universal gravitation unit 8. lesson 1 : newton’s law of universal gravitation in 1687 newton...

Universal Gravitation

Unit 8

Lesson 1 : Newton’s Law of Universal Gravitation

In 1687 Newton published Mathematical Principles of Natural Philosophy

In this work he states :

Every particle in the Universe attracts every other particle with a force that is

directly proportional to the product of their masses and inversely proportional to the

square of the distance between them.

Fg =G m1 m2

r2

product of masses

distance squared

Universal Gravitational Constant

G has been measured experimentally. Its value is

G = 6.673 x 10-11 N.m2 / kg2

Measuring the Gravitational Constant (G)

Henry Cavendish (1798)

F21 = -F12

Gravitational attraction forms an “action-reaction” pair.

The gravitational force exerted by a finite-size, spherically symmetric mass distribution

on a particle outside the distribution is the same as if the entire mass of the distribution

were concentrated at the center.

Force exerted by the Earth on a mass m near

the Earth’s surface.Fg =

G ME m

RE2

Newton’s Test of the Inverse Square Law

The acceleration of the apple has the same cause as the centripetal

acceleration of the moon.

aM

g=

1/rM2

1/RE2

=RE

2

rM2

=(6.37 x 106 m)2

(3.84 x 108 m)2

= 2.75 x 10-4

Centripetal acceleration of the Moon :

(2.75 x 10-4)(9.80 m/s2) = 2.70 x 10-3 m/s2

Newton calculated ac from its mean distance and orbital period :

ac =v2

r=

(2rM / T)2

rM

=42rM

T2

=42(3.84 x 108 m)

(2.36 x 106 s)2

= 2.72 x 10-3 m/s2

Example 1

Three 0.300 kg billiard balls are placed on a

table at the corners of a right triangle. Calculate the

gravitational force on the cue ball (m1)

resulting from the other two balls.

Free-Fall Acceleration and Gravitational Force

Equating mg and Fg :

mg =G ME m

RE2

g =G ME

RE2

g =G ME

(RE + h)2

Free-fall acceleration at a distance h above Earth’s surface :

Example 2

Find the mass of the Earth and the average density of the Earth.

Lesson 2 : Kepler’s Laws of Planetary Motion

Kepler’s First Law

All planets move in elliptical orbits with the Sun at one focus.

r1 + r2 = constant

2a = major axis

a = semimajor axis

2b = minor axis

b = semiminor axis

F1 , F2 are foci located c from center

For planetary orbits, the Sun is at one focus. There is

nothing at the other focus.

For circles, c = 0, e = 0.

e =c

a

Eccentricity (e) of an Ellipse

describes the shape of the ellipse

0 < e < 1

e = 0.25

e = 0.97Earth’s orbit e = 0.017

Kepler’s Second Law

The radius vector drawn from the Sun to a planet sweeps out equal areas in

equal time intervals.

Since = r x F = 0, the angular momentum L is constant.

In the time interval dt radius vector r sweeps out the area dA, which equals half the area (r x dr) of the

parallelogram formed by vectors r and dr.

dA = ½ (r x dr) = ½ (r x v dt) = L

2Mpdt

constant

The radius vector (r) from the Sun to any planet sweeps out equal areas in

equal times.

Kepler’s Third Law

The square of the orbital period of any planet is proportional to the cube of the

semimajor axis of the elliptical orbit.

Since gravitational force provides centripetal force,

G MS MP

r2=

MP v2

r

Since orbital speed = 2r / T,

G MS

r2=

(2r / T)2

r

Solving for T2 ,

T2 =42

GMS

r3

42

GMS

is a constant KS = 2.97 x 10-19 s2/m3

Replacing r with a,

T2 = KS a3

Example 1

Calculate the mass of the Sun using the fact that the period of the Earth’s orbit around the Sun is 3.156 x 107 s and its distance from the

Sun is 1.496 x 1011 m.

Consider a satellite of mass m moving in a circular orbit around the Earth at a constant

speed v and at an altitude h above the Earth’s surface, as shown above.

Example 2

a) Determine the speed of the satellite in terms of G, h, RE (radius of Earth), and ME (mass of

Earth).

b) If the satellite is to be geosynchronous (that is, appearing to remain over a fixed position on the Earth), how fast is it moving through

space ?

Lesson 3 : Gravitational Potential Energy

only valid when the mass is near the Earth’s surface,

where the gravitational force is constant

Ug = mgh

What is the general form of the gravitational potential energy function ?

Gravitational Force is Conservative

Work done by gravitational force is independent of path taken by an object.

As particle moves from point A to B, it is acted upon by a central force

F, which is a radial force.

Work done depends only on rf and ri.

The total work done by force F is the sum of the contributions along the

radial segments.

