A method to evaluate launch speed to orbit the Earth

Consider a body of mass m approaching a large mass M. Every such body (m) has a total energy comprising kinetic (motion) and potential (virtue of position) parts i.e.

E= KE + PE energy = T + V

At any instant mass m is a distance r from M and has a speed v.

Writing KE as T =  m * v * v/2, and PE as V= -G * M * m / r (PE of m increases with distance from M).

A body in orbit with mass m is either closed E<0 (following an elliptical or circular path), unconfined E>0 (hyperbolic path e.g. 1/x) or between E=0 - just escaping the gravitational pull (e.g. a parabolic path x*x) .

T =  m * v * v / 2,   V = -G * M * m / r

For parabolic E=0 = T + V i.e. T = -V

But for a circular orbit about the earth, centripetal (towards M) and gravitational forces equate i.e.

m * vc * vc / r = G * M *m / (r^2) so

m * vc * vc / 2 = 1/2 * G * M * m / r = - V /2

For circular orbit Tc = - V/2

So T is twice as big for parabolic orbit wrt a circular orbit, ie. speed is sqrt(2) times circlular orbit's speed.
Again from circular earth orbit centripetal force equals gravitaional force

m * vc * vc /2 = 1/2 *G * M *m / Re , cancel the 2 and m

vc * vc = G * M / Re so vc =SQRT(G * M / Re).

Circular orbital speed

Values:
G = 6.67408 × 10^(−11)  , Re= 6.3 10 ^ -6 m , Me = 6*10^24 kg
vc = 8*10^3 m/s = 8km/s

Escape velocity

To escape earth's gravity velocity required is = sqrt(2) * vc

Values:
G = 6.67408 × 10^(−11)  , Re= 6.3 10 ^ -6 m , Me = 6*10^24 kg
vc = 8*10^3 m/s = 8kM/s
ve = sqrt(2)*vc  = 11.3 km/s