1
Newton’s Law of Universal Gravitation
Newton was the first to realize that gravity was a universal phenomena that not
only explained how apples fall, but how the moon and the planets move under
the gravitational influence of the Earth and the Sun.
1.1
Universal Gravitation
Newton reasoned (correctly) that the gravitational attraction between two bodies was proportional to the product of their masses and inversely proportional
to the distance between their centres of mass.
Mathematically this is written as
Fg ∝
Mm
r2
(1)
or as
GM m
r2
G is the proportionality constant known as the “gravitational constant”.
Fg =
1.2
(2)
Working Against Gravity
To move “upwards” in a gravitational field, from r1 to r2 the energy required,
W, is given by
Z r=r2
W =
Fg dr
(3)
r=r1
or substituting equation 2
W =
Z
r=r2
r=r1
GM m
dr
r2
(4)
which gives
1
1
W = GM m
−
r2
r1
1.3
(5)
Interesting Cases of Equation 5
There are two cases which are both interesting and useful in orbital mechanics.
The first case deals with launching small satellites (with mass m) into space
with enough energy to escape the gravitational effects of a large body, such as
the Earth (mass M ).
The second case is related to the energy of a satellite which is orbiting the Earth.
1
1.3.1
Case 1
If we wish to separate the two bodies so that their mutual gravitational attraction is zero, we must set r2 = ∞. Equation 5 then becomes
W∞ =
−GM m
r1
(6)
W∞ is known as the gravitational potential energy (or binding enery) and its
value is negative.
If we take M to be the mass of the Earth and r1 its radius, then W∞ is the
energy required to send a spacecraft of mass m from the Earth’s surface to
infinity.
To launch this spacecraft so that it will totally escape the Earth’s gravity we
must give it enough kinetic energy ( 12 mv 2 ) to equal W∞ , which means
GM m
1
mv 2 =
2
r1
(7)
2GM
r1
r
2GM
v=
r1
v2 =
(8)
(9)
The quantity v from equation 9 is called the escape velocity.
Note that the escape velocity is independent of the mass of the satellite.
1.3.2
Case 2
The question is “What is the velocity of a satellite in a perfectly circular orbit
around a large massive object such as the Earth?”
Recall that for a body to move in a circular orbit of radius r one must provide
a centripetal force equal to mv2 /r. In the case of a satellite’s orbit, this force is
provided by the force of gravity.
mv 2
GM m
=
r
r2
GM
r
r
GM
v=
r
v2 =
2
(10)
(11)
(12)
2
Comparing Case 1 and Case 2
The kinetic energy of a mass m is defined as 12 mv 2 therefore we can calculate
the kinetic energy we must impart to an object to:
1. give it orbital velocity vo . From equation 11 we can write
KEorbit =
1
1 GM m
mvo 2 =
2
2 r
(13)
2. give it escape velocity ve . From equation 7 we can write
KEescape =
1
GM m
mve 2 =
2
r
(14)
The interesting discovery is that
KEescape = 2KEorbit
(15)
Equation 15 tells us that the kinetic energy required to launch a spacecraft, so
that it will totally escape from the Earth’s gravitational field, must be exactly
twice the kinetic energy required to maintain a low circular orbit around the
earth.
3
Interpreting the negative sign in
the Gravitational Potential Energy equation
Infinitely far from a large gravitationally attractive mass M, the total energy of
an object mass m, with respect to the large mass M, is zero. That is to say, the
sum of the kinetic energy KE and potential energy PE is zero.
KE + P E = 0
(16)
−GM m
1
mvo 2 +
=0
(17)
2
r
As mass m falls toward the large mass M, the kinetic energy of mass m will
increase. Since energy must be conserved, the gravitational potential energy
must be negative so that the total energy remains zero.
3