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Satellites and projectiles in a gravitational field

A satellite is any object that orbits a planet and this behaviour is caused by gravity.

Examples of satellites are the moon and TV satellites orbiting the Earth.

The body around which the satellite orbits is called its primary.

A geostationary satellite is a satellite that always stands above a fixed point on the surface of the Earth. That means its period $$T$$ is one day (strictly speaking the period of one revolution of the Earth around its own axis).

This can only be achieved by putting the satellite at a specific orbital radius (around $$42000\text{ km}$$ for the Earth). Geostationary satellites can only orbit above the Equator.

A man-made satellite in orbit around the earth.

The orbital speed ($$v_{\text{o}}$$) of a satellite is the speed at which a satellite moves around its primary on a circular orbit, for example the Moon around the Earth.

The escape speed ($$v_{\text{e}}$$) is the initial speed needed for a projectile to leave the gravitational field of another body. A projectile is an object without onboard propulsion. The equations are:

$$$ v_{\text{o}}=\sqrt{G\frac{M}{r_{\text{o}}}} \quad \quad v_e=\sqrt{2G\frac{M}{r_{\ell}}}$$$

If $$r_{\text{o}}=r_{\ell}$$, then $$v_{\text{e}}= v_{\text{o}} \sqrt{2} $$.

In any case, the mass of the satellite is irrelevant to its orbital or escape speed.

$$G=$$universal gravitational constant; $$M=$$mass of the primary (e.g the Earth); $$m=$$mass of the satellite (too small to significantly affect primary), $$r_{\text{o}}=$$orbital radius of a satellite; $$r_{\ell}=$$distance between centre of the Earth and launch position (launch radius).

A man-made satellite in orbit around the earth.

The path of a projectile or a satellite moving in the proximity of a body (e.g. the Earth) depends on the body's speed relative to the orbital speed $$v_{\text{o}}$$ and the escape speed $$v_{\text{e}}$$.

Note that the direction of velocity of the body is variable. The following equations only give the magnitude of velocity (i.e. speed).

$$$v_{\text{o}}=\sqrt{GM/r_{\text{o}}} \quad \quad v_{\text{e}}=\sqrt{2GM/r_{\ell}}$$$

Possible trajectories are:

$$v\lt v_{\text{o}}$$: the object will fall to the primary (e.g. the Earth).

$$v=v_{\text{o}}$$: the object will stay on a circular orbit around the primary.

$$v_{\text{o}}\lt v\lt v_{\text{e}}$$: the object will stay on an elliptical orbit.

$$v\ge v_{\text{e}}$$: the object will leave the gravitational field of the primary.

$$G=$$universal gravitational constant, $$M=$$mass of the primary (e.g the Earth), $$m=$$mass of the satellite (too small to affect primary), $$r_{\text{o}}=$$orbital radius of a satellite, $$r_{\ell}=$$distance between centre of the Earth and launch position (launch radius).

Speed is lower than orbital speed: A (black), Speed is equal to orbital speed: B (red), Speed is greater than orbital speed but smaller than escape speed: C and D (blue), and speed is equal to or greater than escap speed: E (green).

A satellite is kept in its orbit by the gravitational force $$F_{\text{g}}$$ which acts as a centripetal force. For a circular orbit, $$F_{\text{g}}$$ must be exactly equal to the centripetal force $$F_{\text{c}}$$ for the specific orbit. This is the basis for deriving the equation for the orbital speed.

Orbiting speed requirement: $$F_{\text{g}}=F_{\text{c}}$$

Centripetal force (keeping satellite in orbit): $$F_{\text{c}}= ma=mv^2/r$$

Gravitational force (of primary on satellite): $$F_{\text{g}}=GMm/r_{\text{o}}^2$$

From Newton's second law,$$$\begin{align*}m\frac{v^2}{r}= G\frac{Mm}{r^2} \iff & v^2= G\frac{M}{r}\\\iff & v=\sqrt{ \frac{ G M}{r}}\end{align*}$$$

The formula applies to circular orbits and assumes that the primary is stationary.

At orbital speed, the gravitational force (red) is just strong enough to pull the satellite back onto its orbit and the speed along the orbit is just high enough to prevent the satellite from falling down to Earth.

The escape speed can be derived from the definition of potential gravitational energy $$E_{\text{p}}$$. $$E_{\text{p}}$$ of an object is defined as the work done in bringing an object from infinity to its current location.

Escaping a gravitational field means moving to a point where the gravitational field strength is zero (i.e. to infinity).

For most practical purposes, this is the distance at which the field strength at the point is so weak as to be almost zero. This happens when the object is at a great distance from the source of the gravitational field.

The object needs to be given sufficient kinetic energy $$E_{\text{k}}$$ (i.e. it needs to have sufficient speed) to offset the potential energy.

Using this requirement and the equations for potential and kinetic energy, the equation for the escape speed can be derived.

The energy contained within rocket fuel gives the rocket enough kinetic energy to escape the gravitational field of the earth.

Escape speed requirement: $$v=v_{\text{e}}$$ if $$E_{\text{k}}+E_{\text{p}}=0$$

Kinetic and potential energy: $$$E_{\text{k}}=1/2mv^2 \quad \quad E_{\text{p}}=-G\frac{Mm}{r}$$$

Combining and reformulating: $$$\begin{align*}\frac{1}{2}mv_{\text{e}}^2-G\frac{Mm}{r }=0 & \iff v_{\text{e}}^2=G\frac{Mm}{r} \frac{2}{m}\\& \iff v_{\text{e}}^2=2G\frac{M}{r}\\& \iff v_{\text{e}}=\sqrt{2G\frac{M}{r }}\end{align*}$$$

$$G=$$universal gravitational constant; $$M=$$mass of the primary; $$m=$$mass of the satellite; $$r=$$distance between the centre of mass of the primary and the launch position.

The equation assumes no additional propulsion . If for example the spacecraft carries an fuel engine, the energy in the fuel (e.g. chemical energy $$E_{\text{c}}$$) would need to be included in the requirement for escape speed: $$E_{\text{c}}+E_{\text{k}}+E_{\text{p}}=0$$.

The energy contained within rocket fuel gives the rocket enough kinetic energy to escape the gravitational field of the earth.

Kepler's laws are principles of planetary (i.e. satellite) motion first recorded by Johannes Kepler (1571-1630).

Kepler's first law: The orbit of every planet is an ellipse with the primary at one of its foci (an ellipse can be thought of as an elongated circle with two "centres" or foci).

Kepler's second law: A line joining a planet to the sun sweeps out equal areas during equal intervals of time. The implication is that a planet goes much faster in the section of its orbit that is closer to the Sun.

Kepler's third law: The square of the orbital period $$T$$ is proportional to the cube of the semi major axis $$a$$ (one half of the longest diameter of the ellipse) of its orbit: $$$T^2\propto a^3$$$

The Sun is at a focus (f1) of the orbits of planet 1 and planet 2 (First Law). Planet 1 sweeps out section A1 and A2 (which are equal in area) implying a slower speed at A2. The semi-major axis is one half of the lines denoted a1 and a2 in the image (the longest axes of the ellipse).