# What is the formula to calculate how much fuel is needed for a rocket?

I'm theoretically trying to launch a rocket for a school project and therefore I chose Saturn V. I took its weight of 140.000kg (308 647 pounds) and I'm trying to calculate how much fuel it'd take to launch it at two different locations, one on the equator and the other one at the poles.

The gravity acceleration I've already calculated (Equator: $9,797 m/s^2$; Poles: $9,863m/s^2$) as well as the distance, which is $1,414213*10^7$ meters.

Now I'm stuck, for I don't know how to involve the gravitational force declination in my calculation.

I'd like to calculate how many joules (J) are needed to launch a rocket to a height of $1,414213*10^7$ meters.

At first I did that with the formula F_{res}=g$*m$, but that doesn't involve the declination of gravitational force.

As for air resistance, I'd like to calculate that as well, but I think I'll manage on my own.

What formulas or rules of thumb should I use?

• The Tsiolkovsky rocket equation is relevant, but only one part to the puzzle (albeit a very important one). See Is this a correct understanding of Tsiolkovsky's rocket equation? and more generally the rocket-equation tag. See also Wikipedia on the Tsiolkovsky rocket equation. – a CVn Dec 17 '16 at 17:55
• Can you rewrite this so it's just one question? – Organic Marble Jan 21 '17 at 18:58
• One formula you need is $F=G{{M_1 M_2} \over {r^2}}$ where $M_1$ and $M_2$ are masses of Earth and the ship, $G$ is the universal gravitational constant, $r$ is distance between centers of mass of Earth and ship (so $r_0$ is local Earth radius, accounting for the geoid distortion.) Combining this with changing thrust as rocket expels fuel, changing centrifugal acceleration (due to Earth rotation, different at different latitudes) this will result in an absolutely ugly integral. Seriously reconsider your task conditions. – SF. Jan 21 '17 at 20:23
• Why is there so much hate in these responses and question (I mean such negative scores)? – Matthew Oct 7 '17 at 17:11

The problem as asked is impossible to solve. The problem is the energy needed is dependent on your rocket's acceleration and on drag losses.

While I have no proof there is no equation that will answer your question I certainly have never heard of one or seen any indication thereof. You almost certainly have to do a brute force simulation.

Also, while gravity losses will be constant for a rocket going straight up, if your target altitude is high enough it can be more efficient to burn horizontally instead to reduce your gravity loss. As you build horizontal velocity the gravity loss will drop.

• Ah alright, I think I won't involve the drag losses and will take a constant acceleration up to the escape velocity of earth. As for gravitity, The rocket will go straight up all the time. So I'll try to think of a way to calculate the constant gravity losses in my calculation. – Akimoto Dec 18 '16 at 15:22

You haven't provided sufficient information about the specific impulse of the engines you've decided to use, rocket geometry, and flight trajectory. Without these, I cannot help you narrow down on propellant mass to reach final height and velocity (assuming you're simulating an actual Saturn V flight with orbital insertion of payload).

Use rocket equations (I call them momentum equations). $\Delta m$ obtained for imparting the required $\Delta v$ will be the propellant mass required. Realistically, you'd want to add some contingent fuel on top of that to account for drag. If you want a realistic figure, run simulations (using MATLAB?) with changing air density with altitude and flight velocity to compute for instantaneous drag. $(\Delta v + dv)$ will be your new $\Delta v$ for the engine to impart with $dv$ being velocity lost from drag.

You can account for change in gravity based on launch latitude by increasing burn time to achieve same $\Delta v$ because it's not practical to redesign the engine and fuel tanks for greater thrust. The burn time can be computed by writing momentum equations from your inertial frame of reference and then, multiply specific impulse by obtained $\Delta m$ all divided by thrust.

Mathematically, the work done in joules by the engines will be the integral of the thrust curve. Assuming a single-staged rocket with constant linear acceleration, work done will be thrust multiplied by burn time.

Remember, there are multiple ways to mathematically approach this problem. What approach you take depends on the variables you have established and have yet to establish.

Not a simple task, it may take awhile to understand if you don't have prior knowledge to this field.

Assuming you are talking about Rocketdyne F-1 which is the main engine for Saturn V, only calculating the first stage and neglating drag with launch angle of 80 degrees.

Specs:

• 35100 KN in Atm
• $I_{sp}=263 s \text{ (atm)}$
• $I_{sp}=304s \text{ (vac)}$
• Weight with propellant = 5040000 lbs
• Net weight = 287000

For convenience I am just going to take the average specific impulse which is $${(263/304)\over 2}=283.5$$ Mass Flow Rate: 4753000 lb/165 seconds = 212.72 lb/s
Burn Time = 165 seconds.

Now use the formula to determine initial acceleration in y-axis $$(a_0)_y=g_0[F sin \Theta/w)−1]$$ Where $g_0=9.81m/s^2$ or $32.17ft/s^2$
F=force=35100KN
w=weight with propellant

So we get $$32.17ft/s∗[{35100KN∗0.9848\over 22419.03KN}]−1=17.43ft/s^2$$ For the x-axis use the formula $$(a_0)_x=g_0[Fcos\Theta/w]$$ $$32.17ft/s∗[{351000KN∗0.1736\over 22419.03KN}]=87.44ft/s^2$$

For the terminal velocity (where the burn ends): $$(u_p)_y=cIn(m_0/m_f)sin\Theta−t_pg_0$$ c = exhaust velocity
In = natural log
$m_0$ = weight with propellant
$m_f$ = weight after propellant is consumed
$t_p$ = burn time

$$32.17∗283.5In(5040000/287000)∗0.984−165∗32.17= 20409.33 ft/s^2$$ or 6220.76m/s or 13915.5mph

Roughly 0.52 times the earth escape velocity.

(Originally posted on Quora: Finding Delta-V for the First Stage of Saturn V)

• While this may theoretically answer the question, it is preferable to include the essential parts of the answer here, and provide the link for reference. – Nathan Tuggy Jan 21 '17 at 9:22
• @NathanTuggy Since I am extremely new to Stack Exchange, copying formulas down from Quora is difficult for me, do you mind suggest an edit for me? – Raze Jan 21 '17 at 9:27
• @NathanTuggy Actually, I will rewrite the answer on computer tomorrow morning since I have found a major flaw in my calculation. It is 1:30AM in LA – Raze Jan 21 '17 at 9:29
• Look, I'm not saying it's a great question but it does say this: I'd like to calculate how many joules (J) are needed to launch a rocket to a height of 1,414213∗107 meters. Where is that in your answer? I don't think you can just pull some answer from somewhere else and fit the bill on this one. – Organic Marble Jan 21 '17 at 18:27
• I removed the quote formatting. You don't need to use quotes when you are the person who said it in the first place. Stack Exchange normally considers quote-only answers low quality and they are often removed. Welcome to Space Exploration. – kim holder wants Monica back Jan 21 '17 at 21:08