How do you find the propellant mass needed to reach an inputted altitude?(altitude at end of burn plus altitude during coast)
The first thing you should be considering for any rocket mission is the specific energy required. It is conventional in rocketry to use "delta V" ($\Delta V$) as a specific energy metric. You can understand this intuitively as the change in velocity that a propulsion system can impart on its vehicle. A vehicle's demonstrable $\Delta V$ is generally increased by carrying more propellant, utilizing propellants with higher energy density, reducing non-ejected mass (e.g. payload and structural/engine mass), or reducing losses incurred by the trajectory (namely gravity, drag, and steering/cosine losses). The delta V a rocket provides can then be expressed in terms of the losses I described and the required performance:
$$ \Delta V = \Delta V_{id} +\Delta V_g + \Delta V_D + \Delta V_s$$
$\Delta V_{id}$ is the delta V required for the trajectory without losses, while the other subscripted terms account for delta V lost to gravity, drag, and steering. Combining all these contributors gives the actual delta V the vehicle needs to achieve.
So before we calculate any rocket masses, let's focus on computing the delta V for the trajectory you describe. If you can make some assumptions about the vehicles mass fractions, drag coefficient, and engine performance, you can compute the trajectory directly, but it is easier to start with a simple energy balance assuming an instantaneous impulse:
$$KE_{impulse} = PE_{apogee}$$
Here the kinetic energy imparted immediately after the impulse is balanced by the potential energy of the vehicles apex. Plugging in the equations for potential and kinetic energy gives:
$$\frac{1}{2}m_o (\Delta V_{id})^2 = \frac{m_o g_e h r_e}{r_e + h} $$
here $m_o$ is the takeoff mass of the vehicle, $h$ is the height of the vehicles apex, and $r_e$ is the earth's radius (~6,378 km). Let's rearrange to solve for the delta V:
$$ \Delta V_{id} = \sqrt{\frac{2g_ehr_e}{h+r_e}}$$
For your example height of 100,000 ft (30.5 km) this gives an ideal delta V of 770 m/s.
Now let's focus on the losses. These are considerably more challenging to estimate as they depend on the velocity profile. The energy lost to steering is given by:
$$E_{steering} = \int_{0}^{h_{bo}}F \sin\theta_sdh $$
This is the integral of the thrust ($F$) that is not speeding up the vehicle due to a yaw or gimbal angle of $\theta_s$ for the course of the rockets ascent from launch to the burnout height ($h_{bo}$). The other losses can be formulated similarly: by integrating the "lost thrust" with respect to the vehicle trajectory. We can formulate this as a delta V directly with:
$$\Delta V_s = \int_0^{t_b} \frac{F \sin\theta_s}{m}dt $$
Here we integrate the "lost acceleration" with respect to burn time ($t_b$). Now you see the complexity: All the variables in this equation (thrust, steering angle, vehicle mass) are changing with respect to time.
For completeness I will include the formulations for the other two losses:
$$ \Delta V_D = \int_0^{t}\frac{D}{m}dt \quad\quad \Delta V_g = \int_0^{t_b} g \sin\theta_f dt $$
Note here a couple of things:
- the drag force ($D$) is complex to estimate because of the varying atmospheric density throughout the flight and because the drag coefficient (and/or dynamic pressure) vary greatly from the subsonic to hypersonic regimes a rocket may experience.
- I use a generic "time" ($t$) in the drag loss integral bound because drag can continue reducing the vehicles velocity even after burnout.
- the gravitational acceleration ($g$) is distinct from the same at earth's surface I use previously ($g_e$) because if the vehicle is VERY far away from earth, gravity losses may still be incurred, but the magnitude of the acceleration will be reduced according to the inverse square reduction with distance that gravity exhibits.
- $\theta_f$ is the flight path angle; i.e. the angle between the rocket's thrust vector and the direction tangent to a circular orbit. For example, a plane flying straight and level would have a flight path angle of ~0 degrees.
This is getting a bit heavy, eh? If you really wanted to nail down these values, you would need to assume thrust and vehicle mass profiles, know your trajectory, and draw on atmospheric density models and drag coefficient correlations.
You ask for estimations though, and this I can provide. If we assume no steering losses (reasonable for a straight upward trajectory) and an average drag force across the entire ascent of:
$$ \bar{D} = C_D\frac{\bar{\rho}}{8}A(\Delta V)^2$$
then we can formulate an estimate for $\Delta V$. Here $C_D$ is the vehicle drag coefficient, $\bar{\rho}$ is the average atmospheric density throughout the ascent, and $A$ is the reference area used to define the drag coefficient. Of course this formulation for average drag is a BIG assumption, but provides reasonable approximations. Like I say, if you know the particulars of your vehicle, you can attempt to compute drag losses more accurately, but for first approximation there is a notable dearth of correlations. This particular one is very similar to the traditional drag equation excepting the factor of $1/4$ which attempts to account for the use of the delta V in place of an actual velocity. Direct questions on this correlation’s validity to its authors: Heister, Anderson, Pourpoint, and Cassady (authors of this text book). To those who might read through the linked book and complain that the correlation is not there, it is a valid complaint. This particular correlation is from a handwritten homework problem by Heister.
