Derivation of Round Earth Launch Ascent Equations

8/3/2019 Derivation of Round Earth Launch Ascent Equations

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Aero 361 Problem 3

We will simulate the launch from sea level of a multi–stage, rocket powered vehicle.

The vehicle will place a payload into Earth orbit. We will account for variable mass,

variable air density, variable gravitational attraction, as well as some other non–linear

effects. We will consider both a flat Earth assumption and a round Earth assumption.

We will:

1. develop the equations governing this flight for a flat Earth2. develop the equations governing this flight for a round Earth3. develop a numerical solution methodology applicable to solving these equations4. discuss the specification of constants and control parameters required5. develop the density variation with altitude6. include a scenario with multiple stages7. develop the orbit equations and illustrate their use

*******************************************************************1. Flat Earth

Consider the following diagram:

T

W

q

Ï Ï Ï Ï Ï Ï Ï Ï Ï Ï Ï Ï Ï Ï Ï Ï Ï

Ï Ï Ï Ï Ï Ï Ï Ï Ï Ï Ï Ï Ï Ï Ï Ï Ï

x

yV

local horizon

D

Earth

b

1.1. Develop the equations governing this motion

From the above free body diagram, we write:

mr ³@@ + W

³) T

³) D

³

where:r ³+ xi

^) yj^

W ³

+ * mgj^

D³

+ * Dcos(q)i^* D sin(q) j

^

T ³

+ T cos( b ) q)i^) T sin( b ) q) j

^



2

Solving for x@@

and y@@

yields:

x@@ + T

mcos( b ) q) * Dmcosq

y@@ + * g ) T

m sin( b ) q) * Dm sin q

We will now concentrate on the right hand side (RHS) of these equations. If possible,

we wish to write the entire RHS in terms of x, y, x@

and y@

.

1.1.a. Drag

To begin this task, we recognize that aerodynamic drag is given by the equation:

D + C DS

rV 2

2

In this equation, density will vary with altitude (y) and we will use the standard atmo-

sphere to define this variation. Also, velocity varies along the trajectory and we can

write it in terms of its components as:

V 2 + x2@ ) y2@

The reference area generally will be the cross sectional area of our vehicle. Since the

vehicle is a multi–stage design, we will expect that the area of the various stages may

be different. We will account for this effect by allowing S to change when a stage is

dropped off. This occurs when it is out of fuel.

The drag coefficient will be assumed to be a constant for a given launch vehicle.

1.1.b. Gravity

According to Newton, the gravitational acceleration is inversely proportional to the

square of the distance from the center of the body. Introducing the constant, K , we

would write:

g + K

( y ) R E )2

If we assign the sea level value of gravity the symbol, go, then we can evaluate K at

y=0. We substitute this result back into the equation to yield:

g + go R E

( y ) R E )2



3

1.1.c. Flight Path Angle

Further, the flight path angle relative to the local horizon may be expressed in terms

of the velocity components as:

tanq + y

@

x@

1.1.d. Mass

The value of m (vehicle mass) will vary for two reasons: (i) The vehicle will be con-

suming fuel and (ii) stages will be dropping off.

The stage shells will have a mass even when empty of fuel. To account for mass

changes when stages drop off, we simply reduce the vehicle mass by the empty stage

mass at the moment it runs out of fuel.

To account for mass changes due to fuel consumption, we must know how fast the

rocket engine burns fuel. Since this is a multi–stage vehicle, it is possible that the dif-

ferent stages have different rocket motors and thus consume fuel at different rates. In

general, for a given rocket motor, we can write:

m@ + m p@

where m p represents the mass of the propellent. This propellent burn rate is a constant

for a normal mission of this sort and is considered known. Therefore, the vehicle mass

loss rate is also known. We must integrate this mass rate to get the mass, m, at anyinstant in time. We can do this by including this variable in our numerical integration

or by simply using the closed form result:

m + mo ) m p@ t

If the closed form result is used, care must be taken to use the correct value of t in the

various steps of the numerical integration algorithm.

