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Simple orbital mechanics problems

An unmanned satelite orbits the Earth

clear all;
clc;
rp = 7000;                      %[km] Perigee radius
ra = 70000;                     %[km] Apogee radius
mu = 398600;                    %[km^3/s^2]

ecc = (ra-rp)/(ra+rp)           %Eccentricity of the orbit
a   = (ra+rp)/2                 %[km] Semimajor axis
T   = 2*pi/mu^0.5*a^1.5/3600    %[h] Period of the Orbit
E_spec = -mu/(2*a)              %[km^2/s^2]Specific energy
ecc =

0.8182

a =

38500

T =

20.8833

E_spec =

-5.1766

Find true anomaly when Alt = 1000 km;

R_earth = 6378;
theta = acos(a*(1 -ecc^2)/((R_earth+1000)*ecc)-1/ecc)*180/pi %[deg]
theta =

27.6069

Velocity components at this anomaly

hmu = (a*(1 -ecc^2)/mu)^0.5;
vr = 1/hmu *ecc*sin(theta)          %radial component
vt = 1/hmu*(1 + ecc*cos(theta))     %tangential component
vr =

2.8342

vt =

2.0001

Velocities at Apogee and Perigee

vp = 1/hmu*(1 + ecc)
va = 1/hmu*(1 - ecc)
vp =

10.1751

va =

1.0175

The spacecraft is in a 250 km by 300km low eart orbit. How long does it take to coast from perigee to appogee. Semimajor axis of this eliptical orbit

a = R_earth + (250+300)/2; % [km]
% Time which takes to fly from Perigee to Appogee equals to half of the
% Orbit period
Th = pi/mu^0.5*a^1.5/60 %[min]
Th =

45.0045

The altitude of a satelite in an elliptical orbit around the earth is 1600 km at apogee and 600 km at perigee.

ra = R_earth + 1600;
rp = R_earth +600;
% Eccentricity of the Orbit
ecc = (ra - rp)/(ra + rp)
% Semimajor axis;
a = (ra + rp)/2
% The orbital Speed at perigee and apogee
hmu = (a*(1 -ecc^2)/mu)^0.5;
vp = 1/hmu*(1 + ecc)
va = 1/hmu*(1 - ecc)
% The period of the Orbit
T = 2*pi*a^1.5/mu^0.5/60 %[min]
ecc =

0.0669

a =

7478

vp =

7.8065

va =

6.8280

T =

107.2601

A satellite is palced into an orbit at perigee at an altitude of 1270 km with a speed of 9 km/s. Calculate the flight path angle gamma and altitude of the satelite at a true anomaly of 100deg.

altp  = 1270;    %[km]
vp    = 9;       %[km/s]
theta = 100;     %[deg]
rp = altp + R_earth;
h = rp * vp
ecc = h^2/(rp*mu) - 1
gamma = atan( ecc*sind(theta)/(1 + ecc*cosd(theta)));
gamma = 180/pi*gamma    %[deg]
h =

68832

ecc =

0.5542

gamma =

31.1256

Altitude of the satelite at a true anomaly of 100 [deg].

rd = h^2/mu*(1/(1 + ecc* cosd(theta)));
altd = rd - R_earth
altd =

6.7738e+003

A satelite is launched into the orbirt at an altitude of 640 km with a speed 9.2 km/s and flight path angle of 10 deg. Calculate the true anomaly of the lunch poinnt and the period of the orbit

alt = 640;           %[km]
rs  = R_earth + alt; %[km]
vel = 9.2;           %[km/s]
gamma = 10;          %[deg]
vp = vel*cosd(gamma);
vr = vel*sind(gamma);
h  = rs * vp;
ac = (h^2/(mu*rs)-1);
bc = (vr*h/mu);
theta = atan(bc/ac)*180/pi  %Anomaly of the lunch point
%Period of the orbit
%Eccentricity
ecc = ac/cosd(theta);
T = 2*pi/mu^2*(h/(1 - ecc^2)^0.5)^3  %[sec]
theta =

29.7830

T =

1.6075e+004

A satellite has a perigee and apogee altitudes of 250 km and 42000km. Calculate the orbit period, eccentricity and the maximum speed.

rp = R_earth + 250; %km
ra = R_earth + 42000;%km

%Semimajor axis of this eliptical orbit
a =( rp + ra )/2;
T = 2*pi/mu^0.5 * a ^(3/2)/60       %[min]
ecc = (ra - rp)/(ra+rp)
h = (mu*(1 + ecc)*rp)^0.5;
vp = h/rp                           %[km/s]
T =

756.5368

ecc =

0.7590

vp =

10.2852

A rocket lunched from the surface of the earth has a speed 8.85 km/s when powered flight ends at an altitude of 550 km. The flight path angel at this time is 6 degree.Determinine eccentricity

vel  = 8.85 ;           %[km/s]
ra = R_earth + 550;     %[km]
gamma = 6;              %[deg]
vd = vel * cosd(gamma);
h  = ra * vd;
ect = h^2/(mu*ra)-1;
vr = vel* sind(gamma);
est = h*vr/mu;
theta = atan(est/ect);
ecc = ect/cos(theta)
ecc =

0.3742

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