integral calculus. Oh.. and a working knowledge of conics. Obviously.

[glowjet]

I had a feeling that one of our members would have the knowledge to help in some way, just goes to show how helpful SGL can be. Thank you for your input Tiny Small, although I am afraid it is way over my head 

[Ruud]

Use the rocket formula v = s ln(Mfuel/Mpayload - 1) to find the ratio Mfuel/Mpayload in the case that v = 100s. This could correspond to the case of a chemical fuel with exhaust velocity s = 3 km/s and a final velocity after fuel exhaustion of v = 300 km/s which is only 0.1% of light speed. If your rocket weight 100 tons, how much fuel would you need, in tons?

You want to solve  ln( Mfuel / Mpayload + 1) = 100.  (Note the +, I think you made a mistake with the minus)

This gives Mfuel / Mpayload = e^100 - 1

or              Mfuel = 2.688 * 10^43 * Mpayload 

So you would need 2.688 * 10^45 tons of fuel for your 100 ton payload.

[Tiny Small]

Okay, having read into this a little more, there is a much simpler way to solve this than using integration and that is to use the propellent mass fraction:

1-e ^{- delta V/v_e}

Delta V = Total change in Velocity

V_e = constant exhaust velocity

Lets assume that you want to escape earth gravity first. This means you will need a propellent capable of delivering high v_e so a bi-propellent would be the preferred method. These are capable of (2900, 4500] ms^-1. Assume the higher of these. Next, for simplicity, lets assume a single stage burn.

1-e ^-300/4.5 = 1 meaning 100% of your payload would need to be fuel. Actually, you get the same result with a delta V of 200,000 ms^-1, meaning that with a propellent capable of delivering a v_e of 4500 ms^-1, this equation will never work. You either need to reduce your final velocity to around ¹/₃₀ of C, increase your v_e to around 20,000 ms^-1 or increase your starting velocity from 0 ms^-1.

The simplest of these is to reduce your delta V. Assume a delta V of 99,000 ms^-1.

1-e^-99/4.5 = 0.9999999997

This means that 99.99999997% of your rocket mass needs to be fuel, or, for a rocket with a mass of 100,000 kg, it all needs to be fuel apart from 3 one hundredths of a gram. Not really feasible.

Alternatively, if you find some sort of magic fuel with a v_e of 100,000 ms^-1, your 300,000 ms^-1 is feasible.

1-e^-300/100 = 0.9502129316 ~ 0.95, meaning that with your 100,000 kg rocket, 95 tonnes would need to be fuel and you would have 5 tonnes to play with for the engine, fuel housing and payload.

[Astro_Baby]

WOW !

Check out the big brains on display there  Well done you guys......

Its waaaaay beyond my needs for rocketry but then I am nit usually launching 100 ton payloads

[Tiny Small]

In my case, hardly. I constantly struggle with maths. To be honest, I have never really looked into rocketry and just approached it as a general maths problem initially. After having looked into it a little more and discovered the limitations of propellant and then having read through the not insubstantial proofs and derivations and subsequent integrations, it's really peaked my interest. The simple equation of:

Delta V = v_eln(m₀/m₁)

is really quite beautiful in its subtlety. It hides an enormous amount of information and possibilities behind such a small face.

[Astro_Baby]

In my case, hardly. I constantly struggle with maths. To be honest, I have never really looked into rocketry and just approached it as a general maths problem initially. After having looked into it a little more and discovered the limitations of propellant and then having read through the not insubstantial proofs and derivations and subsequent integrations, it's really peaked my interest. The simple equation of:

Delta V = v_eln(m₀/m₁)

is really quite beautiful in its subtlety. It hides an enormous amount of information and possibilities behind such a small face.

Uh Huh yeah right - I'll believe that