Rocket Science 102 : Energy Analysis, Available vs Required ΔV Not in Taylor 1
Available Ignoring Aerodynamic Drag. The available Delta V for a Given rocket burn/propellant load is ( ) V = g I ln 1+ P 0 t sp mf burn General expression for Rocket accelerating along a Horizontal path P mf = m propellant m final ΔV Derived using simple assumptions But application is very general Ignored gravity and drag losses 2
Gravity Losses Propulsive ΔV loss from acting against gravity... ( ΔV )gravity loss T burn 0 = g(t) sinθ dt Applies for rocket accelerating along an ARBITRARY path From previous lecture 3
Drag Losses and This decay Also applies To launch trajectory 4
Aerodynamic Forces Acting on Rocket Lift acts perpendicular to flight path Drag acts along flight path Thrust acts along longitudinal axis of rocket C L = L ift 1 2 ρ (h) V 2 A ref 1 2 ρ (h) V 2 = q "DyamicPressure" C D = D rag 1 2 ρ (h) V 2 A ref A ref reference area ea Typically based on planform or maximum cross section 5
ΔE non conservative t! F non conservative = Drag Losses (2)! F d s! = non path conservative V! dt ΔV 2 loss 2 M path! F non conservative d! s dt dt Lift acts perpendicular to flight path.. Cannot effect energy level of rocket Gravity acts downward (conservative) cannot effect energy level of rocket 6
ΔE drag M For constant C D, M β = Drag Losses (3) ( ) 2 = D rag V = 1 2 ΔV drag equivalent specific energy loss due to drag ΔV drag = ΔV drag = M 2 t 0 C D A ref t 0 D rag V M dt M dt t 1 C 2 D A ref ρ V 3 dt = C A D ref 2 M M 0 "Ballistic Coefficient" t 0 ρ V 3 dt 7
Drag Losses (4) Aerodynamic drag/mass inertial effects can be incorporated into a single parameter. Ballistic Coefficient ( ).. measure of a projectile's ability to coast. = M/C D A ref M is the projectile's mass and C D A ref is the drag form factor. At given velocity and air density drag deceleration inversely proportional to b β β Low Ballistic Coefficients Dissipate More Energy Due to drag 8
Available Delta V Summary ( ) ΔV = g I ln 1+ P tburn 0 sp mf "combustion ΔV " t burn g (t ) t burn sinθ (t ) dt t burn 0 ρv 3 β dt "gravity loss" "drag loss" 9
Required ΔV Need to accelerate from standing still on the ground to orbital velocity, while lifting to orbital altitude, and overcoming gravity and drag losses and insert into proper orbit inclination Factors that Effect Delta V Requirements -Required Final Velocity -Rotational Velocity of Earth -Required Final Altitude - Orbit Inclination Angle 10
ΔV Required (2) V orbital ΔV potential energy ( ΔV ) required total V Launch Site 11
What Happens at Launch? 12
What Happens at Launch? (2) Velocity and Position at Burnout Determine Final Orbit Final Stage Burnout 13
Example 1: Orbital Velocity Isaac Newton explains how to launch a Satellite
Gravitational Physics Now by introducing a bit of "gravitational physics" we can unify the entire mathematical analysis F = G M m M m r 2 i r m "Inverse-square" law "potential" field y M ir F r x Isaac Newton, (1642-1727)
You ve seen it before.. Gravitational Physics (cont'd) Constant G appearing in Newton's law of gravitation, known as the universal gravitational constant. Numerical value of G G = 6.672 x 10 Nt-m -11 2 = 3.325 x 10-11 lbf-ft 2 kg 2 lbm 2
Gravitational Potential Energy Gravitational potential energy equals the amount of energy released when the Big Mass M pulls the small mass m at infinity to a location r in the vicinity of a mass M m Energy of position m P E grav E released = r F dr = M r r G M m r 2 dr = - G M m 1 r - 1 = - G M m r
Orbital Velocity (2) Gravity Centrifugal Force m ω 2 r = m V 2 r
Orbital Velocity (3) µ ~ planetary gravitational constant
ΔV Required (2) V orbital ΔV potential energy ( ΔV ) required total V Launch Site What about the Earth s rotational Contribution? 20
