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Circular Motion and Gravitation7-5 Newton's Law of Gravitation
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.
GRAVITY
The most pervasive force in the universe.
Topic 2.4 ExtendedA – Newton’s law of gravitation
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:
GRAVITYSTRONG ELECTROMAGNETIC WEAK
+
+
nuclearforce
light, heat and charge
radioactivity freefall
Einstein spent years trying to find the “superforce” which has all four of the fundamental forces of nature as special manifestations. This process is called “unification of forces,” and Einstein was not successful.
FYI: After his death, physicists showed that the weak and the electromagnetic forces were manifestations of a single force, called the “electro-weak force.”
ELECTRO-WEAK
WEAKESTSTRONGEST
NEWTON'S LAW OR UNIVERSAL GRAVITATION
Topic 2.4 ExtendedA – Newton’s law of gravitation
It turns out that the force of attraction F between two point masses m1 and m2 is given by
F12 = -Gm1m2
r2
where F12 is the force on m1 caused by m2,
where G is the universal gravitational constant and has the value G = 6.67×10-11 N·m2/kg2,
m1 m2
where r is the distance between the centres of the two masses and is measured in meters.
r
F12
FYI: F21, the force on m2 caused by m1 is equal and opposite (Newton’s 3rd, action-reaction pair)
F21
FYI: The gravitational force obeys an inverse square law. We will find out that next year that the electric force also obeys an inverse square law: F = kq1q2 / r2.
Newton’s Law of Universal Gravitation
Topic 2.4 ExtendedA – Newton’s law of gravitation
If a mass is attracted to more than one other mass, we simply sum all of the forces together (as vectors, of course).
Find the net force acting on m1 (2 kg) caused by m2 (4 kg) and m3 (6 kg)
F12
F13
m1 m2
m3 7 m
3 m
FYI: We do not need to find the force between m3 and m2 because we are interested only on m1.
We simply use Newton’s Law of Gravitation twice:
|F12| = G m1m2
r2= G
(2)(4)72
= 0.163G
F12 = 0.163G x
|F13| = G m1m3
r2= G
(2)(6)32
= 1.333G
F13 = 1.333G y
Fnet = F12 + F13 = 0.163G x + 1.333G y
NEWTON'S LAW OR UNIVERSAL GRAVITATIONFYI: This example illustrates the principle of superposition which means that the total gravitational force on a mass is simply the vector sum of the gravitational forces of all the masses surrounding it.
Topic 2.4 ExtendedA – Newton’s law of gravitation
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 mMr2
But from Newton’s 2nd law we have
F = mag where ag is the acceleration due to gravity.
where r is the distance from the center of the earth.
Equating the two forces we have
mag = G mMr2
so that
ag = GMr2
Gravitational acceleration near surface of planet
GRAVITATION AT A PLANET'S SURFACE
Topic 2.4 ExtendedA – Newton’s law of gravitation
GRAVITATION AT A PLANET'S SURFACEAt the surface of the earth, this reduces to
ag = GMr2
(6.67×10-11)(5.98×1024)(6.37×106)2=
= 9.829878576 m/s2
Question: Is this the expected result?
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 ExtendedA – Newton’s law of gravitation
GRAVITATION AT A PLANET'S SURFACEOur 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 = GMr2
(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 = GMr2
(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 ExtendedA – Newton’s law of gravitation
GRAVITATION AT A PLANET'S SURFACENow, 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.
RE
rEach latitude has a different centripetal acceleration given by
ac = rω2
The earth has a rotational velocity given by
ω =2π rad24 h
1 h3600 s
×
ω = 7.3×10-5 rad/s
Topic 2.4 ExtendedA – Newton’s law of gravitation
GRAVITATION AT A PLANET'S SURFACEA fbd for a mass on the equator like like this:
RE
r
W
ac
FBD mass on equator
We have: ΣFx = maN - W = -mac
N
N = W - mac
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 = – REω2GM
RE2
gapparent = – 6.37×106·(7.3×10-5)2(6.67×10-11)(5.98×1024)
(6.37×106)2
gapparent = 9.795932846 m/s2 = 9.829878576 m/s2
Compare to g for a stationary earth…
FYI: If ag = ac, the apparent weight of the mass becomes zero. Since it has no reason to stay in contact with the planet, it is free to “leave.”At this stage, scientists believe a planet (or star) would disintegrate or “explode.”
Topic 2.4 ExtendedA – Newton’s law of gravitation
NEWTON’S SHELL THEOREMA uniform spherical shell of matter exerts no net gravitational force on a particle located inside it.
m
M
FmM = 0
For a sphere, the net force from opposite conic sections exactly counter-balance one another…
m
M
FmM = 0M
ore
mas
sF
arth
er a
way
Less mass
Closer
No matter where inside the sphere the particle is located.
Force on particle inside spherical shell
FmM = 0
Topic 2.4 ExtendedA – Newton’s law of gravitation
NEWTON’S SHELL THEOREMA 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
FYI: Proof of Newton’s shell theorem required the use of calculus. This is one of the main reasons Newton invented integral calculus!
Force on particle outside spherical shell
FmM =GmMr2
Topic 2.4 ExtendedA – Newton’s law of gravitation
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
inner core Mi
outer core Mo
mantle Mm
crust McFor a point mass m located a distance r from the center of the earth we have
FmM =GmMc
r2+
GmMm
r2
+GmMo
r2+
GmMi
r2
so that
FmM =Gm(Mc+Mm+Mo+Mi)
r2
FmM =GmME
r2
NEWTON’S SHELL THEOREM
Topic 2.4 ExtendedA – Newton’s law of gravitation
GRAVITATIONAL POTENTIAL ENERGYWe’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 celestial-sized spherical masses,
U = - GmMr
Gravitational potential energy
Topic 2.4 ExtendedA – Newton’s law of gravitation
GRAVITATIONAL POTENTIAL ENERGYConsider 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.
m3
m1
m2
r12
r23
r13
U = - + +
Gm1m2
r12
Gm2m3
r23
Gm1m3
r13
Topic 2.4 ExtendedA – 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:
Mm
R
We will use energy considerations to find the escape velocity.
K + U = K0 + U0
If the rocket can reach to r = ∞, and come to a stop there, it has escaped the earth:
+ - = + - GmM
r0
GmMr
12 mv0
212mv2
0 0
=
GmMR
12 mv0
2
2GMR
vesc =escape velocity
Circular Motion and Gravitation7-5 Newton's Law of Gravitation
ESCAPE VELOCITY For us, the escape velocity from the earth is
2(6.67×10-11)(5.98×1024)6.37×106vesc =
2GMR
vesc =
vesc = 11191 m/s
vesc = 25027 mph !
Topic 2.4 ExtendedA – Newton’s law of gravitation
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:
2GMR
vesc =
2GMR
c =
2GMc2 Rs =
2GMR
c2 =
Schwarzchild radius of a 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.
FYI: Laplace used INCORRECT methods to derive his formula. It wasn't until shortly after 1915 and Einstein's publication of his general theory that Schwarzchild postulated the existence of a black hole, and used the general theory to derive the same formula (CORRECTLY). He therefore gets the honor. We have shown Laplace's method.
Topic 2.4 ExtendedA – Newton’s law of gravitation
BLACK HOLES For the earth, RS is given by
2GMc2 Rs =
2(6.67×10-11)(5.98×1024)(3×108)2
Rs =
Rs = 0.00886 m
= 8.86 mm