2-stage rocket - minimum mass necessary

In summary, the conversation discusses the task of designing a two-stage rocket to accelerate a payload of 100kg to a final velocity of 6,000 m/s using the minimum amount of mass for the rocket stages. The rocket must also follow a restriction that the mass of the rocket stage and its fuel tanks must be ten times the amount of fuel. The conversation mentions an equation for the final velocity of a rocket, as well as a brute force method using a Java function to find the minimum mass for the rocket stages. The final answer is (m10 + m20) = 41773.24 kg, with m10 = 39828.264 kg and m20 = 1944.976 kg. The conversation also acknowledges
  • #1
hafman
1
0

Homework Statement



I have to design a two-stage rocket to accelerate a payload of 100kg from rest to a final velocity of 6,000 m/s (no gravity to worry about) using the minimum amount of mass for the rocket stages.

Each stage of the rocket must obey the following restriction:
Whatever the mass of the rocket stage & its fuel tanks, there must be 10x as much fuel. So if rocket stage 1 was 10,000 kg, then there must be 100,000 kg worth of fuel for a total (initial) mass of 110,000 kg in the first stage. Ditto with stage 2.

The rocket has a constant thrust velocity of 1500 m/s

The difficulty is I have to find the 'minimum' initial total mass of both stages required to get the payload up to the required final velocity.


Homework Equations



I know that in general the final velocity of a rocket is:

v = v0 + u * ln(m0/m1), so a two stage rocket would give:

[tex]v2 = u * ln(\frac{m0*m2}{m1*mf})[/tex]

Where:
v0 = initial velocity of the rocket, either 0.0 m/s, or whatever the final speed of the last stage was.

m0 = initial total mass of the rocket before it begins burning.
m1 = final mass of the rocket after burning the stage 1 fuel.
m2 = initial total mass of stage 2 (anything left of stage 1 was ejected)
mf = final mass of the rocket - that's the stage 2 parts minus the fuel, and the payload.

The Attempt at a Solution



I'll use the following:
m10 = initial mass of stage 1, including the fuel.
m20 = initial mass of stage 2, including fuel.
mp = mass of payload.
u = 1500 m/s, exhaust velocity.

[tex]vf = u * ln( \frac{(m10 + m20 + mp) * (m20 + mp)}{(1/11*m10 + m20 + mp)*(1/11*m20 + mp)} )[/tex]

This big mess of ugly comes out from combining the two stages through the properties of logarithms. Stage 2 has an initial velocity which is the final velocity of stage 1.

I get the 1/11*m10 and m20 in the denominator from assuming that all of the fuel is burned, so 10/11 of the mass of that stage is gone based on the restrictions stated earlier.

So this quickly turns into an algebra nightmare and I have to somehow find the minimum value of:

[tex]m10 + m20[/tex] to get a final velocity of 6,000 m/s.

I have to admit I'm kind of lost, I've tried multiplying through all the terms hoping to cancel stuff out, but only wound up back where I started.

However I have solved it by writing a small java function to brute force through all possible values of m10 and m20, and I get the following answer after a few billion calculations:

(m10 + m20) = 41773.24 kg
m10 = 39828.264 kg
m20 = 1944.976 kg

That's "good enough" for me, and to be honest it'll probably be accepted for full credit when I turn it in with a copy of the source code.

However I'd still like to know the "correct" way to derive this. My guess is that I have to rework the equation in terms of the mass as a function of velocity and use some calculus from there. But I honestly could use a few hints.

Here's the code (java) for the function I wrote:

public static void rocket()
{
double payload = 100.0; // The payload the rocket needs to haul.
double v = 6000.0; // The target velocity this rocket needs to reach.
double u = 1500.0; // Thrust velocity.
double delta = 0.001; // The change in mass for m10 and/or m20 for a brute force loop.
double v2 = 0.0; // The calculated guess, to compare to v.

// The minimum mass of the rocket (m10 + m20) - starting at an unrealistically high level so that
// almost anything will be lower.
double minMass = 1000000000.0;
String minLine = ""; // Prints out a formatted line of all the data for the minimum values.

double m10Start = 39800.0;
double m10End = 40000.0;
double m20Start = 1940.0;
double m20End = 1945.0;

double m10 = m10Start;
double m20 = m20Start;

for(m10 = m10Start; m10 < m10End; m10 += delta)
{
for(m20 = m20Start; m20 < m20End; m20 += delta)
{
count++;
v2 = u * Math.log(((m10 + m20 + payload)*(m20 + payload)) / ((1.0/11.0*m10 + m20 + payload)*(1.0/11.0*m20 + payload)));
if(v2 >= 6000.0)
{
if((m10 + m20) < minMass)
{
minMass = m10 + m20;
minLine = m10 + " + " + m20 + " = " + (m10 + m20) + " --- velocity = " + v2;
}
}

}
}

System.out.println(minLine);
System.out.println("Permutations run: " + count);
}

Sorry that it doesn't look great, the tabs are lost in copy/paste and I don't know if you guys have a nice code formatter similar to latex for math.
 
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  • #2
But it should be easy enough to follow. It prints out the following:39828.264 + 1944.976 = 41773.24 --- velocity = 6001.299574651702Permutations run: 3640000Thanks for any help in advance!
 

1. What is a 2-stage rocket and how does it work?

A 2-stage rocket is a type of rocket that is designed to have two separate stages or sections. The first stage usually contains the main engines and is responsible for launching the rocket off the ground. Once the first stage has used up all of its fuel, it is detached and the second stage takes over to continue the journey into space.

2. What is the minimum mass necessary for a 2-stage rocket?

The minimum mass necessary for a 2-stage rocket will vary depending on the specific design and mission requirements. However, in general, the first stage of a 2-stage rocket will need to be heavy enough to provide enough thrust to lift the entire rocket off the ground, while the second stage will need to be light enough to be carried by the first stage but still have enough fuel to continue the journey into space.

3. How does the minimum mass affect the performance of a 2-stage rocket?

The minimum mass of a 2-stage rocket is directly related to its performance. If the rocket is too heavy, it may not have enough thrust to lift off the ground or may not have enough fuel to complete its mission. On the other hand, if the rocket is too light, it may not have enough power to reach its intended destination. Finding the optimal minimum mass is crucial for a successful launch and mission.

4. What factors determine the minimum mass necessary for a 2-stage rocket?

The minimum mass of a 2-stage rocket is determined by several factors, including the desired altitude and speed of the rocket, the payload it is carrying, and the specific fuel and engine technology being used. Other considerations may include the atmospheric conditions and launch location.

5. How is the minimum mass calculated for a 2-stage rocket?

The minimum mass for a 2-stage rocket is typically calculated by engineers using complex mathematical equations that take into account the various factors mentioned above. These calculations are then used to determine the optimal design and fuel requirements for the rocket to achieve its intended mission. Advanced computer simulations and testing are also often used to refine and validate the minimum mass calculation.

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