Orbital Mechanics

The Core Mathematics

The foundation of all orbital mechanics is equating centripetal force to gravitational force.

$$\text{Gravitational Force Provides Centripetal Acceleration}$$ $${\mathrm{F}}_{\text{Gravitational}}={\mathrm{F}}_{\text{Centripetal}}$$ $$\frac{GMm}{r^2}=\frac{m{v}^2}{r}=mr{\omega }^2$$

Key derivations:

1. Orbital Velocity ($v$)

$$v=\sqrt{\frac{GM}{r}}$$

2. Angular Velocity ($\omega$)

$$\omega =\sqrt{\frac{Gm}{r^3}}$$

3. Kepler’s Third Law (Period $T$)

$$T^2=\left(\frac{4{\pi }^2}{GM}\right)r^3⟹T^2\propto r^3$$

Geostationary Orbits

A satellite that remains above the exact same point on the Earth. It requires these conditions:

1. Its period $T$ must be exactly 24 hours.
2. It must orbit from west to East.
3. It must be in an equatorial orbit.

Generalised Conditions

For a satellite orbiting above a planet

1. The orbital period must equal the period of rotation of the planet.
2. The satellite must orbit in the same direction as the planet’s rotation
3. The orbit must be equatorial

Todo

1. What if planet is stationary
2. What happens if orbit is not equatorial

Energies In Orbit

Energy Formulae

Gravitational Potential Energy ($E_p$)

$$E_p=\varphi \times m$$ $$E_p=-\frac{GM}{r}\times m$$ $$E_p=-\frac{GMm}{r}$$

Warning

Gravitational Potential Energy is always negative.

Kinetic Energy ($E_k$)

$$\frac{m{v}^2}{r}=\frac{GMm}{r^2}$$ $$m{v}^2=\frac{GMm}{r}$$ $$\frac{1}{2}m{v}^2=\frac{GMm}{2r}$$ $$E_k=\frac{GMm}{2r}$$

Warning

Kinetic Energy is always positive

Total Energy ($E_T$)

$$E_T=E_P+E_k$$ $$E_T=-\frac{GMm}{r}+\frac{GMm}{2r}$$ $$E_T=-\frac{2GMm}{2r}+\frac{GMm}{2r}$$ $$E_T=-\frac{GMm}{2r}$$

Warning

Total Energy is always negative

Energy Ratio

$$E_T:E_p:E_k$$ $$-\frac{GMm}{2r}:-\frac{GMm}{r}:\frac{GMm}{2r}$$ $$-\frac{1}{2}\left(\frac{GMm}{r}\right):-\left(\frac{GMm}{r}\right):\frac{1}{2}\left(\frac{GMm}{r}\right)$$ $$-\frac{1}{2}:-1:\frac{1}{2}$$ $$1:2:-1$$

Important

The same ratio is applicable for energy changes: $\Delta E_T:\Delta E_P:\Delta E_K$ $1:2:-1$

Warning

This ratio is only applicable when comparing old and new stable orbits. For example, if a satellite is in an orbit $A$ and you fire thrusters or start experiencing drag and finally settles in an orbit $B$, this ratio is only applicable to comparing the energies in orbit $A$ and $B$ not in the time in between.

Scenarios

1. Thrusters are fired (mass assumed to be constant)

Since mass is constant, $GMm$ is constant. Therefore,

$$E_T=-\frac{GMm}{2r}⟹E_T\propto -\frac{1}{2r}$$

• When thrusters are fired kinetic energy increases and total energy increases (becomes less negative)
• For the expression $-\frac{1}{2r}$ to become less negative, $r$ must increase.
• Therefore, orbital radius increases.

2. Atmospheric Drag

When a satellite experiences resistive forces, total energy decreases (becomes more negative). Since,

$$E_T\propto -\frac{1}{2r}$$

For total energy to become more negative, $r$ must decreases. Therefore orbital radius decreases. Since gravitational potential energy is given by,

$$E_p=-\frac{GMm}{r}$$

When radius decreases, gravitational potential energy decreases (becomes more negative) Since kinetic energy is given by,

$$E_K=\frac{GMm}{2r}$$

When radius decreases, kinetic energy increases (becomes more positive)

Binary Star Systems

Two stars ($M_A$ and $M_B$) orbiting a common centre of mass $P$.

A binary star system

Important

• According to Newton’s third law, the gravitational force between the two stars is the same. (star A exerts a force on B, B exerts an equal and opposite force on A)
• Since in circular orbit gravitational force provides centripetal force and the gravitational force is the same, centripetal force is the same.
• They have the same angular frequency $\omega$ and orbital period $T$.
• They remain opposite to each other, with a straight line passing through $P$ joining their centres.

$$\text{Centripetal Force on A}=\text{Centripetal Force on B}$$ $$M_A{r}_A{\omega }^2=M_B{r}_B{\omega }^2$$ $$M_A{r}_A=M_B{r}_B$$ $$\frac{M_A}{M_B}=\frac{r_B}{r_A}$$

Escape Velocity

Definition

Minimum speed an object needs at a point in a gravitational field to escape to infinity, coming to rest with zero total energy.

$$\Delta E_K=\Delta E_P$$

Gravitational Potential Energy is defined as zero at infinity

$$\frac{1}{2}m{v}^2=0-\left(-\frac{GMm}{r}\right)$$ $$\frac{1}{2}m{v}^2=\frac{GMm}{r}$$ $$\frac{1}{2}{v}^2=\frac{GM}{r}$$ $$v=\sqrt{\frac{2GM}{r}}$$

In terms of gravitational potential ($\varphi$)

$$v=\sqrt{2\times \frac{GM}{r}}$$ $$v=\sqrt{-2\varphi }$$