Hunter, KevinHunter, KevinYu, MarcusYu, Marcus

These Next Few StepsThese Next Few StepsUsing the Newton Law of motion and some Using the Newton Law of motion and some outside research, we will derive the basic outside research, we will derive the basic equation that describe the rocket launching equation that describe the rocket launching motion.motion.

The differential equation will give us the rate The differential equation will give us the rate equation modeling the rocket.equation modeling the rocket.

Newton’s 1Newton’s 1stst Law of Motion Law of Motion““A body in uniform motion A body in uniform motion remains in uniform motion and remains in uniform motion and a body at rest remains at rest, a body at rest remains at rest, unless acted on by a non-zero unless acted on by a non-zero net force”net force”

A rocket will stay at rest unlessA rocket will stay at rest unlessacted upon.acted upon.

If a rocket’s thrust can If a rocket’s thrust can overcome Its weight and overcome Its weight and gravity, there will be liftoff.gravity, there will be liftoff.

Newton’s 2Newton’s 2ndnd Law of Motion Law of Motion““The rate at which a body’s momentum changes is equal to the net The rate at which a body’s momentum changes is equal to the net force action on the body”force action on the body”

FFnetnet = dP/dt = dP/dt

Force = change of momentum over change in timeForce = change of momentum over change in time Force = change of mass*velocity over change in timeForce = change of mass*velocity over change in time

F= mass*acceleration (only if mass is constant)F= mass*acceleration (only if mass is constant)

Because rockets burn fuel up, massBecause rockets burn fuel up, massIsn’t constant.Isn’t constant.

We need another way to describe momentum that takes in the factor of We need another way to describe momentum that takes in the factor of changing mass!changing mass!

Newton’s 3Newton’s 3rdrd Law of Motion Law of Motion““If object A exerts a force on object B, then object B exerts an If object A exerts a force on object B, then object B exerts an opposite force of equal magnitude”opposite force of equal magnitude”

Thrust is what makes these rockets fly!Thrust is what makes these rockets fly!

When the thrust of the rocket is greater than the outside force such 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 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 able to travel in the positive direction, while the exhaust of the rocket will travel in the negative opposite direction.will travel in the negative opposite direction.

Theoretical approach to finding Theoretical approach to finding ThrustThrust

For liftoff:For liftoff: Thrust > weight * gravity * drag…Thrust > weight * gravity * drag…

Drag = CDrag = Cdd(pV(pV22/2)A/2)A A = area, CA = area, Cdd = Drag Coefficient = Drag Coefficient P = gas densityP = gas density

Thrust equation:Thrust equation:

Thrust = F = MVThrust = F = MVee+ (P+ (Pee-P-Poo) A) Aee

Where VWhere Vee = Velocity of Exhaust = Velocity of ExhaustPPoo = Atmosphere Pressure = Atmosphere Pressure

PPee = Exhaust Pressure = Exhaust Pressure

M = mass flow rateM = mass flow rate

Lift off equationLift off equation

MMss = Mass of Solids (i.e.: Rocket, payload, navigation, etc.) = Mass of Solids (i.e.: Rocket, payload, navigation, etc.)

MMll = Mass of Liquid Rocket Fuel = Mass of Liquid Rocket Fuel

MMTT = Mass of Rocket System = Mass of Rocket System <M<Mss+M+Mll = M = MTT>>

v= velocity of rocketv= velocity of rocket

Based off of Newton’s 2Based off of Newton’s 2ndnd Law of Motion Law of Motion::Momentum = mass*Momentum = mass*vvelocity (only when mass is constant)elocity (only when mass is constant)When the rockets fire, MWhen the rockets fire, Mll changes. So… changes. So…

Momentum P = (MMomentum P = (Mss + + ΔΔMMll)v)v

……when the rockets fire, the exhaust comes out at some speed vwhen the rockets fire, the exhaust comes out at some speed vee relative to the relative to the speed of the rocket. Therefore the momentum of the exhaust is speed of the rocket. Therefore the momentum of the exhaust is

ΔΔMMll(v-v(v-vee))

Lift off equation:Lift off equation:The Rocket’s mass will eventually drop to MThe Rocket’s mass will eventually drop to Mss..

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

MMss (v + (v + ΔΔv)v)

But with no external force, the rocket system is at a stand still. So we must 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.sum the two parts together to form the initial momentum.

P = (MP = (Mss + + ΔΔMMll)v)v

(M(Mss + + ΔΔMMll)v = M)v = Mss (v + (v + ΔΔv) + v) + ΔΔMMll(v-v(v-vee)) By multiplying and simplifying:By multiplying and simplifying:

F = MF = Ms s ΔΔv - v - ΔΔ M Ml l VVee

F is a force outside of the rocket system (i.e.: gravity)F is a force outside of the rocket system (i.e.: gravity)

Shown DifferentiallyShown Differentially F= MF= Mss(dv/dt) – v(dv/dt) – vexex(dM(dMll/dt)/dt)

Lift off equation:Lift off equation:

F= MF= Mss(dv/dt) – v(dv/dt) – vexex(dM(dMll/dt)/dt) vvexex(dM(dMll/dt) = Thrust/dt) = Thrust So…So… F= MF= Mss(dv/dt) - (dv/dt) - MVMVee+ (P+ (Pee-P-Poo) A) Aee

Our rate equation for momentum of the Our rate equation for momentum of the RocketRocket

dP/dt = MdP/dt = Mss(dv/dt) - [(dv/dt) - [MVMVee+ (P+ (Pee-P-Poo) A) Aee]]

Rocket equation’s application!Rocket equation’s application!

Space Shuttle Columbia’s engine ejects Space Shuttle Columbia’s engine ejects mass at a rate of 30 kg/s with an exhaust mass at a rate of 30 kg/s with an exhaust velocity of 3,100 m/s. The pressure at the velocity of 3,100 m/s. The pressure at the nozzle exit is 5 kPa and the exit area is 0.7 nozzle exit is 5 kPa and the exit area is 0.7 mm22. The mass of the shuttle solids (not . The mass of the shuttle solids (not including fuel) is about 500,000 kilograms. including fuel) is about 500,000 kilograms. Find the acceleration of the rocket.Find the acceleration of the rocket.

VariablesVariables

Given: Given: M = 30 kg/s M = 30 kg/s Ve = 3,100 m/s Ve = 3,100 m/s Ae = 0.7 mAe = 0.7 m2 2

Po = 0Po = 0Pe = 5,000 N/m2 Pe = 5,000 N/m2 MMss = 500,000kg = 500,000kg

CalculationCalculation

Using the rate equation :Using the rate equation :dP/dt = MdP/dt = Mss(dv/dt) - [MV(dv/dt) - [MVee+ (P+ (Pee-P-Poo) A) Aee]]

[MV[MVee+ (P+ (Pee-P-Poo) A) Aee] = 30 x 3,100 + (5,000 - 0) x 0.7 F = 96,500 N ] = 30 x 3,100 + (5,000 - 0) x 0.7 F = 96,500 N

dP/dt = 500,000kg(dv/dt) - 96,500 N dP/dt = 500,000kg(dv/dt) - 96,500 N Since the rocket is in space:Since the rocket is in space:0= 500,000kg(dv/dt) - 96,500 N0= 500,000kg(dv/dt) - 96,500 N96,500 N= 500,000kg(dv/dt) 96,500 N= 500,000kg(dv/dt) DivideDivide

dv/dt = 0.19 m/s^2dv/dt = 0.19 m/s^2

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