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Chapter 13 Outline
Gravitation
• Newton’s law of gravitation
• Weight
• Gravitational potential energy
• Circular orbits
• Kepler’s laws
• Black holes
Gravitation
• Gravity is the main (somewhat understood) driving force
in the universe.
• All masses attract each other.
• Newton’s law of gravitation (1687)
𝐺𝑚1 𝑚2
𝐹g =
𝑟2
Gravitational Force
• Acts on both bodies with equal magnitude.
• When you jump off a diving board, Earth also accelerates towards
you, but your mass is 23 orders of magnitude smaller, so this is
negligible.
• Spherically symmetric masses look like points from the
outside.
• Replace sphere with a single point mass at the center.
Gravitational Constant, 𝐺
• Sir Henry Cavendish used a torsion balance to measure 𝐺
in 1798.
• Very difficult to precisely measure
𝐺 = 6.67 × 10−11 N ∙ m2 /kg 2
Weight
• So far, we have used 𝑤 = 𝑚𝑔.
• While the definition of weight is the total gravitational
force exerted on a body, on the surface of a planet (or
moon, comet…) we can neglect the forces from everything
but that planet.
• At the surface of Earth, the weight of a mass, 𝑚, is:
𝐺𝑚E 𝑚
𝑤 = 𝐹g =
𝑅E2
• So,
𝐺𝑚E
𝑔= 2
𝑅E
Mass of the Earth
• If we measure the local acceleration due to gravity, we
should be able to calculate the mass of the earth:
𝑔𝑅E2
9.81 m/s 2 6.37 × 106 kg 2
𝑚E =
≈
𝐺
6.67 × 10−11 N ∙ m2 /kg 2
𝑚E ≈ 5.97 × 1024 kg
• This matches very well with the accepted value of
𝑚E = 5.972 × 1024 kg
Density of the Earth
• Knowing the mass and radius of
the Earth, we can easily
calculate an average density.
𝑀E
𝜌E−ave = 4 3 = 5.51 g/cm3
3𝜋𝑅E
• But, the rocks we find on the
surface generally have densities
of about 2 to 3.3 g/cm3 .
• The Earth must not have uniform
density.
• The density near the core is larger
than near the surface.
Exam #3
Results
• Average 10.4
• Standard deviation 3.1
Exam #3 Grades
After T-score Adjustment
45
40
35
Frequency
30
25
20
15
10
5
0
50
62
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Grades
85
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More
Gravitational Potential Energy
• Previously, we found that the
gravitational potential energy
was given by:
𝑈g = 𝑚𝑔ℎ
• But, this assumed that 𝑔 was
constant.
• On the surface of a planet, this is a
good approximation.
• In space, it is not!
• Recall our definition of potential
energy:
∆𝑈 = −𝑊 = −
𝑭 ∙ 𝑑𝒍
Gravitational Potential Energy
• As before, we need to set a zero
for the potential energy.
• Now, the convenient zero point is
infinity.
• The potential decreases as the
masses get closer together.
• Keep in mind that this is the
potential energy of the system.
𝐺𝑀𝑚
𝑈g = −
𝑟
Gravitational Potential Energy on Earth
• Near the surface of Earth, this expression should reduce to
the familiar ∆𝑈 = 𝑚𝑔∆ℎ.
• Moving from position 𝑟1 to 𝑟2 , ∆ℎ = 𝑟2 − 𝑟1
1 1
𝐺𝑀𝑚 𝑟2 − 𝑟1
∆𝑈 = −𝐺𝑀𝑚
−
=
𝑟2 𝑟1
𝑟1 𝑟2
• For Earth, 𝑀 = 𝑚E , 𝑟1 𝑟2 ≈ 𝑅E2 , and 𝑔 =
𝐺𝑚E
2 ,
𝑅𝐸
so
𝐺𝑚𝐸
∆𝑈 = 𝑚 2 𝑟2 − 𝑟1 = 𝑚𝑔∆ℎ
𝑅𝐸
Escape Velocity
• If you throw a ball up into the air, what happens?
• It reaches some maximum height, and then comes back down.
• We find this height using conservation of energy.
• What if you threw it with an extremely large initial velocity?
• If the initial kinetic energy is greater than or equal to the magnitude of
the potential energy (with respect to infinity), the ball will not fall
down.
• When equal,
𝐺𝑀𝑚 1
𝑈1 + 𝐾1 = −
+ 𝑚𝑣 2 = 0
𝑟
2
𝑣escape =
2𝐺𝑀
𝑟
Motion of Satellites
• If a projectile is launch horizontally from a great height,
where will it land?
• Depends on the initial velocity.
• For large enough 𝑣0 , the curvature of the Earth is important
• Ignoring air resistance, it may never land.
• In an orbit the object keeps falling towards the Earth, but
never reaches it.
• Closed orbit – ellipse
• Open orbit – doesn’t return
Circular Orbits
• First, we will consider
the simplest orbit, a
circular orbit.
