In any circular orbit, the centripetal force required to maintain the orbit is provided by the gravitational force on the satellite.
Let #r# be the orbital radius as measured from the Earth's center of mass.
#m_s and M_e# mass of satellite and of earth respectively, and
#v# be the orbital velocity of satellite. Equating both forces we have
# F_c = F_g#
#(m_sv^2)/r=G(M_em_s)/r^2#, where #G# is Universal gravitational constant.
#=>v^2=G(M_e)/r# ........(1)
For motion in a circle we know that angular acceleration #omega# is
#omega=(2pi)/T# Also #v=r omega#
#:. v=r((2pi)/T)#
#=> r=v/((2pi)/T)# .......(2)
We are required to calculate the energy required for raising the satellite from earth's surface to GSO, substituting value of #r# from (2) in to (1) and solving for #v# we obtain
#v^2=G(M_e)/(v/((2pi)/T))#
#=>v^3=GM_e((2pi)/T)#
#=>v=[GM_e((2pi)/T)]^(1/3)#......(3)
For a geostationary orbit time period
#T=24# hours #=8.64xx10^4s#.
Mass of Earth, #M_e=5.9736 xx10^24 kg#, and
#G# is #6.6743 xx 10^(−11) m^3 kg^(−1) s^(−2)#. Inserting various values in (3), we obtain
#v=[6.6743 xx 10^(−11)xx5.9736 xx10^24((2pi)/(8.64xx10^4))]^(1/3)#
#=>v=3072ms^-1#
Kinetic energy to be imparted to the satellite
#KE=1/2 m_sv^2#
#KE=1/2 xx3xx(3072)^2#
#=>KE=1.416xx10^7J#, rounded to three decimal places.