Work done by force F is always perpendicular to

displacement.

So, work done along any path between points A and B = 0

W = F(r) drri

rf

U = Uf – Ui = - F(r) drri

rf

Negative of the work done by gravitational

force

F(r) = -GMEm

r2

negative sign means force is

attractive

Uf – Ui = GMEm ri

rf

dr

r2= GMEm [ - ]

1

r ri

rf

Uf – Ui = - GMEm ( - ) 1

rf

1

ri

Taking Ui = 0 at ri = infinity,

U(r) = -GMEm

r(for r > RE)

Because of our choice of Ui, the function U is always negative.

U(r) = -GMEm

r(for r > RE)

In General (for any two particles)

U = -Gm1m2

r

Gravitational potential energy varies as 1/r, whereas gravitational

force varies as 1/r2.

U becomes less negative as r increases.

U becomes zero when r is infinite.

Example 1

A particle of mass m is displaced through a small vertical distance y near the Earth’s

surface. Show that in this situation the general expression for the change in gravitational

potential energy given by

reduces to the familiar relationship U = mgy.

Uf – Ui = - GMEm ( - ) 1

rf

1

ri

Lesson 4 : Energy in Planetary and Satellite Motion

E = KE + U

E = ½ mv2 -GMm

r

Newton’s Second Law applied to mass m :

GMm

r2= ma =

mv2

r

Multiplying both sides by r and dividing by 2,

½ mv2 =GMm

2r

Substituting into E = ½ mv2 -GMm

r,

E =GMm

r

GMm

2r-

E =GMm

2r- (total energy for

circular orbits)

E =GMm

2r-

total energy is negative

KE is positive and equal to half the absolute value of U.

For elliptical orbits, we replace r with semimajor axis a,

E =GMm

2a- (total energy for

elliptical orbits)

Total energy is constant.

Total angular momentum is constant.

In all isolated gravitationally bound two-object systems :

Example 1

The space shuttle releases a 470 kg communications satellite while in an orbit 280 km above the surface of the Earth. A rocket engine on the satellite boosts it into a geosynchronous orbit,

which is an orbit in which the satellite stays directly over a single location on the Earth. How much energy does the engine have to provide ?

Escape Speed

Minimum value of the initial speed needed to allow the object to move infinitely far away from the Earth.

When object reaches rmax, vf = 0, and

½ mvi2 -

GMEm

RE

GMEm

RE

= -

Solving for vi2,

vi2 = 2GME

1

RE

1

rmax

-( )

vi2 = 2GME

1

RE

1

rmax

-( )

If we let rmax infinity,

vesc =2GME

RE

Example 2

Calculate the escape speed from the Earth for a 5000 kg spacecraft, and determine

the kinetic energy it must have at the Earth’s surface in order to move infinitely

far away from the Earth.

Two satellites, of masses m and 3m, respectively, are in the same circular orbit about

the Earth’s center, as shown in the diagram above. The Earth has mass Me and radius Re. In this orbit, which has radius 2Re, the satellites

initially move with the same orbital speed vo, but in opposite directions.

Example 3 (AP 1984 #2)

a) Calculate the orbital speed vo of the satellites in terms of G, Me, and Re.

b) Assume that the satellites collide head-on and stick together. In terms of vo, find the speed v of the combination immediately after the

collision.

c) Calculate the total mechanical energy of the system immediately after the collision in terms of G, m, Me, and Re. Assume that the gravitational potential energy of an object is defined to be zero at an infinite distance from the Earth.

A spacecraft of mass 1,000 kg is in an elliptical orbit about the Earth, as shown above. At point A the spacecraft is at a

distance rA = 1.2 x 107 m from the center of the Earth and its velocity, of magnitude vA = 7.1 x 103 m/s, is

perpendicular to the line connecting the center of the Earth to the spacecraft. The mass and radius of the Earth are

ME = 6.0 x 1024 kg and rE = 6.4 x 106 m, respectively.

Example 4 (AP 1992 #3)

Determine each of the following for the spacecraft when it is at point A.

a) The total mechanical energy of the spacecraft, assuming that the gravitational potential

energy is zero at an infinite distance from the Earth.

b) The magnitude of the angular momentum of the spacecraft about the center of the Earth.

Later the spacecraft is at point B on the exact opposite side of the orbit at a distance rB = 3.6 x 107 m from the center of the Earth.

c) Determine the speed vB of the spacecraft at point B.

Suppose that a different spacecraft is at point A, a distance rA = 1.2 x 107 m from the center of the

Earth. Determine each of the following.

d) The speed of the spacecraft if it is in a circular orbit around the Earth.

e) The minimum speed of the spacecraft at point A if it is to escape completely from the Earth.