1.1.e. Thrust

The thrust of a rocket engine is proportional to the rate at which propellent is con-sumed. Thus, we write:

T + Km p@

The value of K is –go I sp where go is sea level gravity and I sp is the specific impulse

of the engine which is determined primarily by the fuel–oxidizer choice. Typical val-

ues of I sp are around 300–350 seconds for hydrogen derivative fuels with O2 oxidizer.



4

1.2. Write as a First Order System

We reformulate both our second order nonlinear equations above as a system of first

order equations. Consider:

z³+

z1 z2 z3 z4 z5

z1 + x z2 + x

@

z3 + y

z4 + y@

z5 + m

z³@ +

z1@ z2@ z3@ z4@ z5@

+

x@ x@@ y@ y@@m@

+ f ³( z³) +

f 1 f 2 f 3 f 4 f 5

From the above considerations, we can write the function f ³( z³) as:

f ³( z³) +

f 1 f 2

f 3 f 4 f 5

+

z2

1 z5T cos( b ) q) * Dcos q

z4

* g ) 1 z5

T sin( b ) q) * D sinqm p@

Eq 1.

where D + C DSrV 2

2 and V 2 + z2

2 ) z24 , r + r( z3). Eq 2.

Also, T + * go I spm p@ , tan q + z4

z2, and g + go R E

( z3 ) R E )2

. Eq 3.

Next we will develop the round Earth equations.

*******************************************************************



5

2. Round Earth

Consider the following diagram:

W

T

q

x

y V

local horizon

D

Earth

b

Í Í Í Í Í Í Í Í Í Í Í Í Í Í Í Í Í Í Í





f

R

RE

2.1. Develop the equations governing this motion

From the above free body diagram, we write:



R@@

and f@@

6

mr ³@@ + W

³) T

³) D

³

where: R³

+ Ri^

R

W ³

+ * mgi^

R

D³

+ * Dcos(q)i^

f * D sin(q)i^

R

T ³

+T cos( b

)q)i

^

f )T sin( b

)q)i

^

R

R³@

+ R@

i^

R ) R i^

R

@so , but i

^

R

@+ f

@i^

f

thus: R³@

+ R@

i^

R ) R f@

i^

f

Differentiation again gives:

R³@@

+ R@@

i^

R ) 2 R@

f@

i^

f ) R f@@

i^

f * R f2@i^

R since i^

f

@+ * f

@i^

R

Also:

or: R³@@

+ ( R@@

* R f2@)i

^

R ) (2 R@

f@

) R f@@

)i^

f

Eq 4.

Eq 5.

Eq 6.

We substitute these vector expressions into the above statement of Newton’s 2nd law

and collect all terms in the R direction and all terms in the f direction. This yields 2

scalar equations. Solving them for yields:

R@@

+ R f2@ * g ) T m sin( b ) q) * D

m sinq

f@@ + * 2

R@

f@

R) T

mRcos( b ) q) * D

mRcos q

We will now concentrate on the right hand side (RHS) of these equations. If possible,

we wish to write the entire RHS in terms of R, f , R@

and f@

.

2.1.a. Drag

To begin this task, we recognize that aerodynamic drag is still given by the equation:

D + C DSrV 2

2

In this equation, density will vary with altitude ( R–R E ) and we will again use the stan-

dard atmosphere to define this variation. Also, velocity varies along the trajectory and

we can write it in terms of its components as:

V 2 + R2@ ) ( R f@

)2

The reference area variation is as described for the flat Earth case.



7

The drag coefficient will again be assumed to be a constant for a given launch vehicle.

2.1.b. Gravity

The gravitational acceleration is again given by:

g + K ( y ) R E )

2

If we assign the sea level value of gravity the symbol, go, then we can evaluate K at

y=0, which is now R=R E . We substitute this result back into the equation to yield:

g + go R E

R2

2.1.c. Flight Path Angle

Further, the flight path angle relative to the local horizon may be expressed in terms

of the velocity components as:

tanq + R@

R f@

2.1.d. Mass

Vehicle mass is handled as discussed for the flat Earth case.

2.1.e. Thrust

The thrust is handled as discussed for the flat Earth case.

