Chapter 5 Part 2

Chapter 5 Part 2
Newton’s Law of Universal
Gravitation, Satellites, and
Newton’s ideas about gravity
• Newton knew that a force exerted on
an object causes an acceleration.
• Most forces occurred because of contact with
an object.
• The idea of a force without contact met
• Newton theorized the force that holds all
things to the Earth, is the same force that acts
to hold the moon in its nearly circular path.
Force at a distance
• Newton compared the force Earth exerts on
objects at its surface to the force Earth exerts on
the moon.
• On Earth’s surface, acceleration due to gravity, g =
• The centripetal acceleration of the moon is
aR=v2/R. In terms of g, that is about 1/3600g.
• The moon is 60 times farther from Earth’s center
than objects are from the surface.
• Newton concluded Fg on any object decreases
with distance from Earth’s center by 1/d2.
What about mass?
• Newton reasoned, according to his 3rd law of
motion, that the force of gravity on an object
must also be directly proportional to the
masses of BOTH objects.
F 
• mB is the mass of a body, mE is the mass of
Earth, and r is the distance between their
Force vs distance applied
• In analyzing gravity, Newton found, by
examining the orbits of planets around the
sun, the force required to hold different
planets in orbit around the Sun seemed to
diminish as the inverse square of their
distance from the Sun.
Universal Gravitation
• In examining the planets, Newton stepped
further analyzing gravity.
• Newton theorized the gravitational force that
attracts an apple to the Earth and the moon to
the Earth, is the same force that acts between
the Sun and other planets to keep them in
their orbits.
• If gravity acts here, why not between ALL
Finalizing the Universal law of
• Every particle in the Universe attracts every other
particle with a force that is proportional to the
product of their masses and inversely proportional
to the square of the distance between them. This
force acts along the line joining the two particles.
• Universal Gravitational constant, G, (very small
number) was experimentally discovered to
calculate the EXACT force of attraction between 2
objects. G = 6.67 x 10-11 N-m2/kg2 (a constant like
Example 1
• A 50 kg person and a 75 kg person are sitting
on a bench so that their centers are 50 cm
apart. Estimate the magnitude of the
gravitational force each exerts on the other.
Example 1 Solution
• Using the equation for force, F  G
F 
(6.67 x10
(0.50 m )
m1m 2
N * m / kg )( 50 kg )( 75 kg )
• F = 1.0 x 10-2 N
• Which is unnoticeably small unless delicate
instruments are used.
Example 2
• What is the force of gravity acting on a 2000kg
spacecraft when it orbits two Earth radii from
the Earth’s center (a distance rE = 6380 km
above Earth’s surface)? The mass of the Earth
is mE = 5.98 x 1024kg.
Example 2 Solution
• Instead of plugging in all the numbers for our
equation, we could take a simpler approach.
• The spacecraft is twice as far from the Earth’s
center as when at the surface of the Earth.
Therefore, since the force of gravity decreases as
the square of the distance (and ½2 = ¼ ), the force
of gravity on it will be only ¼ its weight at the
Earth’s surface.
• FG = ¼ mg = ¼ (2000 kg)(9.80m/s2) = 4900 N
Geophysical applications of Gravity
near Earth’s surface
• Applying the Universal law between Earth and
objects at Earth’s surface, the FG becomes the
object’s weight.
mm E
• Thus we rewrite the formula: mg  G 2
gG 2
• Cancelling mass we get:
• Acceleration of gravity at Earth’s surface
depends on mE and rE. Once G was known,
the Earth’s mass was determined.
• When dealing with objects at Earth’s surface,
we calculate weight by mg. If we want to
consider objects far from Earth’s surface, we
can calculate the acceleration due to gravity
there by including their mass and distance
from Earth’s surface.
• EX: Estimate the effective value of g on the
top of Mt. Everest, 8848m above the Earth’s
surface. What is the accel due to gravity of
objects that freely fall at this altitude?
Example Solution
• Calling the acceleration due to gravity there,
g’, we replace rE with r = 6380 km + 8.8km =
6389 km = 6.389 x 106 m:
g  G
(6.67 x10
N * m / kg )( 5.98 x10
(6.389 x10 m )
• =9.77m/s2
• Which is a reduction of about 0.3%.
kg )
Satellites and “Weightlessness”
• Artificial satellites are put into Earth orbit by
high accelerations to give them a high
tangential speed, v.
• If tangential speed is too fast, the satellite
escapes Earth orbit. If too slow, it accelerates
downward to Earth due to Earth’s
gravitational pull.
• What keeps a satellite ‘up’ is a combination
between its tangential speed and Earth’s aR.
• Satellites moving in an approx circle have an
acceleration of aR = v2/r. This force is caused
by the force of gravity acting on it.
Newton’s laws on Satellites
• For an orbiting satellite, the only force present
is the force due to gravity.
• Using Newton’s Second law: ΣFR=maR, we find
mm E
• Note that r is the sum of Earth’s radius and the
satellite’s height above Earth: r = rE + h.
Geosynchronous Satellites
• A geosynchronous satellite is one that stays
above the same point on the equator of the
Earth. Such are used for purposes like cable
tv, weather forecasting, and communication
relays. Determine (a) the height above Earth’s
surface a satellite must orbit and (b) such a
satellite’s speed.
• See page 130 for worked out solution.
• Standing on a scale in a stationary elevator, mg
ON the scale is equal to the support force, w, BY
the scale.
• If the elevator accelerates upward, the net force
on you = ma. So we have w – mg = ma.
• Solving for w = mg + ma and you would “weigh”
more than normal.
• In a freely falling elevator, the scale falls at the
same rate as you and cannot push up (support)
your weight, so it would read zero. (Apparent
Weightlessness in satellites
• Since satellites orbiting Earth are essentially
“falling” around Earth, a passenger
experiences the same apparent
weightlessness that you would find in a freely
falling elevator.
Kepler’s Laws (1571-1630)
• Before Isaac Newton, a German astronomer
named Johannes Kepler spent his lifetime
studying planets and their motion. He
developed 3 laws of planetary motion:
• 1st Law: The path of each planet about the Sun
is an ellipse with the Sun at one focus.
• 2nd Law: Each planet moves so that an
imaginary line drawn from the Sun sweeps out
equal areas in equal periods of time.
Kepler’s Third Law
• 3rd Law: The ratio of the squares of the
periods of any two planets (time for one
revolution around the sun) is equal to the
ratio of the cubes of their mean distances
from the Sun.
• If T1 and T2 are the periods for any two planets
 r 
and r1 and r2 aretheir
distances from
T  average
    
the Sun, then T  r 
• If we rewrite this, then r3/T2 should be the
same for each planet.
Your turn to Practice
• Please do Chapter 5 Review pg 141 #s 25-30,
39, and 42. Bonus # 53.

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