# Kepler’s Laws in Satellite Communication

Table of Contents

## Kepler’s Laws of planetary motion

Kepler’s laws apply quite generally to any two bodies in space that interact through gravitation. The more massive of the two bodies is referred to as the primary, the other, the secondary, or satellite.

Kepler’s Laws: Johannes Kepler, based on his lifetime study, gave a set of three empirical expressions that explained planetary motion. These laws were later vindicated when Newton gave the law of gravitation. Though given for planetary motion, these laws are equally valid for the motion of natural and artificial satellites around Earth or for anybody revolving around another body. Here, these laws will be discussed with reference to the motion of artificial satellites around Earth.

### Kepler’s First Law

First Kepler’s Laws: The orbit of a satellite around Earth is elliptical with the center of the Earth lying at one of the foci of the ellipse. The elliptical orbit is characterized by its semi-major axis a and eccentricity e. Eccentricity is the ratio of the distance between the center of the ellipse and either of its foci (= ae) to the semi-major axis of the ellipse a.

A circular orbit is a special case of an elliptical orbit where the foci merge together to give a single central point and the eccentricity becomes zero. Other important parameters of an elliptical satellite orbit include its apogee (farthest point of the orbit from the Earth’s centre) and perigee (nearest point of the orbit from the Earth’s centre) distances. These are described in subsequent paragraphs.

For any elliptical motion, the law of conservation of energy is valid at all points on the orbit.

The law of conservation of energy states that energy can neither be created nor destroyed; it can only be transformed from one form to another. In the context of satellites, it means that the sum of the kinetic and the potential energy of a satellite always remain constant. The value of this constant is equal to −Gm1m2/(2a), where

m1 = mass of Earth

m2 = mass of the satellite

a= semi-major axis of the orbit

The kinetic and potential energies of a satellite at any point at a distance r from the centre of the Earth are given by

Kinetic energy $= \frac{1}{2}(m_{2}\nu ^{2})$

Potential energy $= -\frac{Gm_{1}m_{2}}{r}$

Therefore,

$\frac{1}{2}(m_{2}\nu ^{2})- \frac{Gm_{1}m_{2}}{r}=-\frac{Gm_{1}m_{2}}{2a}$ $\nu ^{2}=Gm_{1}\left ( \frac{2}{r}-\frac{1}{a} \right )$ $\nu = \sqrt{\left [\mu \left (\frac{2}{r}-\frac{1}{a} \right ) \right ]}$

### Kepler’s Second Law

The line joining the satellite and the centre of the Earth sweeps out equal areas in the plane of the orbit in equal time intervals; i.e. the rate (dA/dt) at which it sweeps area A is constant. The rate of change of the swept-out area is given by

$\frac{dA}{dt}=\frac{angular momentum of the satellite}{2m}$

where m is the mass of the satellite. Hence, Kepler’s second law is also equivalent to the law of conservation of momentum, which implies that the angular momentum of the orbiting satellite given by the product of the radius vector and the component of linear momentum perpendicular to the radius vector is constant at all points on the orbit.

The angular momentum of the satellite of mass m is given by mr2ω, where ω is the angular velocity of the satellite. This further implies that the product mr2ω = (mωr) (r) = mvr remains constant. Here v’ is the component of the satellite’s velocity v in the direction perpendicular to the radius vector and is expressed as v cos γ, where γ is the angle between the direction of motion of the satellite and the local horizontal, which is in the plane perpendicular to the radius vector r. This leads to the conclusion that the product rv cos γ is constant.

The product reduces to rv in the case of circular orbits and also at apogee and perigee points in the case of elliptical orbits due to angle γ becoming zero. It is interesting to note here that the velocity component v’ is inversely proportional to the distance r. Qualitatively, this implies that the satellite is at its lowest speed at the apogee point and the highest speed at the perigee point. In other words, for any satellite in an elliptical orbit, the dot product of its velocity vector and the radius vector at all points is constant. Hence,

$\nu _{p}r_{p}=\nu _{a}r_{a}=\nu r \cos \gamma$

where,

• vp =velocity at the perigee point
• rp =perigee distance
• va =velocity at the apogee point
• ra =apogee distance
• v=satellite velocity at any point in the orbit
• r =distance of the point
• γ =angle between the direction of motion of the satellite and the local horizontal

### Kepler’s Third Law

According to the Kepler’s third law, also known as the law of periods, the square of the time period of any satellite is proportional to the cube of the semi-major axis of its elliptical orbit.

The expression for the time period can be derived as follows. A circular orbit with radius r is assumed. Remember that a circular orbit is only a special case of an elliptical orbit with both the semi-major axis and semi-minor axis equal to the radius. Equating the gravitational force with the centrifugal force gives

$\frac{Gm_{1}m_{2}}{r^{2}}=\frac{m_{2}\nu ^{2}}{r}$

Replacing v by ωr in the above equation gives

$\frac{Gm_{1}m_{2}}{r^{2}}=\frac{m_{2}\omega^{2}r^{2}}{r}=m_{2}\omega^{2}r$

which gives $\omega^{2}=Gm_{1}/r^{3}$. Substituting ω = 2π/T gives

$T^{2}=\left ( \frac{4\pi ^{2}}{Gm_{1}} \right )r^{3}$

This can also be written as

$T=\left ( \frac{2\pi}{\sqrt{\mu }} \right )r^{3/2}$

The above equation holds good for elliptical orbits provided r is replaced by the semi-major axis a. This gives the expression for the time period of an elliptical orbit as

$T=\left ( \frac{2\pi}{\sqrt{\mu }} \right )a^{3/2}$

Telemetry Tracking and Command Subsystem