How fast must a satellite leave Earth's surface to reach an orbit with an altitude of 895 km?

Question

How fast must a satellite leave Earth's surface to reach an orbit with an altitude of 895 km?

Answer 1

The earth leaving rocket takes off with zero speed,acceleraTte,then reach about 8.2kilometers  per second  or higher speed inorder to reach an orbit of 895kilometer.

Answer 2

the satellite will be in orbit when the gravitational force is equal to the centrifugal force.

Fg = (G)(M)(m) / (R ^ 2)

Fc = (m)(v ^ 2) / (R + H)

Fg = Fc   condition to remain in orbit

v = [  [(G)(M)(R ^ H)] ^ (0.5)  ] / R

m = mass of satellite (cancelled out)

v = satellite velocity m/s

G = gravitational constant 6.673E-11

M = mass of earth 5.98E+24 kg

R = earth radius 6.37E+6 m

H = orbit high above earth surface in meters 895,000 m

from my excel

v = 8452.58 m/s