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Gravitation
Luis Anchordoqui
Kepler's law and Newton's Synthesis
The nighttime sky with its myriad stars and shinning planets has
always fascinated people on Earth.
Towards the end of the XVI century the astronomer Tycho Brahe
studied the motions of the planets and made accurate observations
Using Brahe's data Johannes Kepler worked out a detailed description
of the motion of the planets about the Sun: three empirical findings
that we now refer to as Kepler's laws of planetary motion
Kepler’s laws provided the basis for Newton's discovery of the law
of gravitation
Luis Anchordoqui
Kepler’s Laws and Newton's Synthesis
Kepler’s laws describe planetary motion.
1.The orbit of each planet is an ellipse with the Sun at one focus.
Luis Anchordoqui
Kepler’s Laws and Newton's Synthesis
(cont’d)
2. An imaginary line drawn from each planet to the Sun
sweeps out equal areas in equal times.
Luis Anchordoqui
Kepler’s Laws and Newton's Synthesis
(cont’d)
The ratio of the square of a planet’s orbital period is proportional
to the cube of its mean distance from the Sun.
Planetary data applied to Kepler’s Third Law
Luis Anchordoqui
Jupiter's orbit
Jupiter's mean orbital radius is 5.2 AU?
What is the period of Jupiter's orbit around the Sun?
τ
J
=11.9 yr
Luis Anchordoqui
Kepler's law and Newton's Synthesis
(cont’d)
The periods of the planets as a function of the mean orbital radii
The square of the periods of the planets versus the cubes
of the mean orbital radii
Luis Anchordoqui
Newton’s Law of Universal Gravitation
If the force of gravity is being exerted on objects on Earth,
what is the origin of that force?
Newton’s realization was that the force must come from the Earth.
He further realized that this force must be what keeps the Moon
in its orbit.
Luis Anchordoqui
Newton’s Law of Universal Gravitation (cont’d)
The gravitational force on you is one-half of a Third-Law pair:
the Earth exerts a downward force on you
and you exert an upward force on the Earth
When there is such disparity in m’s the reaction force undetectable
but for bodies more equal in mass it can be significant
Luis Anchordoqui
Newton’s Law of Universal Gravitation (cont’d)
The gravitational force must be proportional to both masses
By observing planetary orbits
Newton also concluded that
the gravitational force must decrease
as the inverse of the square of the distance between the masses
In its final form the Law of Universal Gravitation reads:
F12= G
m1 m 2
r²
12
^r
12
Luis Anchordoqui
Newton’s Law of Universal Gravitation (cont’d)
The magnitude of the gravitational constant G
can be measured in the laboratory
This is the Cavendish experiment
-11
G = 6.67 x 10
N · m²/kg²Luis Anchordoqui
Every few hundred years most of the planets line up on the same side of the Sun.
Calculate the total force on the Earth due to Venus, Jupiter, and Saturn,
assuming all four planets are in a line.
The masses are
MV = 0.815 M
M J= 318 M
M = 95.1 M
S
their mean d from Sun are 108, 150, 778, and 1430 million km, respectively.
What fraction of the Sun's force on the Earth is this.
17
F Earth/planets = 9.56 x 10 N
22
FEarth/Sun = 3.52 10 N
(Not to scale)
Luis Anchordoqui
Gravity Near the Earth’s Surface: Geophysical Applications
Now we can relate the gravitational constant to the local acceleration
of gravity. We know that, on the surface of the Earth
m mE
mg = G
r²
E
Solving for g gives
g = G
m
r²
E
E
Knowing g and the radius of the Earth, the mass of the Earth can be calculated:
gr²
E
m =
E
G
6
=
(9.80 m/s²)(6.38 x 10 m)²
-11
6.67 x 10
N · m²/ kg²
24
= 5.98 x 10 kg
Luis Anchordoqui
Gravity Near the Earth’s Surface; Geophysical Applications
(cont’d)
The acceleration due to gravity varies over the Earth’s surface due to altitude,
local geology, and the shape of the Earth, which is not quite spherical.
Acceleration Due to Gravity at Various Locations on Earth
Location
Elevation (m)
g (m/s²)
New York
0
9.803
San Francisco
0
9.800
Denver
1650
9.796
Pikes Peak
4300
9.789
Sydney, Australia
0
9.798
Equator
0
9.780
North Pole
0
9.832
Luis Anchordoqui
Falling to Earth
What is the acceleration of an object at the altitude of the space
shuttle's orbit, above 400 km above the Earth surface.
a= 8.70 m/s²
Luis Anchordoqui
Given that the acceleration
of gravity at the surface of
Mars is 0.38 of what it is on
Earth, and that Mars radius
is 3400 km, Pablo, Uniqua
and Austin are trying to
determine the mass of Mars.