• Satellite mass 𝑚; planet
mass 𝑀
• Uniform circular motion:
𝑣2
𝑎rad =
𝑟
• Force provided by gravity:
𝐺𝑀𝑚
𝐹g = 2
𝑟
• Newton’s second law
𝐺𝑀𝑚
𝑣2
=𝑚
2
𝑟
𝑟
Circular Orbital Speed
• Solving for the speed,
𝑣circ =
𝐺𝑀
𝑟
• Two things to note:
• The mass of the satellite
does not matter.
• The speed and the radius
are linked. (Larger radius,
smaller speed)
Circular Orbital Period, 𝑇
• The time it takes for a
full orbit, or the period,
𝑇, can be found from the
speed.
2𝜋𝑟
𝑣circ =
𝑇circ
𝑟
𝑇circ = 2𝜋𝑟
𝐺𝑀
𝑇circ =
2𝜋𝑟 3/2
𝐺𝑀
Total Energy in Circular Orbit
• The total energy (potential and kinetic) of the satellite is:
𝐺𝑀𝑚 1
𝐸 =𝑈+𝐾 =−
+ 𝑚𝑣 2
𝑟
2
• Using 𝑣circ =
𝐺𝑀
,
𝑟
𝐺𝑀𝑚 1
𝐺𝑀
𝐸=−
+ 𝑚
𝑟
2
𝑟
𝐺𝑀𝑚
𝐸=−
2𝑟
• The total energy is negative (bound system) and equal to
one half the potential energy.
• This is an example of the virial theorem.
Kepler’s Laws
• In the 16th century, Nicolaus Copernicus deduced that the
Earth is a planet, and, like all planets, it orbits the Sun.
• In the early 17th century, Johannes Kepler used Tycho
Brahe’s observations of planetary motions to determine
three empirical laws of planetary orbits.
1. Each planet moves in an elliptical orbit, with the sun at
one focus of the ellipse.
2. A line from the sun to a given planet sweeps out equal
areas in equal times.
3. The periods of the planets are proportional to the 32
powers of the major axis lengths of their orbits.
• Newton later discovered the reasons for these laws.
Kepler’s First Law
• Elliptical orbits
• Focal points, 𝑆 and 𝑆′
• Semi-major axis, 𝑎
• Eccentricity, 𝑒, between
zero (circle) and one.
• The sum of the distances
𝑆𝑃 and 𝑆 ′ 𝑃 is constant.
• In our solar system, most
orbits are fairly circular
• Earth: 𝑒 = 0.017
• Most eccentric is
Mercury: 𝑒 = 0.206
Kepler’s Second Law
• The rate at which area is swept out is
called the sector velocity.
𝑑𝐴 1 2 𝑑𝜃
= 𝑟
𝑑𝑡 2 𝑑𝑡
• Triangle area: 𝐴 = 12𝑏ℎ
• The perpendicular component of 𝒗 is
𝑣 sin 𝜙.
• The rate at which 𝜃 is changing is
=𝜔=
𝑣 sin 𝜙
r
𝑑𝐴 1
= 𝑟𝑣 sin 𝜙
𝑑𝑡 2
𝑑𝜃
𝑑𝑡
Kepler’s Second Law
• Recall that angular momentum is 𝑳
= 𝒓 × 𝑚𝒗.
• 𝐿 = 𝑚𝑟𝑣 sin 𝜙
• So, the sector velocity is simply
𝑑𝐴
𝐿
=
𝑑𝑡 2𝑚
• Kepler’s second law is a natural
consequence of the conservation of
angular momentum.
• There can be no torque because the
force is always central (on a line
between the masses.)
Kepler’s Third Law
• For a circular orbit, we already showed that the period was
proportional to the 32 power of the radius.
• This is also true of an elliptical orbit, but the radius is
replaced by the semi-major axis, 𝑎.
𝑇=
2𝜋𝑎3/2
𝐺𝑀
Escape Velocity Revisited
• We already calculated the escape velocity at the surface of
a planet (or any spherical object).
𝑣escape =
2𝐺𝑀
𝑟
• If the sphere becomes more and more dense, the mass will
remain the same, but the radius decreases, and the escape
velocity increases.
• At what point does this become an issue?
Black Holes
• The maximum speed for anything is the speed of light, 𝑐
= 3.00 × 108 m/s.
• If the escape velocity exceeds this, even light cannot
escape.
• This is a black hole.
Schwartzschild Radius
• For a given mass, what will the radius have to be to form a
black hole?
𝑐=
2𝐺𝑀
𝑅S
2𝐺𝑀
𝑅S = 2
𝑐
• This is the Schwartzschild radius, and it defines the event
horizon.
• Note: To properly derive this, we need to use general
relativity, but the result is the same.
What if the Sun collapsed into a black hole?
• What if the Sun collapsed into a black hole?
• Would the Earth’s orbit be affected?
• What would the radius of the event horizon be?
2𝐺𝑚Sun
𝑅S =
≈ 3 km
2
𝑐
• The Sun’s radius now is 6.96 × 108 m.