Transcript Lecture 12.Gravitati.. - Faculty Web Sites at the University of Virginia

Gravitation Applications
Lecturer:
Professor Stephen T. Thornton
Reading Quiz
Astronauts
float in the
space
shuttle
because:
A) They are so far from Earth that Earth’s gravity
doesn’t act any more.
B) Gravity’s force pulling them inward is
cancelled by the centripetal force pushing them
outward.
C) While gravity is trying to pull them inward,
they are trying to continue on a straight-line
path.
D) Their weight is reduced in space so the force of
gravity is much weaker.
Reading Quiz
Astronauts
float in the
space
shuttle
because:
A) They are so far from Earth that Earth’s gravity
doesn’t act any more.
B) Gravity’s force pulling them inward is
cancelled by the centripetal force pushing them
outward.
C) While gravity is trying to pull them inward,
they are trying to continue on a straight-line
path.
D) Their weight is reduced in space so the force of
gravity is much weaker.
Astronauts in the space shuttle
float because they are in “free fall”
around Earth, just like a satellite or
the Moon. Again, it is gravity that
provides the centripetal force that
keeps them in circular motion.
Follow-up: How weak is the value of g at an altitude of 300 km?
Last Time
History of gravitation
Newton’s law of universal gravitation
Kepler’s laws
Free floating in space
Today
Orbital maneuvers
Ocean tides
Geophysical applications
Free floating in space
Satellites and weightlessness
Principle of Equivalence
Black holes
Conceptual Quiz
The Moon
does not crash
into Earth
because:
A) It’s in Earth’s gravitational field
B) The net force on it is zero
C) It is beyond the main pull of
Earth’s gravity
D) It’s being pulled by the Sun as
well as by Earth
E) none of the above is precise
enough
Conceptual Quiz
The Moon
does not crash
into Earth
because:
A) It is attracted to Earth
B) The net force on it is zero
C) It is beyond the main pull of
Earth’s gravity
D) It’s being pulled by the Sun as
well as by Earth
E) None of the above is precise
enough
The Moon does not crash into Earth because of
its high speed. If it stopped moving, it would,
of course, fall directly into Earth. With its
high speed, the Moon would fly off into space
if it weren’t for gravity providing the
centripetal force.
Follow-up: What happens to a satellite orbiting Earth as it slows?
The Global Positioning System
Orbital Maneuvers
Move to
higher orbit
Orbital Maneuvers
Move to
lower orbit
Ocean Tides
Arrows denote force
due to the moon
relative to the force at
the center of Earth.
Newton finally
correctly explained
tides!
Gravitation Force Due To Ring. A mass M is
ring shaped with radius r. A small mass m is placed
at a distance x along the ring’s axis as shown in the
figure. Show that the gravitational force on the
mass m due to the ring is directed inward along the
axis and has magnitude
F
GMmx
x
2
r
2

3
2
.
[Hint: Think of the ring as made up
of many small point masses dM;
sum over the forces due to each dM
and use symmetry.]
Vector Form of Newton’s Universal
Gravitation
If there are many particles, the total force
is the vector sum of the individual forces:
F12
2
1
F13
3
F15
F14
F1
4
5
We can relate the gravitational constant to the
local acceleration of gravity. We know that on
the surface of the Earth:
mm
mg = G
Solving for g gives:
mE
g= G 2
rE
E
2
E
r
g can be measured to 1 part in 109 so that
mineral and oil deposits can be detected using
sensitive gravitometers.
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.
Geosynchronous satellite.
A geosynchronous satellite stays above the same
point on the Earth, which is possible only if it is
above a point on the equator. Such satellites are
used for TV and radio transmission, for weather
forecasting, and as communication relays. They
must have an orbit of precisely 24 hours. In order
to do that, they must be about 22,000 miles
above the Earth and have a precise speed.
2
mM E
mv
G 2
r
r
2
GM
T
3
E
r 
4 2

