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Universal Law of
Gravitation and Orbits
AP Physics C
Newton’s Theory of Gravity
 Legend has it that Newton saw an
apple fall from a tree, and it
occurred to him that the apple
was attracted to the center of the
earth.
 If the apple was so attracted, why
not the moon?
 Newton posited that gravity is a
universal attractive force between
all objects in the universe.
 Unfortunately he had no clue as
to why. That’s what happens
when you stare at the sun, poke
your eyes with needles, and huff
lead.
Universal Law of Gravitation
 Newton’s universal law of gravitation is as follows:
𝑚1 𝑚2
𝐹𝑔 = −𝐺
𝑟2
 G is the gravitational constant. This is a universal constant with the
value G = 6.67 x 10-11 N m2/kg2. This is not the same thing as g!
 The negative sign implies that the force is attractive. This means that
all objects with mass attract one another with a gravitational force.
 The distance that we use is the distance from center of mass to center
of mass.
Inverse-Square
 The force of gravity is an inverse
square relation, meaning that as the
separation distance increases, the
force drops substantially.
 We often are simply concerned with
the magnitude of the force of gravity,
so often the formula is written this
way:
𝑚1 𝑚2
𝐹 =𝐺
𝑟2
 When bringing up gravitational
potential energy we will consider
the force to be negative for
constancy.
But that marker isn’t attracted to
me
 Yes it is.
 Since G is so small, it means that the
attractive force between two 1.0 kg
masses, whose centers are 1.0 m apart,
is 6.7 x 10-11 N.
 This is 100 billion times weaker than the
force of gravity from the earth on either
of the masses!
 Although weak, gravity is a long-range
force.
 Gravity keeps the earth orbiting the sun
and the solar system orbiting the center
of the Milky Way galaxy.
Gravitation

Consider a hammer just above the surface of the earth. The force of attraction
is given by the law of gravitation.
𝐹 =𝐺
𝑚1 𝑚2
𝑟2
 Consider just this cluster of variables:
𝑚𝑒𝑎𝑟𝑡ℎ
𝐺
𝑟2
 Calculate this using the following measurements:
𝑚𝑒𝑎𝑟𝑡ℎ = 5.972 × 1024 𝑘𝑔
𝑟𝑒𝑎𝑟𝑡ℎ = 6.371 × 106 𝑚
2
−11
𝑁𝑚
𝐺 = 6.7 × 10
𝑘𝑔2
Acceleration due to Gravity
 We get 9.8 m/s2!
 Taking the product of those variables on any planet (and
using the mass and radius of that planet) will get you the
acceleration due to gravity on that planet.
 Notice that this means the acceleration is technically not
constant like we have said it is, rather is decreases as we
get farther from the planet.
 Our value of 9.8 works as a great approximation though; it
fluctuates very little until we move very far from the
surface of the earth.
𝑔𝑠𝑢𝑟𝑓𝑎𝑐𝑒
𝐺𝑀
= 2
𝑟
Falling through the Earth
 Suppose we dig a hole through the
center of the Earth.
 If we jump into the hole, we experience
an acceleration of 9.8 m/s2 at the
surface.
 As we fall, the acceleration (and force)
decreases.
 Once we reach the center of the
planet, the net force is zero, but we
continue moving due to our kinetic
energy and momentum.
 After we pass the center, we
experience a growing force towards the
center, slowing us down.
 This means that we will oscillate!
Falling through the Earth
 The period of oscillation would be
the exact same as a satellite orbiting
at the center of the Earth.
 If we approximate that the density
of the Earth is constant (which it
isn’t) and that you would not
incinerate as you pass through the
middle (which you would), we find
that it would take approximately 84
minutes for one full oscillation.
 This also requires the Earth to be
stationary (which it isn’t).
 Regardless, it would work for a
stationary spherically symmetric
rock the size of the Earth.
Gravitational Potential Energy
 For objects near the surface of the Earth, we have used an
approximation to describe the gravitational potential energy.
𝑈 = 𝑚𝑔ℎ
 This is however, just an approximation. To really understand how
gravitational potential energy works, we need to examine the universal
law of gravity. Recall that a conservative force follows the following rule:
𝐹𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒
−𝑑𝑈
=
𝑑𝑥
Gravitational Potential Energy
 We now combine this rule with our formula for the force of gravity.
𝑑𝑈 = −𝐹𝑑𝑟
𝑟
𝐺𝑚𝑀
𝑑𝑟
2
𝑟
∞
𝑟
1
𝑈 𝑟 = 𝐺𝑚𝑀
2 𝑑𝑟
𝑟
∞
−1 1
−𝐺𝑚𝑀
𝑈 𝑟 = 𝐺𝑚𝑀 ∗
+
→𝑈 𝑟 =
𝑟
∞
𝑟
𝑈 𝑟 =−
−
 This result may seem a little bit confusing at
first glance. We choose infinity as our initial
point because it is an easy reference point (a
point from which to measure something).
 At infinity, we can say that the potential
energy is equal to zero, because at that
distance there would be no tendency for the
objects to attract.
Gravitational Potential Energy
 Suppose two masses a distance r1
apart are released from rest.
 How will the small mass move as r
decreases from r1 to r2?
 At r1 U is negative.
 At r2 |U| is larger and U is still
negative, meaning that U has
decreased.
 As the system loses potential energy,
it gains kinetic energy while
conserving mechanical energy.
 The smaller mass speeds up as it falls.
 Remember, changes in potential
energy have physical meaning.
Escape Velocity
 Consider a rocket leaving the earth. It
usually goes up, slows down, and then
returns to earth.
 There exists an initial minimum speed that
when reached the rockets will continue on
forever. Let's use conservation of energy to
analyze this situation!
𝐾𝐸𝑜 + 𝑈𝑜 = 𝐾𝐸 + 𝑈
𝐾𝐸𝑜 = 1 2 𝑚𝑣𝑜2
−𝐺𝑚𝑀
𝑈𝑜 =
𝑅
 In this situation, we will say that the rocket can
only truly “escape” the Earth if it gets to a
distance of infinity. In this case, it would have a
final velocity of zero.
Escape Velocity

