Your spacecraft is traveling at constant velocity V, approaching a distant earthlike planet with mass M and radius R. When activated, your deceleration system provides a constant thrust T until impact with the surface of the planet. During the period of deceleration, your distance x(t) from the center of the planet satisfies the

Define rk2dim()=Prgm
   f(t,x,y):=y
   g(t,x,y):=-x
   n:=50
   t:=0.0
   x:=0.0
   y:=1.0
   t1:=5.0
   h:=(t1-t)/n
   For i,1,n
        t0:=t
        x0:=x
        y0:=y
        f1:=f(t,x,y)
        g1:=g(t,x,y)
        t:=t0+h/2
        x:=x0+(h*f1)/2
        y:=y0+(h*g1)/2
        f2:=f(t,x,y)
        g2:=g(t,x,y)
        x:=x0+(h*f2)/2
        y:=y0+(h*g2)/2
        f3:=f(t,x,y)
        g3:=g(t,x,y)
        t:=t0+h
        x:=x0+h*f3
        y:=y0+h*g3
        f4:=f(t,x,y)
        g4:=g(t,x,y)
        fa:=(f1+2*f2+2*f3+f4)/6
        ga:=(g1+2*g2+2*g3+g4)/6
        x:=x0+h*fa
        y:=y0+h*ga
        Disp t,x,y
   EndFor
EndPrgm

# Program RK2DIM
def F(T,X,Y): return Y
def G(T,X,Y): return -X
N = 50
T = 0.0
X = 0.0
Y = 1.0
T1 = 5
H = (T1-T)/N
for I in range(N):
T0 = T
X0 = X
Y0 = Y
F1 = F(T,X,Y)
G1 = G(T,X,Y)
T = T0 + H/2
X = X0 + H*F1/2
Y = Y0 + H*G1/2
F2 = F(T,X,Y)
G2 = G(T,X,Y)
X = X0 + H*F2/2
Y = Y0 + H*G2/2
F3 = F(T,X,Y)
G3 = G(T,X,Y)
T = T0 + H
X = X0 + H*F3
Y = Y0 + H*G3
F4 = F(T,X,Y)
G4 = G(T,X,Y)
FA = (F1+2*F2+2*F3+F4)/6
GA = (G1+2*G2+2*G3+G4)/6
X = Y0 + H*FA
Y = Y0 + H*GA
print (T,X,Y)
# END

Program title
Define function f
Define function g
No. of steps
Initial t
Initial x
Initial y
Final t
Step size
Begin loop
Save previous t
Save previous x
Save previous y
First f-slope
First g-slope
Midpoint t
Midpt x-predictor
Midpt y-predictor
Second f-slope
Second g-slope
Midpt x-predictor
Midpt y-predictor
Third f-slope
Third g-slope
New t
Endpt x-predictor
Endpt y-predictor
Fourth f-slope
Fourth g-slope
Average f-slope
Average g-slope
x-corrector
y-corrector
Display results
End loop

function xp = f(t,x)
xp = x;
xp(1) = x(2);
xp(2) = -x(1);
function [T,Y] = rkn(t,x,t1,n)
h = (t1 - t)/n;                   % step size
T = t;                            % initial t
X = x′;                           % initial x-vector
for i = 1:n                       % begin loop
   k1 = f(t,x);                   % first k-vector
   k2 = f(t+h/2,x+h*k1/2);        % second k-vector
   k3 = f(t+h/2,x+h*k2/2);        % third k-vector
   k4 = f(t+h ,x+h*k3);           % fourth k-vector
   k = (k1+2*k2+2*k3+k4)/6;       % average k-vector
   t = t + h;                     % new t
   x = x + h*k;                   % new x
   T = [T;t];                     % update t-column
   X = [X;x′];                    % update x-matrix
   end                            % end loop

differential equation

where $G\approx 6.6726\times {10}^{-11}\text{N}\cdot {\text{(m/kg)}}^2$ as in Example 3. Your question is this: At what altitude above the surface should your deceleration systembe activated in order to achieve a soft touchdown? For a reasonable problem, you can take

where $g=GM/R^2$ is the surface gravitational acceleration of the planet. Choose p to be the smallest nonzero digit and q the next-to-smallest nonzero digit in your ID number. Find the “ignition altitude” accurate to the nearest meter and the resulting “descent time” accurate to the nearest tenth of a second.