2.2. Write as a First Order System

We reformulate both our second order nonlinear equations above as a system of first

order equations. Consider:

z³+

z1 z2 z3 z4 z5

z1 + R

z2 + R@ z3 + f

z4 + f@

z5 + m

z³@ +

z1

@ z2@ z3@ z4@ z5@

+

R@

R@@ f@

f@@m@

+ f ³( z³) +

f 1

f 2 f 3 f 4 f 5

From the above considerations, we can write the function f ³( z³) as:



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f ³( z³) +

f 1 f 2 f 3 f 4

f 5

+

z2

z1 z24 * g ) 1

z5T sin( b ) q) * D sinq

z4

* 2 z2 z4 z

1

) 1 z

5

z1

T cos( b ) q) * Dcos qm p@

Eq 7.

where D + C DSrV 2

2 and V 2 + z2

2 ) ( z1 z4)2 , r + r( z1 * R E ). Eq 8.

Also, T + * go I spm p@ , tan q + z2

z1 z4, and g + go R E

z12

. Eq 9.

*******************************************************************

3. Development of a numerical solution scheme

Consider the equation:

z@ + f ( z) , z(0) + z0

A first order integration scheme for this equation is:

zn)

1

+ zn

) dt f ( zn

)

A second order scheme is given by the two step scheme:

z* + zn ) dt f ( zn)

zn)1 + 12 zn ) z* ) dt f ( z*)

A fourth order scheme (Runge–Kutta) is given by the four step scheme:

z* + zn ) dt 2

f ( zn)

z** + zn ) dt 2

f ( z*)

z*** + zn ) dt f ( z**)

zn)1 + zn ) dt 6

f ( zn) ) 2 f ( z*) ) 2 f ( z**) ) f ( z***)



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*******************************************************************

4. Initial Conditions and Thrust Control

Both sets of equations require an initialization of z³

. This amounts to initial position

and velocity and vehicle mass. In addition, the values of go, R E , I sp , m p@ ,C D,S, and

b(t ) must be specified to completely define the problem. Since this is a numerical in-

tegration problem, a value of dt must be specified.

The density as a function of altitude also requires specification of some constants such

as the gas constant for air, sea level values of pressure, temperature, and the so–called

knot values of temperature and the corresponding altitudes on which the standard at-

mosphere is based. More on this is discussed in the next section.

4.1. More on b( t )

The control of this vehicle is the thrust angle. This control function must be specified.

We will assume a piecewise constant variation in which the value of b suddenly

switches from one number to another. We will select the values of b and the times at

which these values become effective. A given value of b will remain in effect until

the next value replaces it. This strategy requires that we choose the number of b' sapriori and then input these along with the corresponding times. More on this is found

in the multiple stage section 6.

*******************************************************************

5. Density Variation With Altitude

The density variation with altitude for use in the drag calculation is handled by assum-

ing a “standard atmosphere” exists. The standard atmosphere is defined and is based

on some average temperature measurements and the hydrostatic equation. The differ-

ential form of the hydrostatic equation is:

dp + * rgdh

The ’h’ in this equation is the geometric altitude. The ’g’ is the local value of the gra-vitational acceleration which is given above and repeated here as:

g + go R E

(h ) R E )2

Substitution of this expression for gravity into the hydrostatic equation yields:



10

dp + * rgo R E

(h ) R E )2

dh

To simplify this equation, we introduce a new altitude, h*, such that:

dh* + R E

(h ) R E )2

dh

which integrates to:

h* + R E

(h ) R E )h

Eq 10.

