circular motion and gravitation 7-5 newton's law of gravitation the galaxy cluster abell 2218...

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


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


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.


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


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.


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?”


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


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.”


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


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


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


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


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


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 !


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.


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

circular motion and gravitation 7-5 newton's law of gravitation the galaxy cluster abell 2218...

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