This means that the total mechanical energy of the system is
equal to zero. Remember that defining the potential as negative
does not carry any significant meaning.
𝐾𝐸𝑜 + 𝑈𝑜 = 𝐾𝐸 + 𝑈
𝐾𝐸𝑜 = 1 2 𝑚𝑣𝑜2
−𝐺𝑚𝑀
𝑈𝑜 =
𝑅
1
2
2 𝑚𝑣𝑜 −
𝐺𝑚𝑀
= 0 + 0 → 𝑣𝑜 =
𝑅
2𝐺𝑀
𝑅
 This is the minimum velocity to escape the gravitational pull of a planet! The
M and R in the equation represent the mass and radius of the planet in
question.
 Keep in mind that this only accounts for the object being launched and the
planet it is launched from. When considering other planets and stars it
becomes more complicated!
 This will also only get an object into orbit and no further.
Kepler’s Laws
 There are three laws that Johannes Kepler
formulated when he was studying the
heavens.
 1st Law – The Law of Orbits: Planets move in
elliptical orbits, with the sun at one focus of
the ellipse.
 2nd Law – The Law of Areas: A line drawn
between the sun and a planet sweeps out
equal areas during equal intervals of time.
 How is it that we know this? Angular
momentum is conserved and thus
constant!
 While further away from the sun, the
planet can move slower, but when
closer to the sun it has to move faster.
 3rd Law – The Law of Periods: The square of
a planet’s orbital period is proportional to
the cube of the semi major-axis length.
Orbital Kinetic Energy
 The velocity of an object in circular orbit is given by the following
equation:
𝑣=
𝐺𝑀
𝑟
 We can substitute this for velocity in our kinetic energy formula to express
the kinetic energy of an orbiting body.
𝑀
= 𝐺 , 𝐾𝐸 = 1 2 𝑚𝑣 2
𝑟
𝐺𝑚𝑀
𝐾𝐸 =
2𝑟
𝑣2
 The question is WHY? Why do we need a new equation for kinetic energy? Well,
the answer is that greatly simplifies the math.
 If we use regular kinetic energy along with potential, we will need both the orbital
velocity AND the orbital radius. In this case, we need only the orbital radius.
Orbital Energy
 We can now create an expression for total mechanical energy of an orbiting
object in terms of only the masses of the objects and the distance between
them (center of mass to center of mass).




𝐸𝑡𝑜𝑡𝑎𝑙 = 𝐾𝐸 + 𝑈
𝐺𝑚𝑀
−𝐺𝑚𝑀
𝐾𝐸 =
,𝑈 =
2𝑟
𝑟
𝐺𝑚𝑀 −𝐺𝑚𝑀 −𝐺𝑚𝑀
𝐸𝑡𝑜𝑡𝑎𝑙 =
+
=
2𝑟
𝑟
2𝑟
So by inspection we see that the kinetic energy function is always positive,
the potential is negative and the total energy function is negative.
In fact the total energy equation is the negative inverse of the kinetic.
The negative is symbolic because it means that the mass “m” is BOUND to
the mass of “M” and can never escape from it. It is called a BINDING
ENERGY.
Notice that we also have a relationship between orbital kinetic and
potential energy:
1
𝐾𝐸 = − 𝑈
2
Orbital Energetics
 The figure shows the
kinetic, potential, and
total energy of a satellite
in a circular orbit.
 Notice how, for a circular
orbit, total energy = U/2.
 It requires positive energy
in order to lift a satellite
into a higher orbit.
Orbital Energetics
 The figure shows the steps
involved to lift a satellite to a
higher circular orbit.
 The first kick increases KE
without increasing U, so KE is
not -U/2, and the orbit is
elliptical.
 The satellite then slows down
as r increases.
 The second kick increases KE
again so that KE = -U/2, and the
orbit is circular.