# Who discovered satellite orbit

## Satellite orbits

### introduction

Many satellites are in stable orbits around the earth, some have left the earth's gravitational field and are moving around other planets, and a few are on their way to leave the solar system. It is therefore an important task of astronomy to determine the orbits of such satellites.

### Track shapes

When calculating the satellite orbits, two speeds \ (v_1, v_2 \) play an important role. They depend on the starting altitude of the satellite (derivation see below).

The first speed \ (v_1 \) is the starting speed, which ensures a stable orbit. If the satellite starts at a higher speed, its orbit deforms into an ellipse. If, on the other hand, a satellite starts at a slower speed, it will probably fall towards the earth and burn up.

The second speed \ (v_2 \) is the starting speed that a satellite needs to escape the earth's gravitational field on a parabolic orbit. If the satellite starts at an even higher speed, it leaves the gravitational field on a hyperbolic orbit.

### Orbit simulation

The following simulation calculates the orbits of satellites for a given starting altitude and speed.

Draw a path

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

### Circular path

Theoretically, a missile needs at least this speed in order to remain in an orbit around a celestial body without driving without hitting its surface. It then moves on a circular path 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 \) = height of the satellite above the surface

In this case, the gravitational force of the earth acts as a centripetal force, which forces the missile on 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}

### ellipse

If the satellite starts at a higher speed, its orbit deforms into an ellipse.

### Escape from the gravitational field

A missile theoretically needs at least the second cosmic speed,
to escape the earth's gravitational field without driving.

\$\$ 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 \) = height of the satellite above the surface

In order to escape the earth's gravitational field, the rocket must theoretically travel infinitely far away. The energy required for this 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 present in the rocket's kinetic energy.

\ 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}

### literature

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