PHY C8: Circular Orbits

We’ll be covering:

  • Circular Orbits
    • Components
    • Kepler’s 3rd Law of Planetary Motion
    • Types of Orbit

Let’s go!

Circular Orbits
We’ve already seen how circular motion works.
Let’s talk about circular orbits caused by gravitational force.

What are the components of circular orbits? | Ask Us - Space Shuttle Speed in Orbit
From Aerospaceweb
Centre of Circular Motion Centre of mass of larger mass (e.g. planet)
Radius, r Radius of planet + altitude of satellite
r = R + h
Period, T Time taken to complete one revolution.

For example:

  • the period of the Earth around the Sun = 365 days
  • the period of the Moon around the Earth = 28 days

You can use these values to calculate:

  • ω = angular speed
  • v = linear speed
Centripetal Force, Fc Gravitational Force
F = GMm/r2
Centripetal Acceleration, ac Gravitational Field Strength
a = g = GM/r2

How are these all related?

Fc = Fg

mv2/r = GMm/r2

Linear speed = circumference of orbit/period

v = 2πr/T

mv2/r = GMm/r2
v2 = GM/rSince g = GM/r2,

v2 = rg

From relation 1:
m(2πr/T)2/r = GMm/r2Rearranging:

T2/r3 = 4π2/GM

From Relation 3:
For different objects orbiting the same body (e.g. planets in a solar system), M is the same (mass of the Sun).Thus, it can be concluded that:
T2 ∝ r3

“The square of the period is proportional to the cube of the orbital radius”

This is Kepler’s 3rd Law of Planetary Motion.


Let’s talk about a few types of Earth orbit:

Zuck Nerds Out On Drones Vs. Satellites For Delivering ...

Low Earth Orbit Satellites orbiting close to the vicinity of Earth.
Their altitude is relatively small enough, such that r ≈ R.Thus, by calculation:

  • v ≈ √(6.4 x 106 x 9.8) ≈ 8000 m/s
  • T = 2πR/v ≈  5000 s
Geostationary Orbit Satellites in equatorial orbits which the same period of rotation as the Earth.

Thus, they are stationary in the sky to an observer on the ground.

Credit: NASA

T = TEarth = 24 hours

This information can be used to calculate other aspects of motion:

  • T2/r3 = 4π2/GM
    Plugging in T = 24 hrs, M = mass of the Earth,
    r = 4.23 x 107
    h = r – R = 3.59 x 107 m
  • v = 2πr/T
    Plugging in T = 24 hrs,
    v = 3070 ms-1

⇐ Previous in Physics: Gravitational Field Strength
⇒ Next in Physics: Gravitational Potential

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