Earth Delta V Boost Need to accelerate from standing still on the ground to orbital velocity, while lifting to orbital altitude, and overcoming drag losses and insert into proper orbit inclination But are we really standing still on ground? No! The earth is rotating 21
Earth Boost (2) Launch Initial Conditions V boost (earth) acts due east Fixed in Inertial Space Earth Rotates Under the Orbit V"boost" Ω = 2 π 23 hrs 3600 + 60 mins 60 + 4.1 sec = 7.29212 10 5 rad sec 22
What is the tangential velocity of the Velocity = Distance Time V equator = earth? 2π 6378 km 23 hrs 3600 + 56 min 60 + 4.1 sec = 0.4651 km /sec Equatorial Radius R e = 6378km inertial equatorial velocity. Actually exceeds the speed Of Sound! why no shockwaves? 23
Earth Radius vs Geocentric Latitude R earth λ R eq = 1 - e2 Earth 1 - e2 Earth cos 2 λ Polar Radius: 6356.75170 km Equatorial Radius: 6378.13649 km e = Earth 1 - b a 2 = a 2 - b 2 a 2 = 6378.13649 2-6378.13649 2 6378.13649 = 0.08181939
Tangential Velocity at various latitudes Latitude cos(lat) velocity velocity (ft/sec) (km/sec) 0 1 0.4638 1521 10 0.98481 0.45675 1497.89259 20 0.93969 0.43583 1429.27248 30 0.86603 0.40166 1317.22464 40 0.76604 0.35529 1165.15360 50 0.64279 0.29812 977.67995 60 0.50000 0.23190 760.50000 70 0.34202 0.15863 520.21264 80 0.17365 0.08054 264.11888 90 0.00000 0.00000 0.00000 V "boost " = ( R + h ) launch Ω cosλ 25
What is the tangential velocity of the earth? (2) 26
What is the mean radius of the earth? 2 da = π [ RE cos ( λ) ] λ R E λ = R eq 1 - e2 Earth 1 - e2 Earth cos 2 λ IAU Convention: Based on Earth's Volume Sphere Volume: RE cos(λ) λ RE λ λ RE dλ λ cos (λ) 4 π 3 R 3 E = V mean E dv = π [ RE cos(λ) 2 ] x RE dλ λ cos (λ)
What is the Earth's Mean Radius? (continued) Based on Volume Ellipsoid Volume: 4 π 3 Sphere Volume: 4 π 3 1- e 2 R 3 eq R 3 sphere R sphere R mean = 1-0.08181939 2 1/6 6378.13649 = 6371.0002 km Mean Radius we have been using is for a Sphere with same volume as the Earth M E = ρ E V E "gravitational radius"
High Inclination Launch 29
Example Launch Delta V Calculation KSC Latitude ~ 28.5 0 to 200 km Orbit.. Due east launch V V earth = orbit = 0.4638 cos 180 28.5 3.986 10 km sec r 5 3 2 Altitude (km) Radius (km) Velocity (km/sec) 200 6578 7.7843 600 6978 7.5579 1000 7378 7.3502 20000 26378 3.8873 35768 42146 3.0753 = 0.4076 km/sec Delta V orbit = V orbit V earth = 7.7843-0.4076 = 7.7377 km/sec 30
High Inclination Launch (2) KSC Latitude ~ 28.5 0 to 200 km Orbit.. Due east launch V earth = = 0.4076 km/sec But for a high inclination launch.. We don t get all of the boost V boost Earth Rotational Velocity Fixed in Inertial Space V boost acts due east (earth) V"boost" V "boost " = ( R + h launch ) Ω cosλ sin Az launch = ( R + h launch ) Ω cosi 50 o Inclination launch from KSC ~ V boost ~ 0.4638 = 0.2660 km/sec cos π 180 55 31
How do we account for the change in potential energy due to lifting the vehicle 200 km? 32
Gravity Losses Accounting for Potential Energy 33
Potential Energy Revisited r Gravitational Potential Energy r U grav W performed on m = F dr G M m r 2 Gravitational potential energy equals the amount of work energy released when a mass m at infinity is pulled by gravity to the location r in the vicinity of a mass M dr Energy of position = - G M m 1 r - 1 = - G M m r = M r m m But in High School Physics you learned That gravitational Potential energy was (per unit mass) was Just g h how do these models reconcile? 34
R = r e 6371 km Potential Energy Revisited (cont d) h P.E. h = µ R + h P.E. = µ R P.E. h = µ R + h P.E. = µ R ΔP.E.= P.E. h P.E. ΔP.E.= µ R + h µ R ReR (( )) ( ) ΔP.E.= µ R ( R + h) R µ R + h R R + h = µ R R + h ( ) h 35
Potential Energy Revisited (cont d) Check acceleration of gravity F grav = mmg r 2 i _ r F grav m = g(r) = µ r 2 F grav = M m H r 2 i! r g( r) F grav m = M G r 2! i r = µ r 2! i r g = 1 h R +h R g( r) dr = 1 h R +h R µ r dr = 1 2 h µ r R +h = 1 R h µ R + h ( ) µ ( R ) = µ R + h ( ) R ΔP.E.= g = µ R R + h µ R + h ( ) R ( ) h ΔP.E. h = g h Just like you learned in 12 th grade physics! 36