This results in:

dp + * rgodh*

Now from averaged experimental temperature data, it is known that the temperature

variation in the atmosphere is piecewise linear. We take the function T(h*) to be a

piecewise linear function defined completely by the (temperature, altitude) pairs at the

intermediate (knot) points. On each linear segment of this function, we may write;

T(h*)=T i + ai[h*–(h*)i] Eq 11.

where ’ai’ is called the lapse rate of the ith linear segment and it represents the slope

of the straight line. Generally, the values of (T i ,hi) are given for the various segmentsof the atmosphere. The following table is acceptable:

i

hi(km)

Ti(_K)

1

0

288.16

2 3 4 5 6 7 8 9

11 25 47 53 79 90 105 +

216.66 216.66 282.66 282.66 165.66 165.66 225.66 225.66

Thus, the lapse rate for the segment from i=1 to i=2 would be called a1 and would becomputed as:

a1 + T 2 * T 1h*

2 * h*1

Likewise, the lapse rate for the segment from i to i+1 would be called ai and would be

computed as:



11

ai + T i)1 * T ih*

i)1 * h*i Eq 12.

Note that the denominators of these equations require h* which must be computed

from the h given in the table. Also note that several of the intervals have a zero lapse

rate which implies a constant temperature (isothermal) region.

In general, the density, r, is related to the pressure, p, and temperature, T by the equa-

tion of state: p=rRT , whereR is the gas constant for air. Solving this relation for rand substituting into the hydrostatic equation yields:

dp p + * go

Rdh*

T

In the isothermal regions, (say the ith region), this equation integrates to:

p + pie* go

RT ih**h*

i

Eq 13.

In the regions of non–zero lapse rate, we substitute the linear equation for temperature

into the differential equation to get:

dp p + * go

Rdh*

T i ) ai(h* * h*

i )

which integrates to:

p + piT T i*

go

Rai

Eq 14.

In order to apply Eq 13. and Eq 14., we need values for pressure at the knot points.

These are determined as follows:

Set p1 to sea level pressure [ 1.01325(10)5 N/m2]

Compute p2 from Eq 14. with i=1.

Compute p3 from Eq 13. with i=2 and h*=(h*)3Compute p4 from Eq 14. with i=3.Compute p5 from Eq 13. with i=4 and h*=(h*)5....

etc.

Thus given a value for h, the steps required for computation of density are:

1. compute h* from Eq 10.



12

2. test h and find which interval it is in ( what value of i satisfies:

hi < h < hi+1 )

3. determine if this interval is isothermal or non–zero lapse rate

4. If isothermal, use Eq 13. to compute the pressure, p

5. If non–zero lapse rate, use Eq 14. to compute the pressure, p

6. In either case, compute the temperature, T with Eq 11.

7. With p and T known, compute density, r with the equation of

state: p=rRT .

*******************************************************************

6. The Multiple Stage Scenario

This problem involves a multiple stage vehicle. Each stage carries fuel, including the

final payload stage. We will denote each of Ns stages by k. The k th stage empty hullmass is denoted by mhk , and the fuel mass for this stage is mfk . We will adopt a conven-

tion that the payload stage is the Ns stage. The total vehicle lift–off mass is then given

by:

mlo + N s

k +1

(mhk ) m fk )

Eq 15.

We know that the vehicle mass will decrease as the mission proceeds. We also know

that at certain values of vehicle mass, namely those that coincide with fuel depletion

in a given stage, the vehicle mass will decrease suddenly as an empty hull is dropped

off and a new stage is ignited. We will organize these mass “breakpoints” in a se-

quence as:

{m1, m2,m3, AAA, mhN s } Eq 16.

We can calculate the values in this sequence as:

m1 + mlo * m f 1

m2 + m1 * mh1 * m f 2

m3 + m2 * mh2 * m f 3 Eq 17.

@@@

m N s+ mhN s

+ m N s*1 * mh( N s*1) * m fN s



13

If m denotes the current vehicle mass, we might handle stage drop–off as follows:

do k=1,N sif(m.lt.mk )then

m=m–mhk

mk =0goto 100

endif

enddo

100 continue

This approach is somewhat wasteful in that for later stages, all stages are checked even

though we already discarded them. Since the number of stages is typically small, this

is not really a serious consideration. The only way the computer knows we discarded

them is that we set mk to 0 once that mass threshold has been reached. The smallest

value that vehicle mass should ever have is then the payload mass.

6.1. The Input Data

It is possible in a multiple stage scenario for different rocket engines to be used for

different stages. The rocket engine choice is reflected in the values of I sp and m p@ .

