Question

Find the energy necessary to put 7 kg , initially at rest on Earth's surface, into geostationary orbit.

Answer 1

It is not clear what you want to put into the geosynchronous (gs) orbit.
I suppose you want to put a 7 kg mass into the gs orbit.

The energy of the mass m in such an orbit is
E = P.E. + K.E.
...= GM m / (r - R) + m v^2 /2 ........(1)
v is the linear velocity of the mass m in gs orbit of radius r
This given by ,
GM m / r^2 = m v^2 / r
v^2 = GM / r ..........................(2)
Put this value in (1)
E = GM m / (r - R) + GM m / 2 r = ........(3)

For a gs orbit the angular velocity of the mass m must be same as that of the earth's rotation on its axis.
ω = 2π / T = 2 π / 24 x 3600 = 7.272 x10^-5 rad/s

From (2),
GM / r = r^2 ω^2
r^3 = GM / ω^2
G = 6.67x10^-11
M = 6x10^24 kg
That gives r , the radius of the gs orbit from the center of the earth, as
r = 4.23x10^7 m

R = 6.4 x10^6 m (radius of the earth)
r - R = (42.3 - 6.4)x10^6 = 35.9 x10^6 m

From (3) you can obtain the energy of the mass m = 7 kg as ,
E = 11.1143 x 10^6 J

This is the minimum energy required to put the object of mass 7 kg on the earth surface into a gs orbit.