Satellite Re-entry. Differential Equations. Need Matlab/Octave.

The aim of this paper is to explain why it is inherently di cult to predict the point of reentry for a de-orbiting satellite. Ultimately, your explanation may be as simple as that satellites travels an enormous distance over a short period of time, leaving room for big errors, or it may involve a more subtle reason. Whatever the case, create a di erential equations model of a satellite’s position that includes reasonable physical assumptions. Solve the model equations numerically and use numerical experimentation (or even better, analysis) to support your conclusions. Two to three single-spaced pages is an appropriate length.

Where to start? The equations of motion in vacuum are

m1x00 =  Gm1m2x;


where G is the gravitational constant, m1 is the body’s mass and m2 is the mass of the earth. We know that all orbits are conic sections and so you may as well assume that x(t) = hx1(t); x2(t)i is a vector in the plane. Consider that the earth radius is around 6000 km and that the International Space Station circles the earth in 92 minutes. This suggests expressing length in thousands of km (megameters, Mm) and time in minutes. Put in all the units and make the appropriate cancelations to end up with the relevant dimensionless equations. They take the form

x001 = f1(t; x1; x2; x01; x02);                x002 = f2(t; x1; x2; x01; x02);

which you then convert in the usual fashion to a system of rst order equations by introducing the new variables v1(t) = x01(t) and v2(t) = x02(t):

The atmosphere high above the earth’s surface is not completely vacuum and so satellites eventually come out of orbit due to atmospheric drag, a process called orbital decay. (Solar storms are another factor.) To account for atmospheric drag, add the drag force1

FD =     12 CDA jvjv;    v = hx01; x02i;

to the force due to gravity in the equations of motion. Here CD = 2:2 is a good value for the drag coe cient, A is the cross-sectional area (hundreds of meters squared) and is atmospheric density. For density ; we can use the approximation

1 2 0   8:5  103m   0  
(x ; x ) = exp R  r   ;   = 1:3 kg m 3: (SEE NEXT PAGE!)


Here r = x21 + x22 is the distance of the satellite to earth’s center and R = 6:4 Mm (megameters). This way, atmospheric density is equal to sea-level density 0 whenever r = R and decays exponentially as r ! 1:


Corrigendum. The form of the prior expression for is adequate for altitudes up to about 100 km but experimental models show that the decay rate actually changes at higher altitudes. In fact, the rate of decay for higher altitudes is less than for lower altitudes. The following modi ed expression accounts for the two di erent rates and is a more accurate representation of the experimental values. Let a = r R be altitude (still a function of x1 and x2) and take

(a) =  0 exp(f(a))


where the piecewise function

f(a) = ( 19:28  17:63(a   0:14); 0:14   a <  
  137:71a;     0 a < 0:14


accounts for the di erent rates.

A Matlab/Octave code that executes this is

z = rho0*exp(-137.71*a.*(a<0.14) + (-19.28-17.63*(a-0.14)).*(a>=0.14));


For a custom paper on the above or a related assignment, place your order now!

Use  “20” as the discount code to redeem your offer!

What We Offer:

• Affordable Rates – (15 – 25% Discount on all orders above $50 )
• 100% Free from Plagiarism
• Masters & Ph.D. Level Writers
• Money Back Guarantee