Consider a satellite in elliptical orbit around a planet of mass M, and suppose that physical units are so chosen that $GM=1$ (where G is the gravitational constant). If the planet is located at the origin in the xy-plane, then the equations of motion of the satellite are

Let T denote the period of revolution of the satellite. Kepler’s third law says that the square of T is proportional to the cube of the major semiaxis a of its elliptical orbit. In particular, if $GM=1,$ then

(For details, see Section 11.6 of Edwards and Penney, Calculus: Early Transcendentals, 7th ed., Hoboken, NJ: Pearson, 2008).) If the satellite’s x- and y-components of velocity, $x_3=x^{\text{′}}=x_1^{\text{′}}\text{ and }{x}_4=y^{\text{′}}=x_2^{\text{′}},$ are introduced, then the system in (2) translates into a system of four first-order differential equations having the form of those in Eq. (22) of this section.

1. Solve this $4\times 4$ system numerically with the initial conditions

   $$x(0)=1,y(0)=0,x^{\text{′}}(0)=0,y^{\text{′}}(0)=1$$

   that correspond theoretically to a circular orbit of radius $a=1,$ so Eq. (3) gives $T=2\pi .$ Is this what you get?

2. Now solve the system numerically with the initial conditions

   $$x(0)=1,y(0)=0,x^{\text{′}}(0)=0,y\text{′}(0)=\frac{1}{2}\sqrt{6}$$

   that correspond theoretically to an elliptical orbit with major semiaxis a $a=2,$ so Eq. (3) gives $T=4\pi \sqrt{2.}$ Is this what you get?

Halley’s comet last reached perihelion (its point of closest approach to the sun at the origin) on February 9, 1986. Its position and velocity components at that time were

(respectively), with position in AU (astronomical units, in which the unit of distance is the major semiaxis of the earth’s orbit) and time in years. In this system, the threedimensional equations of motion of the comet are

where

Solve the equations in (4) numerically to verify the appearance of the yz-projection of the orbit of Halley’s comet shown in Fig. 7.7.15. Plot the xy- and xz-projections as well.

Figure 7.7.16 shows the graph of the distance r(t) of Halley’s comet from the sun. Inspection of this graph indicates that Halley’s comet reaches a maximum distance (at aphelion) of about 35 AU in a bit less than 40 years and returns to perihelion after about three-quarters of a century. The closer look in Fig. 7.7.17 indicates that the period of revolution of Halley’s comet is about 76 years. Use your numerical solution to refine these observations. What is your best estimate of the calendar date of the comet’s next perihelion passage?

The night before your birthday in 2007 you set up your telescope on a nearby mountaintop. It was a clear night, and you had a stroke of luck: At 12:30 A.M. you spotted a new comet. After repeating the observation on successive nights, you were able to calculate its solar system coordinates ${\mathbf{p}}_0=(x_0,y_0,z_0)$ and its velocity vector ${\mathbf{v}}_0=(v_{x0},{\text{ v}}_{y0},v_{z0})$ on that first night. Using this information, determine the following:

• the comet’s perihelion (point nearest the sun) and aphelion (point farthest from the sun),

• the comet’s velocity at perihelion and at aphelion,

• the comet’s period of revolution around the sun, and

• the comet’s next two dates of perihelion passage.

Using units of length in AU and time in earth years, the equations of motion of your comet are given in (4). For your personal comet, begin with random initial position and velocity vectors with the same order of magnitude as those of Halley’s comet. Repeat the random selection of initial position and velocity vectors, if necessary, until you get a plausible eccentric orbit that ranges well outside the earth’s orbit (as most real comets do).