Transcript Topic 2_4_Ext A__Newton`s Law of Gravitation

GRAVITY
Circular Motion and Gravitation
7-5 Newton's Law of Gravitation
The most pervasive force in the
universe.
The galaxy cluster Abell 2218 is so
densely packed that its gravity bends
light passing through it. The arc-shaped
structures are stars that lie five times
further away than the cluster itself.
Einstein spent years trying to find the “superforce” which has all four of
Topic
Extended
the fundamental forces
of nature2.4
as special
manifestations. This
process isAcalled
“unification of forces,”
and Einstein
was not successful.
– Newton’s
law of
gravitation
FYI: After his death, physicists showed that the weak and the
electromagnetic forces were manifestations of a single force, called the
“electro-weak
force.”
Any piece
of matter in the universe will attract all
other pieces of matter in the universe.
The gravitational force is the weakest of all the
forces:
ELECTRO-WEAK
STRONG
ELECTROMAGNETIC
WEAK
GRAVITY
+
+
nuclear
force
STRONGEST
light,
heat and
charge
radioactivity
freefall
WEAKEST
Topic 2.4 Extended
F21, the
on m2 causedlaw
by m1 of
is equal
and opposite (Newton’s
A force
– Newton’s
gravitation
FYI:
3rd, action-reaction pair)
NEWTON'S LAW OR UNIVERSAL GRAVITATION
FYI: The gravitational force obeys an inverse square law. We will find
outIt
thatturns
next year
that
the electric
force also
an inverse
out
that
the force
of obeys
attraction
F square
between
twoF =point
m1 and m2 is given by
law:
kq1q2 /masses
r2 .
Newton’s Law of
m1m2
Universal Gravitation
r2
is the force on m1 caused by m2,
F12 = -G
where F12
where G is the universal gravitational constant and
has the value G = 6.67×10-11 N·m2/kg2,
where r is the distance between the centres of the
two masses and is measured in meters.
m1
F12
F21
r
m2
Topic 2.4 Extended
A – Newton’s law of gravitation
N
EWTON'S
LAW ORillustrates
UNIVERSAL
GRAVITATION
FYI:
This example
the principle
of superposition which means
If a mass is attracted to more than one other mass,
that the
total
gravitational
force
on
a mass
is simply
themvector
sum we
of
FYI:
do not
need all
to find
force
between
m3 and
because
2 (as
weWe
simply
sum
ofthethe
forces
together
vectors,
the
forces
are
interested
only
on mof1. all the masses surrounding it.
ofgravitational
course).
on m1 (2 kg) caused by m2
(4 kg) and m3 (6 kg)
We simply use Newton’s
Law of Gravitation twice:
m3
7 m
F13
m1
F12
3 m
Find the net force acting
m2
|F12| = G
m1m2
r2
= G
(2)(4)
= 0.163G
2
7
F12 = 0.163G x
|F13| = G
m1m3
r2
= G
(2)(6)
= 1.333G
32
F13 = 1.333G y
Fnet = F12 + F13
= 0.163G x + 1.333G y
Topic 2.4 Extended
A – Newton’s law of gravitation
GRAVITATION AT A PLANET'S SURFACE
Consider a mass m located near the earth’s surface:
It will feel an attractive force caused by the earth’s
mass M, given by
F = G
mM
r2
where r is the distance from the center of the earth.
But from Newton’s 2nd law we have
F = mag
where ag is the acceleration due to gravity.
Equating the two forces we have
mag = G
mM
r2
so that
ag =
GM
r2
Gravitational acceleration
near surface of planet
Topic 2.4 Extended
Question:AIs–thisNewton’s
the expected result?
law of gravitation
GRAVITATION AT A PLANET'S SURFACE
At the surface of the earth, this reduces to
ag
GM
= 2
r
=
(6.67×10-11)(5.98×1024)
(6.37×106)2
= 9.829878576 m/s2
As an aside, you might be curious as to how the
various constants were found.
The radius of the earth RE is an easy-to-find value,
and it was known to a good approximation by the Greek
astronomer and mathematician Eratosthenes (3 B.C.).
The value of the universal gravitational constant G
was considerably more difficult to find. In 1798, an
experimental physicist by the name of Henry Cavendish
performed a very delicate experiment to determine G.
Even Newton did not know the value of G.
Finally, knowing the value of freefall acceleration,
one can indirectly calculate the mass of the earth
knowing G and RE.
Topic 2.4 Extended
A – Newton’s law of gravitation
GRAVITATION AT A PLANET'S SURFACE
Our formula for ag works for any spherical mass, such
as the moon, or the sun, or any other planet,
satellite, or star.
For an astronaut on the surface of the moon, for
example,
ag
GM
= 2
r
=
(6.67×10-11)(7.36×1022)
(1.74×106)2
= 1.62 m/s2.
FYI: This is 9.8/1.6 = 1/6th that of the freefall acceleration on the
surface of earth.
For an astronaut in the space shuttle orbiting at an
altitude of 200 km,
ag
GM
= 2
r
=
(6.67×10-11)(5.98×1024)
(6.37×106 + 200000)2
= 9.24 m/s2 .
FYI: This is nearly that of the freefall acceleration on the surface of
earth. Why are the astronauts considered to be “weightless?”
Topic 2.4 Extended
A – Newton’s law of gravitation
GRAVITATION AT A PLANET'S SURFACE
Now, it just so happens that g at the equator is less
than g at the poles.
The reason for this is that the earth is rotating.