Can you help them?
23
M = 6.4 x 10 kg
Luis Anchordoqui
Kepler’s Laws from Newton's Law
Kepler’s laws can be derived from Newton’s laws.
Irregularities in planetary motion led to the discovery of Neptune,
and irregularities in stellar motion have led to the discovery
of many planets outside our Solar System.
Luis Anchordoqui
Derivation of Kepler Second Law
In a time dt the planet moves a distance vdt and the radius vector r
sweeps out the area shaded in the figure
This is half the area of the paralelogram
form by vectors r and v dt
|r x mv|
dA = ½ |r x v dt| =
dt
2m
dA L
― = ―
dt 2m
L is the angular momentum of the planet around the Sun.
Because the force of the planet is along the line from the planet
to the Sun, it exerts no torque about the Sun
L is constant
The rate at which area is swept out is the same for all parts of the orbit
Luis Anchordoqui
Derivation of Kepler Second Law (cont’d)
L = constant
rv sin Φ = constant
At aphelion and perihelion Φ = 90º
ra v a = rp vp
Luis Anchordoqui
Newton’s Law of Gravitation implies
Kepler’s Third Law for circular orbit
Planet Earth moves with speed v in an approximate circular orbit of
radius r about the Sun
The gravitational force on the Earth by the Sun provides the
centripetal acceleration v²/r
GM Mʘ = M
r²
v² =
GMʘ
v²
r
r
Because the Earth moves a distance 2π r in time τ its speed is related
to the period by
2π r
v =
τ
Substituting
4π²
τ² =
GM
ʘ
3
r
Luis Anchordoqui
Gravitational Potential Energy
The general definition of the potential energy is
dU = -F • dl
where F is a conservative force acting on a particle and dl is a general
displacement of the particle
GM m
dU = -F • dl = -(-F g r)
ˆ • dl =
dr
r²
GM m
U = GM m ∫r¯²dr=­
+ U0
r
Luis Anchordoqui
Gravitational Potential Energy (cont’d)
We can set U0 = 0 and then U = 0 at r
GM m
U = r
∞
Gravitational potential of the Earth starting at the surface of the planet
GM m
U = R
and increasing with rising r
Luis Anchordoqui
Satellites
Satellites are routinely put into orbit around the Earth.
The tangential speed must be high enough so that the
satellite does not return to Earth, but not so high that it
escapes Earth’s gravity altogether.
Luis Anchordoqui
Satellites (cont’d)
The satellite is kept in orbit by its speed – it is continually
falling, but the Earth curves from underneath it.
Luis Anchordoqui
Escape speed
K f + U f = K i + Ui
The speed near the Earth’s surface corresponding to zero
total energy is called the escape speed ve
GM m
0 = ½ mv²
e
R
(
v e = 2GM
R
½
) =11.2km/s
Luis Anchordoqui
Escape speed (cont’d)
The kinetic energy of an object at a distance r from the center
of the Earth is
K = E – U (r)
When E < 0 (E1 )
K = 0 at r = rmax
When E > 0 (E2 )
the object can escape the Earth
Luis Anchordoqui
Gravitational field
The gravitational field at point P is determined by placing a point particle
of mass m and calculating the force on it due to all other particle
Fg
g = lim
m 0
m
The gravitational field at a field point due to masses of a collection of
point particles is the vector sum of the fields due to individual masses
g = ∑ gi
i
The locations of these point are called source point
To find the gravitational field of a continuous object we finf the field dg
due to a small element of volume with mass dm and integrate over the
entire mass distribution of the object
g = ∫ dg
The gravitational field of the Earth at a distance r≥ R
points
towards the Earth and has a magnitude
GM
g (r) =
r²
Luis Anchordoqui
Gravitational field of two point particles
Two point particles, each of mass M, are fixed in position
on the y axis at y = +a and y = -a.
Find the gravitational field at all points in the x axis as a function of x.
g = -
2GMx
(x²+a²)
3/2
ˆi
Luis Anchordoqui
A gravity map of Earth
The Gravity Recovery and Climate Experiment (GRACE) is the first mission
in NASA's Earth System Science Pathfinder project, which uses satelliteborne instrumentation to aid research on global climate change.
Twin satellites launched in March 2002
are making detailed measurements of
Earth gravitational field. They are in
identical orbits with one satellite
directly in front of the other by about
220 km. The distance between the
satellites is continuously monitored with
micrometer accuracy using onboard
microwave telemetry equipment.
How does the distance between the two
satellites changes as the satellites
approach a region of increased mass?
Luis Anchordoqui
A gravity map of Earth (cont’d)
As the twin satellites approach a region where there is an excess of
mass the increased gravitational field strength due to the excess mass
pulls them forward (toward the excess of mass).
