Question

Find the orbital velocity ,time period , height and total energy of a satellite?

Answer 1

Satellites: Orbital velocity, Time period and Energy 

A body moving in an orbit around a planet is called satellite. The moon is the natural satellite of the Earth. It moves around the Earth once in 27.3 days in an approximate circular orbit of radius 3.85 � 105 km. The first artificial satellite Sputnik was launched in 1956. India launched its first satellite Aryabhatta on April 19, 1975.

Satellites A body moving in an orbit around a planet is called satellite. The moon is the natural satellite of the Earth. It moves around the Earth once in 27.3 days in an approximate circular orbit of radius 3.85 � 105 km. The first artificial satellite Sputnik was launched in 1956. India launched its first satellite Aryabhatta on April 19, 1975.Orbital velocity Artificial satellites are made to revolve in an orbit at a height of few hundred kilometres. At this altitude, the friction due to air is negligible. The satellite is carried by a rocket to the desired height and released horizontally with a high velocity, so that it remains moving in a nearly circular orbit. The horizontal velocity that has to be imparted to a satellite at the determined height so that it makes a circular orbit around the planet is called orbital velocity. Let us assume that a satellite of mass m moves around the Earth in a circular orbit of radius r with uniform speed vo. Let the satellite be at a height h from the surface of the Earth. Hence, r = R+h, where R is the radius of the Earth. The centripetal force required to keep the satellite in circular orbit is F = mv02/r = mv02 / R+hThe gravitational force between the Earth and the satellite isF = GMm / r2 = GMm/(R+h)2For the stable orbital motion,v0 = root(gR2/R+h)If the satellite is at a height of few hundred kilometers (say 200 km), (R+h) could be replaced by R.orbital velocity, vo = rt(gR)If the horizontal velocity (injection velocity) is not equal to the calculated value, then the orbit of the satellite will not be circular. If the injection velocity is greater than the calculated value but not greater than the escape speed (ve = 2 vo), the satellite will move along an elliptical orbit. If the injection velocity exceeds the escape speed, the satellite will not revolve around the Earth and will escape into the space. If the injection velocity is less than the calculated value, the satellite will fall back to the Earth. Time period of a satellite Time taken by the satellite to complete one revolution round the Earth is called time period.Time period, T = circumference of the orbit / orbital velocityT = 2πr / v0  = 2π(R +h) / v0where r is the radius of the orbit which is equal to (R+h).v0 = root[GM/R+h] soT = 2π (R+h) root[(G+h)/GM]As GM = gR2, T = 2π root[(R+h)3/gR2]If the satellite orbits very close to the Earth, then h << RT = 2π root[R/g]Energy of an orbiting satelliteA satellite revolving in a circular orbit round the Earth possesses both potential energy and kinetic energy. If h is the height of the satellite above the Earth's surface and R is the radius of the Earth, then the radius of the orbit of satellite is r = R+h.If m is the mass of the satellite, its potential energy is,EP = - GMm/rwhere M is the mass of the Earth. The satellite moves with an orbital velocity of vo = root[GM/(R+h)]Hence, its kinetic energy is,EK = 1/2 mvo2  EK = GMm / 2(R+h)The total energy of the satellite is, E = EP + EKE = − GMm / 2(R+h)The negative value of the total energy indicates that the satellite is bound to the Earth.Geo-stationary satellites A geo-stationary satellite is a particular type used in television and telephone communications. A number of communication satellites which appear to remain in fixed positions at a specified height above the equator are called synchronous satellites or geo-stationary satellites. Some television programmes or events occuring in other countries are often transmitted 'live' with the help of these satellites. For a satellite to appear fixed at a position above a certain place on the Earth, its orbital period around the Earth must be exactly equal to the rotational period of the Earth about its axis. Consider a satellite of mass m moving in a circular orbit around the Earth at a distance r from the centre of the Earth. For synchronisation, its period of revolution around the Earth must be equal to the period of rotation of the Earth (ie) 1 day = 24 hr = 86400 seconds. The speed of the satellite in its orbit isv = Circumference of orbit / Time periodv = 2π r / TThe centripetal force is F = mv2/rF = 4mπ2r /T2The gravitational force on the satellite due to the Earth isF = GMm/r2For the stable orbital motion 4mπ2r / T2  = GMm / r2We know that, g = GM/R2r3 = gR2T2 / 4 π2The orbital radius of the geo- stationary satellite is, r = [gR2T2 / 4 π2]1/3Substituting T = 86400 s, R = 6400 km and g = 9.8 m/s2, the radius of the orbit of geo-stationary satellite is calculated as 42400 km. The height of the geo-stationary satellite above the surface of the Earth is h = r - R = 36000 km.