Hohmann Transfer Orbit: Calculating Delta V for Transfers between Orbits, Assignments of Aerospace Engineering

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700 km

REarth

Transfer Ellipse

+2 pts. EXTRA

for plotting

Figure 1: Transfer Orbit between the Surface of the Earth and a 700km Orbit with Earth

plotted to scale.

500 km REarth

10000 km

Transfer Ellipse

Figure 2: Transfer Orbit between 500km and a 10000km Orbit With Earth plotted to scale.

+ 3pts.

for discussion

Discussion:

This strategy, Hohmann Transfer Orbit, for calculating ∆V is primarily used for calculating

the required velocity change for transferring from one orbit to another, as in Figure 2. The

conditions at the beginning and end of the transfer orbit are determined completely by the

velocity of the s/c on the initial and final orbit. Assuming a launch from the surface of the

Earth brings many additional factors into account, such as the effects the atmosphere has

on the s/c trajectory. However, using this Hohmann Transfer approximation provides a

∆VEarth= 8.04 [km/s]. The method of assuming a direct insert into a 700km Orbit provides a

∆V700km= 7.49 [km/s].

N:\School\TA\251_TA_Williams\HW\HW3\Hohmann_transfer.m

Page

February 11, 2004

PM

Stephanie (^) Van (^) Y

.m (^) file (^) for (^) orbit (^) graphing (^) in (^) cartesian (^) coordinates clear (^) all close (^) all R_Earthclc = 6378; %[km] (^) Radius (^) of (^) the (^) Earth H_initial (^) = (^) 500; %[km] (^) Altitude (^) of (^) Initial (^) Orbit H_final (^) = (^) 10000; %[km] (^) Altitude (^) of (^) Final (^) Orbit

R_IN

(^) R_Earth (^) + (^) H_initial;

R_OUT

(^) R_Earth (^) + (^) H_final; theta (^) = (^) [0:0.001:2*pi];

Elliptical (^) Transfer (^) Orbit ra (^) = (^) R_OUT; %[km] (^) Radius (^) of (^) Periapsis rp (^) = (^) R_IN; %[km] (^) Radius (^) of (^) Apoapsis a = .5*(rp+ra); %[km] (^) Semimajor (^) axis e = (^1)

(rp/a); % Eccentricity (^) of (^) Transfer (^) ellipse Pe (^) = (^) a(1-(e^2)); %km Semi-Latus (^) Rectum (^) (Parameter) x_Trans (^) = (^) [[Pe./(1+e.cos(theta))].cos(theta)]; y_Trans (^) = (^) [[Pe./(1+e.cos(theta))].*sin(theta)];

Earth (^) to (^) scale x_Earth (^) = (^) R_Earth.cos(theta); y_Earth (^) = (^) R_Earth.sin(theta);

Inner (^) Orbit (^) of (^) Hohmann (^) Transfer x_Inner (^) = (^) R_IN.cos(theta); y_Inner (^) = (^) R_IN.sin(theta);

Outer (^) Orbit (^) of (^) Hohmann (^) Transfer x_Outer (^) = (^) R_OUT.cos(theta); y_Outer (^) = (^) R_OUT.sin(theta); title('Ellipticalplot(x_Earth,y_Earth,'k',x_Trans,y_Trans,'m-.',x_Inner,y_Inner,'b--',x_Outer,y_Outer,'r')figure(1) (^) Hohmann (^) Transfer (^) Orbit (^) 500km (^) to (^) 10000km (^) Circular Orbits') xlabel('x (^) (km)') ylabel('y (^) (km)') axis (^) equal

Feasible

Region

Infeasible Region

Figure 1: For the Staging conditions stated in Problem 2 of Homework 3 the mass of each

stage was plotted for a range of ∆V ratios for different payload masses (1-1000[kg]) to

determine the feasible region of analysis for the ∆V ratio.

+3 pts. for figures 1 and/or 2

The student must demonstrate (somehow) that they

understand that part of the original graph (Figure 1) is

infeasible. Only providing Figure 2 information is okay.

Figure 2: Using the results for Figure 1 this plot shows the feasible range of (∆V 1 /∆Vtot)

ratios for which the rocket can be designed. This graph also gives and idea of the

magnitude of initial masses that will be under investigation (mpay = 1-1000 [kg] in this

graph)

Figure 9: Calculating the cost for the rocket based on the propellant mass and inert mass

shows that the optimal design point for this particular rocket. The optimal (∆V 1 /∆Vtot) is the

same for both cost and mass (since the cost was based on mass!) within a reasonable

amount of error.

+ 5pts.

Full credit is awarded for comparing (and discussing) the

(∆V 1 /∆Vtot) for the minimum mass and minimum cost.

Begin Extra graphs which describe the problem.

mpay = 1000 [kg]

mpay = 500 [kg]

mpay = 100 [kg]

Figure 4: Total mass to (∆V 1 /∆Vtot) ratio form mpay = 100- 1000 [kg]. This plot simply shows

how the total mass of the vehicle scales with payload mass. For 10x the payload mass the

overall vehicle mass goes up an order of magnitude. Is this typical for launch vehicle?

Figure 6: Propellant ratios fractions for

each stage. The crossing point dentifies the

(∆V 1 /∆Vtot) ratio at which each stage would

carry the same ratio of propellant.

The (mp/mtot) for each stage is 0.38 and 0.14 for

each stage respectively. This means that the

first stage would have more overall propellant

than the second stage. This is consistent with

previous findings. Also, the first stage has

much more mass to move than the second

stage.

Figure 7: This graph of (mp/mf) for each stage shows that for (∆V 1 /∆Vtot)=0.23 the minimum

amount of propellant is used, and therefore the minimum total mass. The graph shows

that for the first stage mp1 ~ m^ f1. While for the second stage mp2 ~ 3m^ f2.