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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.