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# Tag Archives: apogee

## 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

Published with MATLAB® 7.10