Delta V gravity r e = launch altitude 200 km orbit from KSC r e ~ 6371 km DV gravity 5 2 ( 3.9860044 10 ) 200 0.5 = 1.9516 km/sec 6371 ( 6371 + 200) 37
Total Delta V Required (cont d) Root Sum Square of Required Kinetic Energy (Horizontal) + Potential Energy (Vertical) ( ΔV ) required total = V orbital V "boost " earth 2 + ΔV gravity 2 = V orbital V "boost " earth 2 + 2 µ h orbit ( ) R R + h orbit ( ) Ω cosi R + h launch ( ) Ω cosλ sin Az launch = V "boost " = R + h launch ( + ) Ω 38
ΔV tburn = g 0 I sp ln( 1+ P mf ) g 1+ P mf =1+ M propellant M dry = M dry M dry + M propellant M dry Energy Summary Available DV Path Dependent Propulsive Energy gravity loss energy dissipation = M initial M final! t burn (t ) t sin( θ (t ) ) dt g( t)= Required DV Path Independent µ R + h t ( )! 1 ρ V 3 dt (t ) (t ) t burn β (t ) β (t ) = m (t ) C D(t ) A ref ΔV required = ( V orbit V boost ) 2 2 + ΔV gravity V orbit µ R + h V boost = ( R + h launch ) Ω cos( i orbit ) ΔP.E.= ΔV 2 gravity 2 = µ h R + h t ( ) R = g h 39
Homework 1 Space Shuttle has the following mass fraction characteristics 1) Calculate the actual propellant mass fraction as the shuttle sits on the pad P mf = M propellant M "dry" + M payload P mf = M initial M final 1 40
Homework 1 (cont d) Assume that Shuttle is being launched on a Mission to the International Space Station (ISS) ISS orbit altitude is approximately 375 km above Mean sea level (MSL), assume that Shuttle Pad 41A altitude approximately Sea level, Latitude is 28.5 deg., ISS Orbit Inclination is 51.6 deg. Assume that the Earth is a perfect sphere with a radius of 6371 km µ = M G = 3.9860044 10 5 km 3 Calculate 1) The required Orbital Velocity 2) The Boost Velocity of the Earth sec 2 at the Pad 41A launch site (along direction of inclination) 3) Equivalent Delta V required to lift the shuttle to altitude 4) Total Delta V required to reach the ISS orbit 41
Homework 1 (cont d) The 2 SRB s each burn for approximately 123 seconds and produce 2,650,000 lbf thrust at sea level The 3 SSME engines each burn for ~509.5 seconds and each produces 454,000 lbf thrust at sea level Each of the SSME s consume 1040 lbm/sec of propellant 6) Calculate the average specific impulses of the SRB s, the SSME s, and the Effective specific impulse of the Shuttle Launch System as a whole during the First 123 seconds of flight (ignore altitude effects) Hint: I sp = 0 T burn F thrust m propellant dt ( 0 T burn = Impulse) total M g g 0 propellant 0 dt 42
Homework 1 (cont d) 7) Based on the calculated Delta V requirements for the mission, what would be the required propellant mass fraction For the space shuttle to reach orbit in a single stage assuming the mean launch specific impulse? -- base this calculation on the mean I sp for the system during the first 123 seconds after launch 8) How does the shuttle manage to reach orbit?.? 43
Homework 1 (cont d). Next evaluate estimate launch conditions by breaking calculation into two stages.. That is i) Stage 1 first 123 seconds SRB s and SSME s burning ii) Stage 2 after SRB s jettisoned.. Only SSME s burning stage 1 stage 2 : ΔV total = ΔV stage1 + ΔV stage2 + ΔV stage3... = # ofstages i=1 ΔV stage _ i 44
Homework 1 (cont d) i) Stage 1 first 123 seconds SRB s and SSME s burning -- Assume shuttle flys ~ vertically during Stage 1 flight. stage 1 Flight is vertical 9) Calculate Available Delta V for stage 1 Based On Mean I sp, and P mf (ignore altitude effects) -- Include gravity losses and assume an 8% drag loss in the available propulsive Delta V assume g(t) ~ g 0 = 9.8067m/sec 2 ( ΔV ) available = g 0 I sp ln M initial ΔV M final ( ΔV ) drag 0.10 g 0 I sp ln M initial M final ( )gravity loss ( ΔV ) drag 45
Homework 1 (cont d) 46
Questions?? 47