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 1

st

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 2

nd

Law of Motion

“The rate at which a body’s momentum changes is equal to the net force action on the body”  F net = 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 3

rd

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 = C d (pV 2 /2)A  A = area, C d = Drag Coefficient  P = gas density

Thrust equation:

Thrust = F = MV e + (P e -P o ) A e  Where V e = Velocity of Exhaust P o = Atmosphere Pressure P e = Exhaust Pressure M = mass flow rate

Lift off equation

M s = Mass of Solids (i.e.: Rocket, payload, navigation, etc.) M l = Mass of Liquid Rocket Fuel M T = Mass of Rocket System  v= velocity of rocket

Based off of Newton’s 2 nd Law of Motion

: Momentum = mass*velocity (only when mass is constant) When the rockets fire, M l  Momentum P = (M s changes. So… + ΔM l )v …when the rockets fire, the exhaust comes out at some speed v e relative to the speed of the rocket. Therefore the momentum of the exhaust is  ΔM l (v-v e )

Lift off equation:

The Rocket’s mass will eventually drop to M s .

At the same time, its speed increases a small amount (because of a lighter load).

 M s (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 = (M s (M s + ΔM l )v + ΔM l )v = M s (v + Δv) + ΔM l (v-v e ) By multiplying and simplifying: F = M s Δv - Δ M l V e  F is a force outside of the rocket system (i.e.: gravity) Shown Differentially  F= M s (dv/dt) – v ex (dM l /dt)

Lift off equation: F= M

s

(dv/dt) – v

ex

(dM

l

/dt)

  v ex (dM l /dt) = Thrust So…   F= M s (dv/dt) MV e + (P e -P o ) A e Our rate equation for momentum of the Rocket dP/dt = M s (dv/dt) - [ MV e + (P e -P o ) A e ]

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 m 2 . 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 m 2 Po = 0 Pe = 5,000 N/m2 M s = 500,000kg

Calculation

Using the rate equation : dP/dt = M s (dv/dt) - [MV e + (P e -P o ) A e ] [MV e + (P e -P o ) A e ] = 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