Each latitude has a different
centripetal acceleration
given by
ac = rω2
The earth has a rotational
velocity given by
2π rad
ω =
× 1 h
24 h
3600 s
ω = 7.3×10-5 rad/s
r
RE
Topic
2.4
Extended
FYI:
If astage,
the apparent
weight
of the
mass
zero. Since
g = ac,scientists
At this
believe
a planet
(or
star)becomes
would disintegrate
or it
ac to “leave.”
has
no reason
stay in contactlaw
with the
planet,
it is free
A –toNewton’s
of
gravitation
“explode.”
N
GRAVITATION AT A PLANET'S SURFACE
W
like like this:
ΣFx = ma
We have:
N - W = -mac
N = W - mac
FBD mass on
equator
A fbd for a mass on the equator
N = m(ag – ac)
The normal force is the
apparent weight of the
mass, so that
Wapparent = m(ag – ac)
mgapparent = m(ag – ac)
gapparent = ag – ac
gapparent = GM2 – REω2
gapparent
r
RE
Compare to g for a
stationary earth…
RE
(6.67×10-11)(5.98×1024)
=
– 6.37×106·(7.3×10-5)2
6
2
(6.37×10 )
gapparent = 9.795932846 m/s2
= 9.829878576 m/s2
Topic 2.4 Extended
A – Newton’s law of gravitation
NEWTON’S SHELL THEOREM
A uniform spherical shell of matter exerts no net
gravitational force on a particle located inside it.
M
FmM = 0
For a sphere, the
net force from
opposite conic
sections exactly
counter-balance
one another…
m
FmM = 0
Less mass
Closer
m
More mass
Farther away
M
No matter where inside the
sphere the particle is located.
FmM = 0
Force on particle
inside spherical
shell
Topic
2.4 required
Extended
FYI: Proof of Newton’s
shell theorem
the use of calculus. This
is one of the
main
reasons Newton
invented
integral calculus!
A –
Newton’s
law
of gravitation
NEWTON’S SHELL THEOREM
A uniform spherical shell of matter exerts a net force
on a particle located outside it as if all the mass of
the shell were located at its center.
M
m
r
FmM
GmM
=
r2
Force on particle
outside spherical
shell
Topic 2.4 Extended
A – Newton’s law of gravitation
NEWTON’S SHELL THEOREM
Even though the earth is not homogeneous, we can use
Newton’s shell theorem to prove that we were justified
in treating the earth as a point mass located at its
center in all of our calculations
For a point mass m
crust Mc
located a distance r from
mantle Mm
the center of the earth we
have
outer core Mo
GmMc + GmMm
FmM =
inner core Mi
r2
r2
GmMo + GmMi
+
r2
r2
so that
Gm(Mc+Mm+Mo+Mi)
FmM =
r2
GmME
FmM =
r2
Topic 2.4 Extended
A – Newton’s law of gravitation
GRAVITATIONAL POTENTIAL ENERGY
We’ve already discussed gravitational potential energy
U = mgy.
The problem with this formula is that it is a local
formula: It works only in the vicinity of the surface
of the earth.
It is beyond the scope of this course to prove the
following formula, but for point masses, or celestialsized spherical masses,
U = -
GmM
r
Gravitational potential
energy
Topic 2.4 Extended
A – Newton’s law of gravitation
GRAVITATIONAL POTENTIAL ENERGY
Consider the three charges
shown here, “assembled from
infinity.” How much
potential energy is stored in
the configuration?
Find the potential energies in
pairs, then sum them up. Since
U is a scalar, this is easy.
U = -
m2
r12
m1
Gm1m2
Gm2m3
Gm1m3
+
+
r12
r23
r13
r23
r13
m3
Topic 2.4 Extended
A – Newton’s law of gravitation
ESCAPE VELOCITY
Escape velocity it the minimum velocity required to
escape the gravitational force of a planet.
Consider a rocket of mass m on the surface of a
planet of mass M, say the earth:
We will use energy considerations to find the escape
velocity.
If the rocket can reach to r = ∞, and come to a stop
there, it has escaped the earth:
K + U = K0 + U0
0
0
1
1
GmM
mv02 +
mv2 + =
M
R
m
2
r
1
mv02 =
2
vesc =
2GM
R
2
GmM
R
escape
velocity
-
GmM
r0
Circular Motion and Gravitation
7-5 Newton's Law of Gravitation
ESCAPE VELOCITY
For us, the escape velocity from the earth is
vesc =
vesc =
2GM
R
2(6.67×10-11)(5.98×1024)
6.37×106
vesc = 11191 m/s
vesc = 25027 mph !
FYI: Laplace used INCORRECT methods to derive his formula. It
wasn't until shortly after
1915 and
Einstein's
publication of his general
Topic
2.4
Extended
theory that Schwarzchild postulated the existence of a black hole, and
A – Newton’s law of gravitation
used the general theory to derive the same formula (CORRECTLY). He
therefore gets the honor. We have shown Laplace's method.
BLACK HOLES
Soon after publication of Newton’s law of gravity a
mathematician by the name of Laplace postulated the
existence of a black hole – a body so massive the
even light cannot escape from it:
2GM
vesc =
R
2GM
c =
R
2GM
c2 =
R
Schwarzchild
2GM
radius of a
Rs = 2
c
black hole
FYI: The Schwarzchild radius tells us what the radius of an object of
mass M would have to be in order to become a black hole.
Topic 2.4 Extended
A – Newton’s law of gravitation
BLACK HOLES
For the earth, RS is given by
Rs =
2GM
c2
2(6.67×10-11)(5.98×1024)
Rs =
(3×108)2
Rs = 0.00886 m
= 8.86 mm