Applying Newton's laws to rocket motion is not only exciting, but also quite instructional. Problems involving rocket motion illustrate how to use Newton's 2nd law when mass is not constant. Here we will look at how high a single stage rocket will go under the influence of gravity. To make the problem manageable, a few simplifying assumptions are made:
- Motion in the y direction only
- Drag is neglected
- Constant burn rate for rocket
- force due to gravity is constant
- Rocket does not escape Earth's gravity
- Ideal rocket
- Lots of other miscellaneous terms
To familiarize the reader with what is involved in this calculation, we will start with the answer and then derive the equation. So the max altitude the rocket will achieve is given by
![{\displaystyle y_{max}=-{\frac {gt_{b}^{2}}{2}}+{\frac {v_{e}m_{f}t_{b}}{m_{i}-m_{f}}}ln\left({\frac {m_{f}}{m_{i}}}\right)+v_{e}t_{b}+{\frac {v_{b}^{2}}{2g}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/404b56bcf13a8733f14071e3549dc2bd677f5def)
Description of variables:
: Time that the rocket is burning fuel (given) \\
: acceleration of Gravity (given) \\
: Final mass of rocket after burn (given) \\
: Initial mass of rocket (given) \\
: velocity of rocket at burnout (calculated) \\
: Exhaust velocity, velocity of the fuel as it is ejected out of the Rocket (given)\\
The problem is best approached by breaking it into two parts. First, we calculate the altitude that the rocket reaches when all its fuel is burned. After burnout, the rocket still climbs to a higher altitude until gravity finally brings its velocity to zero. Think of it like shooting a bullet into the sky, after the initial thrust of the gun, the bullet still goes higher (duh!). So the second calculation is to add the distance traveled after burnout.
The goal is to apply Newton's 2nd law, so let us start there. For our one dimensional case
![{\displaystyle F=ma=-mg}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2a2cd5dafcce171b0c8db332519779b53baa5675)
Since we do not have constant mass throughout the rocket burn, we also have
![{\displaystyle F={\frac {d}{dt}}\left(mv\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6f8195e576b37ee413cbe80627341c225e3faefa)
Using the chain rule and setting (2) equal to (3)
multiply by dt
![{\displaystyle -mgdt=v_{e}dm+mdv}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d06ec50350e5221696aefc09b193b3b80e103e3b)
We still have three differentials, so we cannot directly integrate this equation. Since we are assuming a constant burn rate k and it is positive
so
plug this into (4) to get
Separate variables to setup the integration
divide by m
integrate
The initial velocity of the rocket is zero, so carrying out the integration gives us the velocity at a given mass
simlify using properties of the log function
![{\displaystyle v(m)={\frac {g}{k}}\left(m-m_{i}\right)-v_{e}ln({\frac {m}{m_{i}}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7ff7542390e60eff110f7e51462f04b2687ebe0f)
It is time to take care of the constant k.
rearrange
integrate
which gives
Note the sign, we have a positive k, since
will always be bigger than m (the rocket is ejecting mass). Plug this into (5).
cancel terms to get
While this equation gives us the velocity of the rocket at burnout, we also want altitude. As usual, integrate velocity to get position
To integrate the last term, we need to replace dt with dm, since m is a function of t
All but the last term are simple. For the lazy, an integral table can be used to solve it.
Use typical integral substitution technique for the last term, so set
substituting this in leaves us with
Integrating the logarithm function is done through integration by parts
![{\displaystyle \int udv=uv-\int vdu}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1e734628c64af3e2f4661c5c74259fe96d811265)
setting
then differentiating and integrating yields
plugging these into (6) gives
which is equal to
so going back to the original integral and evaluating the limits we have
simplfying and using
plug in k and expand
finally, except for the last term, we get equation (1)
![{\displaystyle y=-{\frac {gt_{b}^{2}}{2}}+{\frac {v_{e}m_{f}t_{b}}{m_{i}-m_{f}}}ln\left({\frac {m_{f}}{m_{i}}}\right)+v_{e}t_{b}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/762c9b8c9d10fa4202f68a141e1b85141c6356bc)
To find how much more altitude is gained after burnout, we note that there is no more thrust and rocket is under constant acceleration, g, so we go back to Newton's 2nd law
with a constant force due to gravity
integrate to get velocity
integrate again to get distance traveled
so the distance traveled after burnout is
combining this with the distance traveled during the burn (7), yields equation (1), the total distance rocket traveled
![{\displaystyle y_{max}=-{\frac {gt_{b}^{2}}{2}}+{\frac {v_{e}m_{f}t_{b}}{m_{i}-m_{f}}}ln\left({\frac {m_{f}}{m_{i}}}\right)+v_{e}t_{b}+{\frac {v_{b}^{2}}{2g}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/404b56bcf13a8733f14071e3549dc2bd677f5def)