GM E  2 r 
v 


r
 T 
T  24 hr
2
2
Lagrange point
The mathematician Joseph-Louis Lagrange
discovered five special points in the vicinity of the
Earth’s orbit about the Sun where a small satellite
(mass m) can orbit the Sun with the same period T as
Earth’s (= 1 year).
One of these “Lagrange Points,”
called L1, lies between the Earth
and Sun on the line connecting
them.
Several satellites are being placed
in Lagrange points. We probably
will not be able to service them like
we have done with the Hubble.
Sun
Satellites and “Weightlessness”
Objects in orbit are said to experience
“weightlessness”. They do have a gravitational force
acting on them, though! The satellite and all its
contents are in free fall, so there is no normal force.
This is what leads to the experience of
weightlessness.
More properly, this effect is called apparent
weightlessness, because the gravitational force
still exists. It can be experienced on Earth as
well, but only briefly:
Gravitational Field
The gravitational field is the gravitational
force per unit mass:
F
g=
m
The gravitational field due to a single mass
M is given by:
GM
g = - 2 rˆ
r
Principle of Equivalence
Inertial mass: the mass that appears in
Newton’s second law.
Gravitational mass: the mass that appears
in the universal law of gravitation.
Principle of equivalence: inertial mass and
gravitational mass are the same.
We can do no experiment to tell the difference
between gravitational and inertial mass.
Fundamental tenet of the General Theory of
Relativity.
One way to
visualize the
curvature of space
(a twodimensional
analogy):
If the gravitational field is strong enough,
even light cannot escape, and we have a
black hole. Einstein predicted in 1915 that
light should be attracted by gravity to mass.
At rest
Light should be deflected by a massive
object. On the right side, the person can
not tell whether acceleration caused the
light to bend or whether gravity did it.
This bending has been
measured during total
solar eclipses.
Gravitational Lensing
Massive stars can collapse under the
gravitational force. They can become
black holes, and nothing can escape
even light.
Einstein showed gravity even bends
light!
Gravity Assist
http://www.youtube.com/watch?v=I3F88
w3LkiI
Milky Way Galaxy. The Sun rotates about the center
of the Milky Way Galaxy (see figure) at a distance of
about 30,000 light-years from the center (1 ly = 9.5 x
1015 m). If it takes about 200 million years to make one
rotation, estimate the mass of our Galaxy. Assume that
the mass distribution of our Galaxy is concentrated
mostly in a central uniform sphere. If all the stars had
about the mass of our Sun (2 x 1030 kg), how many
stars would there be in our Galaxy?
Conceptual Quiz
Two satellites A and B of the same
mass are going around Earth in
concentric orbits. The distance of
satellite B from Earth’s center is
twice that of satellite A. What is the
ratio of the centripetal force acting
on B compared to that acting on A?
A) 81
B) 41
1
C) 2
D) it’s the same
E) 2
Conceptual Quiz
1
8
1
4
A)
Two satellites A and B of the same
B)
mass are going around Earth in
concentric orbits. The distance of
C)
satellite B from Earth’s center is
twice that of satellite A. What is the D) it’s the same
E) 2
ratio of the centripetal force acting
1
2
on B compared to that acting on A?
Using the Law of
Gravitation:
Note the
1/r2 factor
Mm
F G 2
R
1
4
we find that the ratio is .
Conceptual Quiz
A planet of mass m is a distance d from
Earth. Another planet of mass 2m is a
distance 2d from Earth. Which force
vector best represents the direction of
the total gravitation force on Earth?
2d
E
Earth
D
d
A B
m
C
2m
Conceptual Quiz
A planet of mass m is a
distance d from Earth.
Another planet of mass 2m
is a distance 2d from
Earth. Which force vector
best represents the
direction of the total
gravitation force on Earth?
2d
D
d
A
The force of gravity on the
Earth due to m is greater than
the force due to 2m, which
means
that
the
force
component pointing down in
the figure is greater than the
component pointing to the
right.
2m
E
B
C
m
F2m = GME(2m) / (2d)2 =
1
2
GMm / d 2
Fm = GME m / d 2 = GMm / d 2