I will spare the derivation from energy balancing: the resulting approximation for delta V required for a vertical ascent to a particular altitude is:
$$\Delta V = \frac{g_et_b + \sqrt{(g_et_b)^2 + \left(2-\frac{C_D\bar{\rho}Ah}{2m_o}\right)\left(\frac{g_er_eh}{r_e+h}\right)}}{1-\frac{C_D\bar{\rho}Ah}{4m_o}} $$
This is a useful equation to evaluate the individual effects of the varied inputs on delta V, an exercise I will leave to the reader.
If we assume an average drag coefficient of 0.75, a vehicle cross sectional area of 80 square cm, an average atmospheric density of 0.5 kg/m3, a vehicle launch mass of 200 kg, and a burn time of 10 seconds (reasonable values for a sounding rocket), your proposed 100,000 apex requires 940 m/s.
Phew. That was a lot of hard work just to account for 170 m/s of drag and gravity losses. Of course you don't know your burn time without knowing your propellant mass, etc. So making use of this correlation would best be done with some iteration (i.e. assume burn time/launch mass, compute delta V, compute vehicle mass, compute burn time, repeat).
Ah but how will you compute vehicle mass?
This is the easy part. It is finally time to whip out the rocket equation:
$$\Delta V = g_eI_{sp}ln\left(\frac{m_o}{m_f}\right)$$
Here we have some new terms: the specific impulse ($I_{sp}=\frac{F}{\dot{m}}$) which ratios thrust to propellant mass flow, and the burnout mass ($m_f$). As you have noted in your other questions, it will be appropriate to formulate this in terms of the total mass ratio ($r = \frac{m_o}{m_f}$). Note: there is little agreement on the symbols used for the various mass ratios.
$$r = \exp\left(\frac{\Delta V}{g_eI_{sp}}\right)$$
Now, in order to figure out how much propellant you need, you need to know how much extra stuff you'll be carrying. The mass breakdown of a rocket is:
$$m_o = m_p + m_f = m_p + m_i + m_{pl}$$
You have correctly noted in your other questions that the difference between inital and final masses is the propellant mass ($m_p$), but the piece I think you missed is that you must define the final mass. This is composed of the inert mass ($m_i$) (i.e. rocket structure and engine mass) and the payload mass ($m_{pl}$).
The reason we separate the inert and payload masses is because the inert mass scales with the launch mass; if we need to carry more propellant we will need bigger, heavier tanks, beefier engines, etc. Luckily, the ratio of inert mass to initial mass has well reported values. We define it as the inert mass fraction:
$$ \delta = \frac{m_i}{m_o}$$
Here is a figure from a NASA brief detailing inert mass fractions for a variety of rockets.
Note that across all these varied vehicles and propellant combinations, the inert mass fraction is roughly bound between 3% and 15%. The brief also identifies a weak positive correlation with thrust to weight ratio. Smaller sounding rockets will typically have higher inert mass fractions just by virtue of the cube-square relationship between propellant volume and tank surface area.
We are very near the end now. I need to introduce one more term: the payload mass fraction:
$$ \lambda = \frac{m_{pl}}{m_o}$$
Note now that the total mass fraction can be expressed in terms of the inert and payload fractions:
$$ r = \frac{1}{\lambda + \delta}$$
Rearranging and plugging gives the following expressions for the vehicle initial mass and the vehicle propellant mass. Note that these will necessarily be a function of payload mass.
$$ m_o = \frac{m_{pl}}{\frac{1}{r}-\delta} \quad\quad m_p = m_{pl}\left(\frac{1}{\frac{1}{r}-\delta}-\frac{\delta}{\frac{1}{r}-\delta}-1\right)$$
And there you have it. So to summarize, you might assume the various vehicle parameters I mention, compute the delta V, compute r, compute the vehicle masses, use these masses to update your assumptions and recompute, repeat until the values match. I can't speak to the specific impulse offered by cherry limeade, but you can get some info on how to compute expected Isp in this thread. Assuming an Isp of 200 seconds, an inert mass fraction of 25%, and a 10kg payload for your 100,000 foot ascent gives a total mass fraction of 1.61, an initial rocket mass of 27 kg, a propellant mass of 10 kg, and an inert mass of 7 kg. Now of course I assumed 200kg and an 80 square cm cross section to compute the required delta V. With these masses, I can refine those values and recompute the delta V. After one or two more calculations, the mass value should converge.
