motion in space

Requirements for orbit:

  

Orbital velocity and orbital period:

 The requirements to orbit an object can be determined by considering the example of a body of mass m, circling the earth, at a constant angular velocity w, at a radiusR from the center of the earth. The body in the rotating frame of reference will see the force of attraction due to the earth and the pseudo centrifugal force mw²R. From the universal law of gravitational force, the force that the body experiences due to the earth is seen to towards the center of the earth.

 We get   

                                           GmM/R²=mw²R

                                             W= (GM/R³)^1/2

                                         V = wR= (GM/R)^1/2

At a height of h the orbital velocity will be    

                                           V = {GM/(R+h)}^1/2

          

        The orbital velocity about the earth is typically a few km/s. It decreases as the radius of the orbit increases. It is seen to decrease from 7.9 km/s at earth`s surface to about 2.66km/s at a height of 50,000km.

The time period of revolution for one orbit is:

                                               T =2(3.14)R/V

                                           T = 2(3.14)[R³/GM]^1/2

                                          T=0.304*10^-6 R^3/2sec

Geosynchronous and Geostationary Orbits:

          If a body is in circular orbit with an angular velocity equal to that of earth and also rotates in the same direction as that of earth from east to west, its movement is synchronous with the rotation of the earth and it is said to be in geosynchronous orbit.  It will complete one rotation in a day.

               (GM/R)^1/2 = W(7.273*10^-5) rad/s

The radius of the geosynchronous orbit Rfrom the above equation is:

                             R³ = GM/w²

By substituting the values of G,M and w the value of R =42,164km.

With the radius of the earth being 6378km, the geosynchronous altitude would therefore be 42164-6378= 35,786km  above the earth.

            If the plane of the geosynchronous orbit is in the equatorial plane of the earth and the body rotates in the same direction as the earth rotates , the orbiting body will always appear to be stationary to an observer standing on the earth`s surface . It is then said to be in geostationary orbit. The orbiting body will appear to be a fixed point in the sky for an observer on the ground. Arthur Clarke, author of several science fiction books, had put forward this concept of a geostationary orbit in 1945. The orbit is therefore also referred to as Clarke orbit.

  The first satellite to be placed in the geostationary orbit was syncom-2 on July 26, 1963.

Eccentricity and Inclination of orbits:

          Orbits are not always circular and in the equatorial plane. An orbit could be elliptic and the departure from a circular orbit is known as eccentricity of the orbit. It is defined as the ratio of the distance between the foci to the length of the major axis of the ellipse. This is shown below.

                                        

      Here L_c is the distance between the foci and 2a is the length of the major axis giving the eccentricity e=L_c/2a. In the limit of circular orbit, L_c=0 and the eccentricity is zero. A body gets into an elliptic orbit when the orbital velocity provided to it is not equal to the orbital velocity required for a circular orbit.

Highly elliptic orbits with large inclinations are provided when the space craft has to remain over certain restricted zones over long durations.

Polar and Retrograde Orbits:

                                   

      The inclination of an orbit is the angle between the orbital plane and the equatorial plane of the planet about which the body is orbiting. An inclination of zero denotes that the orbit is in the equatorial plane, i.e the orbital plane and the equatorial plane coincide. An inclination of 90⁰ represents the orbit from pole to pole and is known as polar orbit. An inclination of 180⁰represents the orbiting body to move in the equatorial plane. It is known as a retrograde equatorial orbit.

Molniya orbit:

                    A particular orbit known as Molniya orbit is highly elliptic with the half major and half minor axis being about 46000km and 6800km respectively. The inclination of the orbit is 63.4⁰. Space crafts in the Molniya orbit remain in the northern hemisphere for as much as 11hrs in an orbit having time period of 12hrs, I.e; it takes only 1hr to travel the southern hemisphere.

Transfer orbit :

                       Instead of directly taking a spacecraft to geosynchronous orbit, the spacecraft is initially placed in an elliptical orbit about the earth. The orbit is known is known as the geosynchronous transfer orbit about the earth and is schematically shown in below figure. 

                                              

       The farthest point of the spacecraft from the center of the earth Is called Apogee. The point of the spacecraft closest to the earth is called perigee. The apogee and perigee are shown in fig. the apogee of the elliptic orbit is so chosen that it is near to the radius of the geosynchronous orbit of 42164km. the perigee is small at about 6630km, the choice being governed such that the drag effects at this altitude will be negligible. The elliptic transfer orbit is circularized to give geosynchronous orbit.

Escape velocity:
          If a body is provided with sufficient velocity to overcome the gravitational attraction, it would reach infinite radius and escape the attractive force of planet. This velocity is referred to as Escape velocity

                                             V_e=(2GM/R)^1/2

Velocities: Means of Providing The Required

           Velocities of several km/s are seen to be necessary for orbiting bodies about the earth. Velocities required by the bodies to escape from the earth are even higher. Large values of momentum or impulse are therefore essential. Jules Verne, the famous science fiction author, in his book from earth to the moon – de la terra a la Lune in 1874, suggested the use of a canon gun. If a velocity of 10km/s is considered to be provided impulsively to a body of say 1000kg over a period of 1ms, the energy 5*10⁷kJ and the power is 5*10¹⁰KW. This continues a large quantity of power. Even if the very high velocities could somehow be provided to the body, frictional heating in the atmosphere would destroy it . The high velocities are therefore achieved by gradually adding velocity to the body by the rocket as the body is being pushed up to the required orbit.