Many satellites move in stable orbits around the earth, some have left the gravitational field of the Earth and move around other planets and a couple are on the way to leave the solar system. An important task of astronomy is therefore to determine the orbits of these satellites.

In the calculation of satellite orbits two speeds \( v_1, v_2 \) play a major role. They depend on the starting altitude of the satellite (see below for derivation).

The first velocity \( v_1 \) is the launching speed, which ensures a stable circular orbit. If the satellite is launched at a higher speed its orbit is deformed into an ellipse. If a satellite, however, is launched with a smaller velocity it will probably fall to the ground and explode.

The second velocity \( v_2 \) is the launching speed, which empowers a satellite to escape the gravitational field of the Earth on a parabolic path. If a satellite is launched with even higher speed it will leave the gravitational field on a hyperbolic path.

The following simulation calculates the trajectories of satellites at a given starting altitude and velocity.

Altitude

\( \rm km \)

Launch velocity

\( \rm \frac{m}{s} \)

ResetStart

Semi-major axis

\( \approx \) \( \rm km \)

Num. eccentricity

A missile theoretically requires at least this speed to remain in an orbit around a celestial body, without falling onto the ground. It then moves on a circular orbit around the celestial body.

$$ v = \sqrt{G \, \dfrac{M}{R+h}} $$
\( M \) = Mass of the celestial body \( R \) = Radius of the celestial body,

\( G \) = Gravitational constant, \( h \) = Altitude of the satellite over the ground

The gravitational force of the earth in this case acts as a centripetal force, which forces the missile onto a circular path.

\begin{aligned} F_\rm{Z} &= F_\rm{G} \\ \\ \dfrac{\cancel m \cdot (v_1)^2}{\cancel{R+h}} &= G \, \dfrac{\cancel m \cdot M}{(R+h)^{\cancel 2}} \\ \\ (v_1)^2 &= G \, \dfrac{M}{R+h} \\ \\ v_1 &= \sqrt{G \, \dfrac{M}{R+h}} \\ \\ \end{aligned}If a satellite is launched at a higher speed its orbit deforms into an ellipse.

A missile theoretically requires at least the second cosmic velocity

to escape the gravitational field of the Earth.

$$ v_2 = \sqrt{2 \, G \, \dfrac{M}{R+h}} $$
\( M \) = Mass of the celestial body \( R \) = Radius of the celestial body,

\( G \) = Gravitational constant, \( h \) = Altitude of the satellite over the ground

In theory to escape the gravitational field of the Earth, a missile has to move infinitely far from the Earth. The required energy can be calculated:

\begin{aligned} \Delta W &= G \cdot M \cdot m_\rm{R} \cdot \left( \dfrac{1}{r_1} - \dfrac{1}{r_2} \right) \\ \\ &= G \cdot M \cdot m_\rm{R} \cdot \left( \dfrac{1}{R+h} - \dfrac{1}{\infty} \right) \\ \\ &= G \cdot M \cdot m_\rm{R} \cdot \dfrac{1}{R+h} \\ \end{aligned}This work must be available in the kinetic energy of the missile.

\begin{aligned} E_\rm{kin} &= \Delta W \\ \\ \dfrac{\cancel m_\rm{R}}{2} \cdot (v_2)^2 &= G \cdot M \cdot \cancel m_\rm{R} \cdot \dfrac{1}{R+h} \\ \\ (v_2)^2 &= 2 \cdot G \cdot M \cdot \dfrac{1}{R+h} \\ \\ v_2 &= \sqrt{ 2 \,\, G \dfrac{M}{R+h} } \\ \\ \end{aligned}- Wikipedia: Article about "Escape velocity"

- Deutsche Version: Artikel über "Satellitenbahnen"