Projection of Satellite

Launching an artificial satellite into a stable orbit requires giving it a very specific COMBINATION of speed and direction -- something a single-stage rocket cannot achieve, which is why a minimum of a TWO-stage rocket is always used. When the first stage's fuel is ignited at the Earth's surface, it lifts the satellite VERTICALLY, straight upward, imparting a vertical velocity of projection. If this vertical (upward) velocity is less than the escape velocity $v_e$, the satellite would simply fall back to the Earth if released with only this vertical motion; equalling or exceeding $v_e$ in the purely vertical direction would instead send it escaping outward to infinity -- neither of which gives a satellite that ORBITS the Earth. This is precisely why a single-stage vertical launch alone can never place a satellite into orbit.

Instead, once the first rocket stage has carried the payload to the desired height, the launcher is reoriented through 90 degrees (using remote control) into the HORIZONTAL direction, the first stage is detached, and a second rocket stage then imparts a specific HORIZONTAL velocity $v_h$ to the satellite, so that it can circle the Earth rather than simply falling back or flying off. The precise horizontal velocity of projection, at a given height, needed to make the satellite move in a stable CIRCULAR orbit at that height is called the critical velocity or orbital velocity, $v_c$.

For a satellite of mass m orbiting at height h (so its orbital radius is $R+h$), the necessary centripetal force for its circular motion is supplied entirely by the Earth's gravitational pull on it: $\text{Centripetal force}=\text{Gravitational force}\quad\Longrightarrow\quad\dfrac{mv_c^2}{R+h}=\dfrac{GMm}{(R+h)^2}.$ Cancelling m and one factor of $(R+h)$ from both sides gives $v_c^2=\dfrac{GM}{R+h}\quad\Longrightarrow\quad v_c=\sqrt{\dfrac{GM}{R+h}}=\sqrt{g_h(R+h)},$ where $g_h$ is the acceleration due to gravity at that height. This critical speed depends only on the mass of the Earth and the height of the orbit (or, equivalently, the local gravitational acceleration there) -- it is completely INDEPENDENT of the satellite's own mass, and it DECREASES as the orbital height increases (a higher orbit needs a smaller circular speed).

A special case worth remembering: for a satellite orbiting very close to the Earth's surface, $h\ll R$ so $R+h\approx R$, giving $v_c\approx\sqrt{GM/R}=\sqrt{gR}\approx7.92$ km/s -- the MAXIMUM possible critical speed for any Earth-orbiting satellite, and, notably, at least 25 times faster than the fastest passenger aircraft. …