Therefore we will allow for these to vary from stage to stage. Furthermore, a reference

area, S, is needed in the Drag equation. Often, successive stages in a multiple stage

vehicle get smaller as the payload end is approached. We will account for this possibil-

ity also.

A possible input data structure for this problem is outlined below.

1. filenames for files written by the program

2. upper limit on flight time allowed (Tmax)

3. flag to determine what to do after all fuel is exhausted (burnout)

possibilities are:

(a) quit program

(b) compute orbit parameters and quit

(c) continue integration until t=Tmax or h=0.4. number of stages (Ns)

5. loop on stages: for each stage, input the following

I sp, m p@ , S, C D , mh , m f , dt

nbetas

time1 , beta1



14

time2 , beta2

@@@

timenbetas , betanbetas

Please note that in the above, values of C D and dt are input for each stage. This is done

just to keep the algorithm as general as possible. It is quite possible that different dt’s

are desirable for different stages. At lift off, initial velocity is 0. Things begin rather

slowly so a larger time step is possible to get things going. However, after a little time,

velocities reach large values thus suggesting a smaller time step. The value of C D will

be the same for all stages for this problem but might be different if the stage shape

changes and yawing were accounted for.7. The Orbit Equations

The strategically guided multi–stage rocket eventually runs out of fuel. This point in

the trajectory is called burn–out. At this point in the trajectory, the vehicle has a posi-

tion and a velocity which may be used to determine the characteristics of the orbit tra-

jectory which the vehicle will follow after burnout. This section documents the theory

used and the equations developed for determining these orbit characteristics.

7.1. Governing Equations

The equations governing the orbit phase of the trajectory are derived from first prin-ciples. Newton recognized that the force exerted on one body by another acts toward

the other and is proportional to the product of the two masses and inversely proportion-

al to the square of the distance between them. Thus, we write the force on the vehicle

by the earth as:

F + GmM e R2

where R is the distance from the center of earth to the vehicle, m is the vehicle mass,

M e is earth mass, and G is the universal gravitational constant. In order to assign adirection to this force, we recognize that it acts on the vehicle in a direction toward the

center of the earth. This direction can be written as:

* R³

R

The vector force expression is then:



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F ³

+ *GmM e

R3 R³

Now, Newton’s 2nd law asserts that this force must equal the vehicle mass times its

acceleration. This allows us to write:

F ³

+ *GmM e R3

R³

+ mR³@@

Eq 18.

The acceleration expression is given in Eq 6. and repeated here for convenience:

R³@@

+ ( R@@

* R f2@)i

^

R ) (2 R@

f@

) R f@@

)i^

f

From this, and Eq 18.,we may write the following two scalar equations.

R

@@

* R f

2@

+ *G M e

R2 2 R

@

f

@

) R f

@@

+ 0. Eq 19.

From the position and velocity vector expressions in Eq 4. and Eq 5. and repeated here

for convenience:

R³

+ Ri^

R R

³@+ R

@i^

R ) R f@

i^

f

we can construct the angular momentum, h. It is given by:

h + h³

@ k ^+ ( R

³ R

³@) @ k

^+ R2 f

@Eq 20.

The time derivative of this angular momentum, h, is given by:

dhdt

+ R2 f@@

) 2 RR@

f@

+ R( R f@@

) 2 R@

f@

)

Thus, with reference to Eq 19., we find that:

dhdt

+ R( R f@@

) 2 R@

f@

) + 0

so that:

h + R2 f@ + constant Eq 21.

This constant may be determined by inserting the burn–out conditions.

The first of Eq 19. can also be integrated. We let:

u + 1 R



16

Then:

R + 1u Eq 22.

so, from Eq 21.:

f@ + hu2 Eq 23.

Also, time differentiation of Eq 22. yields:

R@

+ * R2u@

R@@

+ * 2 RR@

u@

* R2u@@

But:

u@ + dud f

f@ + u f@ + uu2h

where prime denotes differentiation with respect to f and substitution from Eq 23. has

been made. So:

R@

+ * uh Eq 24.

Also:

R@@

+d

dt R@

+d

d f( R@

) f@

+ *hu

f@

+ *u

u2h2 Eq 25.

