10.2.6 Orbital Motion, Speed & Energy

• Since most planets and satellites have a near circular orbit, the gravitational force F_G between the sun or another planet provides the centripetal force needed to stay in an orbit
  • Both the gravitational force and centripetal force are perpendicular to the direction of travel of the planet
• Consider a satellite with mass m orbiting Earth with mass M at a distance r from the centre travelling with linear speed v

F_G = F_circ

• Equating the gravitational force to the centripetal force for a planet or satellite in orbit gives:

• The mass of the satellite m will cancel out on both sides to give:

• Where:
  • v = linear speed of the mass in orbit (m s^-1)
  • G = Newton's Gravitational Constant
  • M = mass of the object being orbited (kg)
  • r = orbital radius (m)

• This means that all satellites, whatever their mass, will travel at the same speed v in a particular orbit radius r
  • Since the direction of a planet orbiting in circular motion is constantly changing, the centripetal acceleration acts towards the planet

A satellite in orbit around the Earth travels in circular motion

Time Period & Orbital Radius Relation

• Since a planet or a satellite is travelling in circular motion when in order, its orbital time period T to travel the circumference of the orbit 2πr, the linear speed v is:

• This is a result of the well-known equation, speed = distance / time and first introduced in the circular motion topic
• Substituting the value of the linear speed v from equating the gravitational and centripetal force into the above equation gives:

• Squaring out the brackets and rearranging for T² gives the equation relating the time period T and orbital radius r:

• Where:
  • T = time period of the orbit (s)
  • r = orbital radius (m)
  • G = Newton's Gravitational Constant
  • M = mass of the object being orbited (kg)

• The equation shows that the orbital period T is related to the radius r of the orbit. This is also known as Kepler’s third law:

For planets or satellites in a circular orbit about the same central body, the square of the time period is proportional to the cube of the radius of the orbit

• Kepler’s third law can be summarised as:

Graphical Representation of T² ∝ r³

• The relationship between T and r can be shown using a logarithmic plot
  • T² ∝ r³
  • 2 × log(T) ∝ 3 × log(r)
• The graph of log(T) in years against log(r) in AU (astronomical units) for the planets in our solar system is a straight line graph:

• The graph does not go through the origin since it has a negative y-intercept
• Only the log of both T and r will produce a straight line graph

Energy of an Orbiting Satellite

• An orbiting satellite follows a circular path around a planet
• Just like an object moving in circular motion, it has both kinetic energy (KE) and gravitational potential energy (GPE) and its total energy is always constant
• An orbiting satellite's total energy is calculated by:

A graph showing the kinetic, potential and total energy for a mass at varying orbital distances from a massive body

Total energy = Kinetic energy + Gravitational potential energy 

• This means that the satellite's KE and GPE are also both constant in a particular orbit
  • If the orbital radius of a satellite decreases its KE increases and its GPE decreases
  • If the orbital radius of a satellite increases its KE decreases and its GPE increases

At orbit Y, the satellite has greater GPE and less KE than at at orbit X

• A satellite is placed in two orbits, X and Y, around Earth
• At orbit X, where the radius of orbit r is smaller, the satellite has a:
  • Larger gravitational force on it
  • Higher speed
  • Higher KE
  • Lower GPE
  • Shorter orbital time period, T

• At orbit Y, where the radius of orbit r is larger, the satellite has a:
  • Smaller gravitational force on it
  • Smaller speed
  • Lower KE
  • Higher GPE
  • Longer orbital time period, T

ANSWER:    D

• • Since B is at a larger orbital radius (3R instead of R) it has a longer time period since T² ∝ R³ for an orbiting satellite
  • However, satellite B will travel much slower than A
  • Its larger orbital radius means the force of gravity will be much lower for B than for A

• A synchronous orbit is:

When an orbiting body has a time period equal to that of the body being orbited and in the same direction of rotation as that body

• These usually refer to satellites (the orbiting body) around planets (the body being orbited)
• The orbit of a synchronous satellite can be above any point on the planet's surface and in any plane
  • When the plane of the orbit is directly above the equator, it is known as a geosynchronous orbit

Geostationary Orbit

• Many communication satellites around Earth follow a geostationary orbit
  • This is sometimes referred to as a geosynchronous orbit

• This is a specific type of orbit in which the satellite:
  • Remains directly above the equator
  • Is in the plane of the equator
  • Always orbits at the same point above the Earth’s surface
  • Moves from west to east (same direction as the Earth spins)
  • Has an orbital time period equal to Earth’s rotational period of 24 hours

• Geostationary satellites are used for telecommunication transmissions (e.g. radio) and television broadcast
• A base station on Earth sends the TV signal up to the satellite where it is amplified and broadcast back to the ground to the desired locations
• The satellite receiver dishes on the surface must point towards the same point in the sky
  • Since the geostationary orbits of the satellites are fixed, the receiver dishes can be fixed too

Low Orbits

• Some satellites are in low orbits, which means their altitude is closer to the Earth's surface
• One example of this is a polar orbit, where the satellite orbits around the north and south pole of the Earth
• Low orbits are useful for taking high-quality photographs of the Earth's surface. This could be used for:
  • Weather
  • Military applications

Geostationary satellite in orbit

Orbital Motion

• When a satellite is moving in orbital motion, the velocity of the satellite will be given by:

• Where
  • v_satellite = velocity of the satellite (m s^-1)
  • M = mass of the Earth (or large body being orbited) (kg)
  • G =  The gravitational constant = 6.67 x 10^-11 Nm²kg^-2
  • r = the orbital radius of the satellite (m)
• Therefore the kinetic energy of the satellite will be:

• Where
  • E_K = kinetic energy (J)
  • v = velocity (m s^-1)
  • M = mass of the Earth (or large body being orbited) (kg)
  • m = mass of satellite or orbiting object (kg)
  • G =  The gravitational constant = 6.67 x 10^-11 Nm²kg^-2
  • r = the orbital radius of the satellite (m)
• The gravitational potential energy of an orbiting object is:

• Where
  • E_P = potential energy (J)
  • M = mass of the Earth (or large body being orbited) (kg)
  • m = mass of satellite or orbiting object (kg)
  • G =  The gravitational constant = 6.67 x 10^-11 Nm²kg^-2
  • r = the orbital radius of the satellite (m)

• The total energy for any satelite orbiting a massive body is found from the combination of potential energy and kinetic energy
  • The result is show in the equation below: 

• Where
  • E_T = The total energy of the satellite (J)
  • E_P = The potential energy of the satellite (J)
  • E_K = The kinetic energy of the satellite (J)
  • v = the velocity of the satellite (m s^-1)
  • M = mass of the Earth (or large body being orbited) (kg)
  • m = mass of satellite or orbiting object (kg)
  • G =  The gravitational constant = 6.67 x 10^-11 Nm²kg^-2
  • r = the orbital radius of the satellite (m)