Transcript The Rocket Equation!!

Hunter, Kevin
Yu, Marcus
These Next Few Steps
Using the Newton Law of motion and some
outside research, we will derive the basic
equation that describe the rocket launching
motion.
The differential equation will give us the rate
equation modeling the rocket.
Newton’s 1st Law of Motion
“A body in uniform motion
remains in uniform motion and
a body at rest remains at rest,
unless acted on by a non-zero
net force”
A rocket will stay at rest unless
acted upon.
If a rocket’s thrust can
overcome Its weight and
gravity, there will be liftoff.
Newton’s 2nd Law of Motion
“The rate at which a body’s momentum changes is equal to the net
force action on the body”

Fnet = dP/dt
Force = change of momentum over change in time

Force = change of mass*velocity over change in time
F= mass*acceleration (only if mass is constant)
Because rockets burn fuel up, mass
Isn’t constant.
We need another way to describe momentum that takes in the factor of
changing mass!
Newton’s 3rd Law of Motion
“If object A exerts a force on object B, then object B exerts an
opposite force of equal magnitude”
Thrust is what makes these rockets fly!
When the thrust of the rocket is greater than the outside force such
as gravity, drag, and the weight of the rocket, then the rocket will be
able to travel in the positive direction, while the exhaust of the rocket
will travel in the negative opposite direction.
Theoretical approach to finding
Thrust
For liftoff:

Thrust > weight * gravity * drag…
Drag = Cd(pV2/2)A


A = area, Cd = Drag Coefficient
P = gas density
Thrust equation:
Thrust = F = MVe+ (Pe-Po) Ae

Where Ve = Velocity of Exhaust
Po = Atmosphere Pressure
Pe = Exhaust Pressure
M = mass flow rate
Lift off equation
Ms = Mass of Solids (i.e.: Rocket, payload, navigation, etc.)
Ml = Mass of Liquid Rocket Fuel
MT = Mass of Rocket System

<Ms+Ml = MT>
v= velocity of rocket
Based off of Newton’s 2nd Law of Motion:
Momentum = mass*velocity (only when mass is constant)
When the rockets fire, Ml changes. So…

Momentum P = (Ms + ΔMl)v
…when the rockets fire, the exhaust comes out at some speed ve relative to
the speed of the rocket. Therefore the momentum of the exhaust is

ΔMl(v-ve)
Lift off equation:
The Rocket’s mass will eventually drop to Ms.
At the same time, its speed increases a small amount (because of a lighter
load).

Ms (v + Δv)
But with no external force, the rocket system is at a stand still. So we must
sum the two parts together to form the initial momentum.

P = (Ms + ΔMl)v
(Ms + ΔMl)v = Ms (v + Δv) + ΔMl(v-ve)

By multiplying and simplifying:
F = Ms Δv - Δ Ml Ve

F is a force outside of the rocket system (i.e.: gravity)
Shown Differentially
 F= Ms(dv/dt) – vex(dMl/dt)
Lift off equation:
F= Ms(dv/dt) – vex(dMl/dt)




vex(dMl/dt) = Thrust
So…
F= Ms(dv/dt) - MVe+ (Pe-Po) Ae
Our rate equation for momentum of the
Rocket
dP/dt = Ms(dv/dt) - [MVe+ (Pe-Po) Ae]
Rocket equation’s application!
Space Shuttle Columbia’s engine ejects
mass at a rate of 30 kg/s with an exhaust
velocity of 3,100 m/s. The pressure at the
nozzle exit is 5 kPa and the exit area is 0.7
m2. The mass of the shuttle solids (not
including fuel) is about 500,000 kilograms.
Find the acceleration of the rocket.
Variables
Given:
M = 30 kg/s
Ve = 3,100 m/s
Ae = 0.7 m2
Po = 0
Pe = 5,000 N/m2
Ms = 500,000kg
Calculation
Using the rate equation :
dP/dt = Ms(dv/dt) - [MVe+ (Pe-Po) Ae]
[MVe+ (Pe-Po) Ae] = 30 x 3,100 + (5,000 - 0) x 0.7 F = 96,500 N
dP/dt = 500,000kg(dv/dt) - 96,500 N
Since the rocket is in space:
0= 500,000kg(dv/dt) - 96,500 N
96,500 N= 500,000kg(dv/dt)
 Divide
dv/dt = 0.19 m/s^2
End