Substitution of Eq 23. and Eq 25. into the first of Eq 19. yields:

* u u2h2 * h2u3 + *G M eu2

Dividing this by –u2h2 yields:

u ) u + G M eh2

+K

7.2. The Orbit Equation

The solution to this 2nd order ODE is u( f) = uparticular + uhomogeneous.

where obviously, uparticular =K. uhomogeneous is found from the characteristic equation

of the homogeneous problem which is l2 + 1 = 0, the roots of which are: "i. This

solution is:

uhomogeneous + C 1ei f ) C 2e*i f



17

But since: ei f + cos( f) ) i sin( f)

uhomogeneous + (C 1 ) C 2)cos( f) ) i(C 1 * C 2)sin( f)

Thus:

u( f) + 1 R( f)

+K) (C 1 ) C 2)cos( f) ) i(C 1 * C 2)sin( f)

We will require C 1 and C 2 to be real numbers and since R( f) is real, we must require

that:

C 1 = C 2 = C

Following rearrangement, this results in:

R( f) +1

K1 ) 2C

Kcos( f)

The quantity 2C K

may be shown to be the eccentricity, e. Eccentricity is a convenient

parameter for determining the nature of the orbit. It can easily be shown that:

e = 0 represents a circular orbit

0 < e < 1 represents an elliptic orbit

e = 1 represents a parabolic orbit

e > 1 represents an hyperbolic orbit

This equation is thus rewritten as:

R( f) +1K

1 ) ecos( f)

Also, the derivative of this equation is:

R( f) +eK

sin( f)

[1 ) e cos( f)]2

Eq 26.

From this, it is clear that R( f) + 0 where f = 0. This implies that the f in this equation

has a built in reference orientation and is therefore a different f than the one used in

the figure in Section 2. For this reason, the f in the present case will be denoted as

f*. These two f’s are different only by some constant. Therefore we write:

f = f* + Df



18

This implies that:

f@

+ f@

*

We rewrite our orbit equation as:

R( f*) +1K

1 ) ecos( f*)Eq 27.

7.3. The Velocity Equation

Next, we develop an equation for velocity. Since:

V 2 + R2@ ) ( R f@

)2Eq 28.

and:

R@ + R( f*) f@ + R( f*) h R2

Eq 29.

where:

f@

+ h R2

Eq 30.

Substitution of Eq 29., Eq 26., Eq 27., and Eq 30. into Eq 28., yields:

V 2 + h2K21 ) 2e cos f* ) e2Eq 31.

7.4. Some Significant Results

The point of closest approach of an earth orbit is called perigee. This is clearly the

point at which R in Eq 27. is a minimum. This will occur when f* = 0. So:

R perigee +1K

1 ) e

The velocity at this point will be a maximum and from Eq 31. is seen to be:

V + hK(1 ) e)

The point of maximum distance of an elliptical earth orbit is called apogee. This is

clearly the point at which R in Eq 27. is a maximum. This will occur when f* = p.

So:

Rapogee +1K

1 * e



19

The velocity at this point will be a minimum and from Eq 31. is seen to be:

V + hK(1 * e)

7.5. Orbit Determination Procedure

We have developed all of the necessary equations. We now will take our vehicle burn–out conditions and determine the orbit parameters.

At burn–out, we know R and f@

from which we can compute the angular momentum,

h from:

h + R2 f@

We then can compute K from:

K+ G M eh2

Also at burn–out we can determine velocity from:

V 2 + R2@ ) ( R f@

)2

We can solve Eq 27. for: e cos( f*)

e cos( f*) + 1K R

* 1

Then by re–arranging Eq 31. we can solve for e from:

e + V 2

h2K2* 1 * 2e cos f*

Of course f* is now determined from:

f* + cos*1e cos( f*)e

Now, the relationship between f and f* may be determined:

Df = f - f*

The point of closest approach is then given by:

R perigee +1K

1 ) e



which must be greater than the radius of the earth if your orbit is a successful one.

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Derivation of Round Earth Launch Ascent Equations

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