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ACCELERATED MOTION AND NEWTONS LAWS

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					THE “SATELLITE EQUATION” AND KEPLER’S
THIRD LAW REVISITED

The “Satellite Equation”
A satellite may be considered any object which orbits a
much larger object by virtue of the gravitational
interaction between the two. The Earth has both a very
large “natural satellite” (the Moon) and a multitude of
much smaller manufactured satellites (communication
satellites, the Hubble telescope and the international
space station, for example). At the same time, the Earth
is itself a satellite of the Sun, as are the other planets in
the solar system, along with numerous other objects such
as asteroids and comets.

Satellite orbits may be either perfect (or nearly perfect)
circles or ellipses (highly or only slightly elongated). For
simplicity here, however, we will only consider satellites
whose orbits are circular.

Consider a satellite of mass m in a circular orbit around
a planet of mass M (such as the Moon orbiting the
Earth). Since the satellite is traveling around a circle at
constant speed, it’s centripetal acceleration is
                              v2
                         a
                               r ,
                            2




and, by Newton’s Second Law, the force exerted on the
satellite by the planet can therefore be expressed as
                                 mv 2
                      F  ma 
                                  r .
However, because this force is due to the gravitational
pull of the planet, it can also be written in another way,
according to Newton’s Law of Universal Gravitation, i.e.
                             GMm
                        Fg  2 .
                               r
Equating these two expressions for the same force, we
can obtain
                      mv 2 GMm
                            2 ,
                        r      r
and cancelling m and one r from both sides leaves
                             GM
                        v2 
                              r .
Finally, taking the square root of both sides, we obtain
the “Satellite Equation”, namely:
                             GM
                        v
                               r .
Here, it must be remembered that M is the mass of the
planet, r is the radius of the orbit and v is the orbital
speed of the satellite.
                                3




The Satellite Equation holds the key to the theoretical
behaviour of satellites, and it has a number of extremely
important applications.

Application 1
Problem:
Find the speed of the international space station as it
orbits at a distance of 340 km above the Earth’s surface.

Solution:
First, we find the radius of the orbit to be
        r  6.38  10 6  340,000  6.72  10 6 m.
Then, from the Satellite Equation,

 v
    GM
        
                            
             6.67 10 11 5.98 10 24
                                      
                                       
     r              6.72 10 6                       m/s.
     (Note that this is about 27,700 km/h!)

Essentially, the satellite equation tells us the exact speed
to which we must accelerate a satellite by the time it
reaches its orbital radius, and as long as we “shut off the
rockets” at that precise radius and speed, a perfect
circular orbit will automatically result! In other words,
this equation is of critical importance every time a
satellite is launched.
                           4



Application 2
On several occasions already in the course, we have used
the fact that the mass of planet Earth is about
5.98  1024 kg. Have you ever wondered how we know
the mass of our own planet? In fact, it is thanks to the
Satellite Equation and our knowledge of the Moon’s
orbit.

Problem:
Use the Satellite Equation and knowledge of the Moon’s
orbit to determine the Earth’s mass.

Solution:
We know that the accepted radius of the Moon’s orbit is
3.84  108 m. Moreover, we have already used this
radius and the time for one complete orbit to calculate
the orbital speed of the Moon, which is about
1.02  103 m/s.

Squaring and rearranging the Satellite Equation, we
have
                 GM              v2r
            v       , i.e. M  G .
             2

                  r
Hence,

  M
                     2
     v 2 r 1.02 103 3.84 108
                              
                         11                       kg !
      G         6.67 10
                             5



Note that the mass of any astronomical object, which has
a satellite whose orbital radius and orbital speed are
known, can be calculated in exactly the same way. For
example, knowing the Earth-Sun distance and the fact
that it takes 365 days for Earth to orbit the Sun is also
how we were first able to determine the Sun’s mass.
(Since the formula above involves the universal
gravitational constant G, however, none of these masses
could be determined until after the numerical value of G
was found around 1800!) Needless to say, these
calculations would be impossible without the Satellite
Equation.

Kepler’s Third Law Revisited
As a final application, recall that Kepler devised and
published his three laws of planetary motion well before
Newton had invented calculus, and developed Newton’s
Laws and the Universal Law of Gravitation.

In particular, Kepler’s Third Law states that the square
of the orbital period of a planet is directly proportional
to the cube of the orbital radius, a result which seemed
truly remarkable given that he was able to determine
this solely by analysing astronomical data on the
movement of the five inner planets. Now, with the help
of the Satellite Equation, we can actually prove that
Kepler’s Third is correct (at least for circular orbits).
                              6



The key is to realize that we now have two ways to
express the orbital speed of a planet. First, if we call the
orbital period T, the speed is just the circumference (i.e.
the distance traveled during one orbit) divided by the
time, or
                          2r
                       v
                           T .
Second, from the Satellite Equation,
                         GM
                   v
                           r .
Equating and inverting these two expressions,
                        T     r
                           
                       2r   GM .
Squaring both sides,
                      T2         r
                             
                    4 r
                       2 2
                               GM , i.e.
                     4 2 r 3  4 2  3
                T 
                 2
                              GM r .
                                     
                      GM            
      4 2
Since GM is just a constant for any planet, this proves
that, exactly as Kepler proposed, T2 is indeed
proportional to r3!

				
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posted:3/21/2010
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