MSG-4 Launch into GEO

Category: Applications

The fourth and last satellite of the Meteosat Second Generation (MSG-4) was succesfully launched from French Guiana on July 15, 2015 aboard an Ariane 5 ECA rocket. MSG-4 will be placed, as its predecessors, in a geosynchronous also called Geosynchronous Equatorial Orbit (GEO). A geosynchronous orbit has an orbital period matching the rotation rate of the Earth. This is a sidereal day, which is 23h 56m 4s in length, and represents the time taken for the Earth to rotate once about its polar axis relative to a distant fixed point. Lets see what has to be done to get the satellite there ...

For many kinds of orbits including the geostationary one, launches are conducted from near the equator in an eastern direction as the radial velocity of earth is greatest there so less kinetic energy has to be delivered by the Ariane 5 rocket.

After lift off and separation of the first two stages the satellite will be placed in a high elliptical Geostationary Transfer Orbit (GTO). One option of describing orbits in general is by their kinetic and potential energy

$E_{Tot}&space;=&space;E_{Kin}&space;+&space;E_{Pot}$

$\mu&space;=&space;GM_{Earth}$

with $$\mu$$ defining the gravitational parameter of the earth. $$a$$ is the semi-major axis and $$0 \le e \le 1$$ the excentricity. r represents the true anomaly of the satellite, e.g. its actual position in orbit and EC defines the Earth Center located in one focal point of the ellipse. A is the apogee and P the perigee, e.g. where the distance of the satellite is a maximum w.r.t. the EC or a minimum respectively. Converting the equation gives the velocity on an elliptical orbit and hence on a GTO

$v_{ell}&space;=&space;\sqrt{\mu&space;\;&space;\frac{2}{r}-\frac{1}{a}}$

For circular orbits $$a$$ is equal to r and therefore the equation simplifies to

$v_{circ}&space;=&space;\sqrt{\frac{\mu&space;}{a}}$

That means that for an elliptical orbit the the orbit velocity $$v_{ell}$$ depends on the current position r of the satellite in orbit. So if the orbit is not a circle, hence the satellite does not move at constant velocity, it appears to oscillate east-and-west at a rate of two cycles per sidereal day. A geosynchronous orbit also has to lie in the equatorial plane (inclination of 0 degree) because otherwise the satellite does not remain at a fixed point in the sky - it would appears to oscillate north-and-south at a rate of one cycle per sidereal day.

In order to get the satellite from GTO into the circular GEO a so called in-plane maneuver has to be performed. A in-plane maneuver is defined as a maneuver that changes the shape of an orbit. The shape of an orbit is basically described by two parameters, e.g. the semi-major axis, and the eccentricity. That means if for example a in-plane maneuver is performed on a circular orbit, the resulting orbit will be always an elliptical orbit due to the change in shape. If we consider now impulsive maneuvers where the burn duration is much smaller than the orbital period only the kinetic energy is changed at the location of the maneuver, but for all other points along the orbit the altitude is changed. Usually the altitude of the perigee or agogee. For MSG-4 the perigee of the GTO has to be lifted to GEO altitude. This means that once the satellite reached the apogee it has to increase its velocity. The reason is that the velocity in the apogee of an elliptical orbit inside the circular target orbit is always samller compared to the required circular velocity - otherwise it would not 'fall back' to an elliptical orbit.  In order to know the velocity increment required at apogee to get the satellite from a GTO to a GEO following the following calculations have to be done. By knowing the subsequent parameters calculated and determined during initial orbit determination

$r_{apo}&space;~&=&~&space;42,&space;164~km$

$r_{GTO}&space;~&=&~&space;24,&space;400~km$

$r_{GEO}&space;~&=&~&space;42,&space;164~km$

and using the equations for the respective velocities for circular (GEO) and elliptical (GTO) orbits the velocities for the apogee $$v_{GTO}$$ and the required circular velocity $$v_{GEO}$$can be calculated. First we calculate the velocity in apogee of the GTO orbit and then the velocity for the circular orbit. Subtracting the higher GEO velocity from the GTO velocity gives the necessary in-plane velocity increment.

$\Delta&space;v_{IPL}&space;=&space;v_{GEO}&space;-&space;v_{GTO}&space;=&space;3.075&space;-&space;1.063&space;=&space;1.472~km~s^{-1}$

In practice this increase of velocity is done in a step by step approach to finally place the satellite into its proposed target box, e.g. its final position in the sky.

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