ENERGY IN ORBIT

In this paper Energy means Energy per Unit Mass (Specific Energy). Energy is a scalar which has several forms: heat, chemical, nuclear, kinetic, potential. Though energy is conserved in our universe, it may change from one form to another, usually eventually becoming heat. We will be concerned only with kinetic and potential energy, though when a rocket fires it turns chemical energy into kinetic energy. For simplicity, we have always assumed that this takes place instantaneously, though in reality that is not possible.

Kinetic energy is the energy an object has because of its speed. It is calculated as: ke = 1/2 v.v In two dimensions this

becomes ke = .5*(v1^2 + v2^2). Since v is calculated at each time-step in the orbits, it is no trouble to add an extra formula to the loop to calculate the ke the object has at each time-step.

Potential energy is energy stored in a field. A simple example is stretching a spring. One must use energy to stretch a spring and this energy is stored in the spring and may be recovered (for example the potential energy stored in a spring once ran most wristwatches). To calculate the potential energy (pe) stored in a field, one needs its potential function, which is usually denoted by phi. In two dimensions if f(x,y) = [ f1(x,y),f2(x,y) ] is a force field, then its potential function is a function phi from two-space into the real line (one-space) such the d phi /dx = - f1(x,y) and d phi / dy = - f2(x,y). The minus sign is necessary because the object is losing other forms of energy, which it is storing in the field. See the example below.

> restart:with(stats):with(statplots):with(linalg):with(plots):

Warning, new definition for norm

Warning, new definition for trace

> phi(x,y) = x^2*y^2; Diff(phi(x,y),x) = -f1(x,y);f1(x,y)=-diff(x^2*y^2,x);Diff(phi(x,y),y) = -f2(x,y);f2(x,y)=-diff(x^2*y^2,y);f(x,y)=[-f1(x,y),-f2(x,y)];f(x,y)=[-2*x*y^2,-2*x^2*y];fieldplot([-2*x*y^2,-2*x^2*y],x=-10..10,y=-10..10);

The pe which the object has is a function of its position and in this case would be phi(s1, s2). DE books and calculus books show how to determine the potential function from a forcefield (Not all forcefields have a potential). In the case of the earth's gravitational field the potential is phi(x,y) = - MG/sqrt(x^2 + y^2). The following does not show how this is obtained, but shows it to be correct. Maple suffers some difficulty writing this in good form.

> phi(x,y)=-MG/sqrt(x^2+y^2);Diff(-MG/sqrt(x^2+y^2),x)=diff(-MG/sqrt(x^2+y^2),x);f1(x,y)=-rhs(%);Diff(-MG/sqrt(x^2+y^2),y)=diff(-MG/sqrt(x^2+y^2),y);

> f2(x,y)=-rhs(%);

>

> `Thus the forcefield of the earth's gravity is`;f(x,y)=[-MG*x/(x^2+y^2)^1.5,-MG*y/(x^2+y^2)^1.5];` `;`And the potential energy of an object moving in this field is`;pe=-MG/sqrt(x^2+y^2);

Thus since at each time-step of the orbit s1 and s2 are calculated, the pe may be calculated as pe = -MG/sqrt(s1^2 + s2^2).

For the 38000 km/h orbit, first a large amount of chemical energy is converted into ke and pe.

> Ke:=.5*38000*38000; Pe:=-5.166E12/sqrt((-6600)^2+0^2);`Te(total energy)`:=pe+ke;

After the initial burn, no further energy is put into the object and its energy must remain constant at -6.07E7. However, as the object gets further from the earth, it slows down as its kinetic energy is converted to potential energy. After it reaches its apogee (point furtherest from the earth), the potential energy stored in the field is converted back to kinetic energy and it begins increasing speed. Energy has no natural zero point. This is set up so that an object in closed orbit has negative energy. One can compute escape velocity by determining the velocity that would make the total energy zero (parabolic orbit). An object escaping in a hyperbolic orbit has positive energy.

> t[0]:=0.0;s1[0]:=0.0;s2[0]:=-6600.0;v1[0]:=38000.0;v2[0]:=0.0;t[0]:=0.0;MG:=5.166E12;DELt:=.001;Te:=-6.07E7;

> `COMPUTATIONS OMMITED BECAUSE THIS IS AN ASSIGNED PROBLEM`;`for i from 1 to 30000 do;`;`Compute t, A, V, S, ke, pe, te`; `od:`;

>

> `ke and pe are computed from the formulas`;ke[i]:=.5*(v1[i]^2+v2[i]^2); pe[i]:=-MG/sqrt(v1[i]^2+v2[i]^2);

S1 and S2 must be put into lists before using scatterplot since it operates on type list, list .

> S1:=[seq(s1[1000*i],i=0..24)]:

> S2:=[seq(s2[1000*i],i=0..24)]:

> scatterplot(S1,S2,symbol=circle,scaling=constrained,color=red,title=`v1(0)=38000 km/h. P = 24.253 h. Points 1 h apart. Note slow speed near apogee as ke is converted to pe`);

Put ke, pe and t into lists before scatterplot.

> KE:=[seq(ke[i],i=0..30000)]:PE:=[seq(pe[i],i=0..30000)]:

> T:=[seq(t[i],i=0..30000)]:

> scatterplot(T,KE,symbol=point, color=red,labels=[`Time`,`Energy`],title=`Kinetic energy --- red Potential energy --- blue Total energy --- green`):p1:=%:

> scatterplot(T,PE,symbol=point,color=blue):p2:=%:

> TE:=[seq(ke[i]+pe[i],i=0..30000)]:

> scatterplot(T,TE,symbol=point,color=green):p3:=%:

> display(p1,p2,p3);

Note that ke (red) decreases as the object approaches apogee (t = 12) while pe (blue) increases. The opposite occures as the object "falls" back towards perigee (t = 24). But the total energy (green) remains constant. This graph contains a few hours of the second cycle.

>

> scatterplot(T,TE,symbol=point,color=green,title=`Total energy`,labels=[`Time`,`Energy`]);p3:=%:

The object is subject to a small virtual force that shows up in the graph above of te. The slight variation in te from -6.12E7 to -6.02E7 is due to roundoff error in the computer and the use of Euler's Method for computing the orbit. This force is, of course not real. The roundoff could be decreased by using double precision, while the error of Euler's Method could be decreased by using a more advanced computation method (Take Numerical Analysis). Evaluation of models is often hard to do. The above graph serves as somewhat of an evaluation of this model. Clearly it is not good enough for use with real spacecraft.

In circlular orbit (v1[0] := 28000 km/h) the ke is constant since the speed is constant, and also pe as well as te are constant.

In a v1[0] := 42000 km/h orbit, the object has positive energy and escapes Earth in a hyperbolic orbit.