The pull on the leading satellite is greater than the pull on the trailing
satellite because the leading satellite is closer to the excess mass.
Consequently the leading satellite is gaining speed more rapidly than is
the trailing satellite.
This results in an increase in
separation between the satellites
Luis Anchordoqui
A gravity map of Earth (cont'd)
Gravity anomaly maps show how much the Earth's actual gravity field differs
from the gravity field of a uniform, featureless Earth surface.
The anomalies highlight variations in the strength of the gravitational force
over the surface of the Earth
The gal (sometimes called galileo) is defined as
1 centimeter per second squared
1 gal = 1 cm/s²
Luis Anchordoqui
The gravitational slingshot effect
The figure shows the planet Saturn moving in the negative x direction at its orbital
26
speed (with respect to the Sun) of 9.6 km/. The mass of Saturn is 5.69 x 10 kg.
A spacecraft with mass 825 kg approaches Saturn. When far from Saturn, it moves
in the plus x direction at 10.4 km/s. The gravitational attraction of Saturn
(a conservative force) acting on the spacecraft causes it to swing around the planet
(orbit shown as dashed line) and head off in the opposite direction.
Estimate the final speed of the spacecraft after it is far enough away to be
considered free of Saturn's gravitational pull.
vf = - 29.6 km/s
Luis Anchordoqui
Ocean tides
The ocean tides have long been of interest to humans
The Chinese explained the tides as the brething of the Earth
*Around the sun once a year
Galileo tried unsuccessfully to explain the tides
effect of Earth´s motion
* On its own axis once a day
(could not account for the timing of the approximately two high tides each day)
Mariners have known for at least 4000 yr that tides are related to Moon’s phases
Exact relationship
hidden behind many complicated factors
Newton finally gave an adequate explanation
Luis Anchordoqui
Ocean tides (cont’d)
To the moon
Ocean tides are caused by the gravitational attraction of the ocean
Calculation is complicated
To the sun
Surface of the Earth is not an inertial system!
Earth’s rotation
Timing of tidal events is related to
Revolution of the moon around the Earth
If the moon was stationary in space
tidal cycle would be 24 hours long
However
The moon is in motion revolving around the Earth
1 revolution takes about 27 days
Adding about 50 minutes to the tidal cycle
Luis Anchordoqui
Ocean tides (cont’d)
Second factor controlling tides on Earth’s surface is then sun’s gravity
The height of the average solar tide is
MʘR²
moon
≈ 0.46 the average lunar tide
M moon R²
ʘ
At certain times during the moon’s revolution around the Earth
The direction of its gravitational attraction is aligned with the sun’s
During these times the two tide producing bodies act together
Creating the highest and lowest tides of the year
Spring tides occur every 14-15 days during full and new moons
Luis Anchordoqui
Ocean tides (cont’d)
When gravitational pull of moon and sun are right angles to each other
Daily tidal variations on the Earth are at their least
Neap tides occur during the first and last quarter of the moon
Luis Anchordoqui
g of a spherical shell of a solid sphere
Consider a uniform spherical shell of mass M and radius R
GM
g = ˆr r > R
r²
g = 0
r < R
To understand this last result consider the shell
segments with masses m 1and m2 that are proportional
to the areas A and A which in turn are proportional
1
2
to the radii r1 and r2
m1
A1
=
=
m2
A2
m1
m2
r²1 = r²2
r²
1
r²2
Because the gravitational force falls off inversely as the square of the
distance, the force on m due to the smaller mass m on the left, is exactly
0
1
balanced by that due to the more distant but larger mass m 2 on the right
Luis Anchordoqui
g inside a solid sphere
Consider a uniform solid sphere of radius R and mass M
As we have seen the field inside a spherical shell is zero
The mass of the sphere outside r exerts no force inside r
Only the mass M’ whithin the radius r contributes to the
gravitational field at r
Luis Anchordoqui
g inside a solid sphere (cont’d)
The mass inside r produces a field equal to that of a point
mass M’ at the center of the sphere
For a uniform sphere
M’
4
―π
= 34
―π
3
r3
R3
M
The gravitational field at a distance r < R is then
g
GM
=- GM’ =r
r²
r²
r3
GMr
= R3
R3
Luis Anchordoqui
Radially Dependent Density
A solid sphere of radius R and mass M is spherically symmetric but not
uniform. Its density ρ is proportional to the distance r from the center
for r < R. That is, ρ = Cr, for r < R, where C is a constant.
(a) Find C (b) Find g for all r > R. (c) Find g at r = R/2.
C =
g = -
GM
r²
M
4
πR
ˆ
r
(r > R)
GM ˆ
g = r (@r = R/2)
Luis Anchordoqui
4R²