Download Section 7.2 Using the Law of Universal Gravitation

Mr. Borosky
Physics Section 7.2 Notes
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Section 7.2 Using the Law of Universal Gravitation
Objectives
Solve orbital motion problems.
Relate weightlessness to objects in free fall.
Describe gravitational fields.
Compare views on gravitation.
Read intro paragraph p. 179
ORBITS OF PLANETS AND SATELLITES
Read Section.
Go over Cannonball example p. 179
The curvature of the projectile would continue to just match the
curvature of Earth, so that the cannonball would never get any
closer or farther away from Earth’s curved surface. The cannonball
therefore would be in orbit. This is how a satellite works.
Thus a cannonball or any object or satellite at or above this
altitude could orbit Earth for a long time.
A satellite in an orbit that is always the same height above Earth
move in uniform circular motion.
ac = v2 / r
Fc = mac = mv2 / r
Combining the above equation with Newton’s Law of Universal
Gravitation yields the following equation
(GmEm / r2) = (mv2 / r)
Speed of a Satellite Orbiting Earth – is equal to the square root of
the universal gravitation constant times the mass of Earth divided
by the radius of the orbit.
v = √ (GmE) / r
Period
square
of the
T
of a Satellite Orbiting Earth – is equal to 2Π times the
root of the radius or the orbit cubed divided by the product
universal gravitation constant and the mass of Earth.
= 2Π √ r3 / (GmE)
Physics Principals and Problems © 2005 Started 2006-2007 School Year
Mr. Borosky
Physics Section 7.2 Notes
Page 2 of 4
The equations for the speed and period of a satellite can be used
for any object in orbit about another object.
The mass of the central body will replace mE in the equations and r
will be the distance between the centers of the orbiting body and
central body.
Since the acceleration of any mass must follow Newton’s 2nd Law (F =
ma) more force is needed to launch a more massive satellite into
orbit. Thus the mass of a satellite is limited to the capability of
the rockets used to launch it.
Do Example Problem 2 p. 181
v = √ (GmE) / r
v = √ (6.67 * 10-11)(5.97 * 1024) / (6.38 * 106 + .225 * 106)
v = √ (3.98199 * 1014) / (6.605 * 106)
v = √ (6.029 * 107)
v = 7.765 * 103 m/s
T
T
T
T
T
T
=
=
=
=
=
=
2Π √ r3 / (GmE)
2Π √ (6.605 * 106)3 / (6.67 * 10-11)(5.97 * 1024)
2(3.14) √ (2.8815 * 1020) / (3.98199 * 1014)
6.28 √ (7.236 * 105)
6.28 (850.647)
5342.062 s
(If convert = 89 min or 1.5 hours
Do Practice Problems p. 181 # 12-14
ACCELERATION DUE TO GRAVITY
Read Section.
F = GmEm / r2 = ma
a = GmE / r2
a = g and r = rE so
g = GmE / rE2
mE = grE2 / G
a = g (rE / r)2
This last equation shows that as you move farther from Earth’s
center (r becomes larger) the acceleration due to gravity is reduced
according to the inverse square relationship.
Physics Principals and Problems © 2005 Started 2006-2007 School Year
Mr. Borosky
Physics Section 7.2 Notes
Page 3 of 4
Weightlessness – an object’s apparent weight of zero that results
when there are no contact forces pushing on the object. This is
also called Zero g.
There is gravity in space. Gravity is what causes the shuttle and
satellites to orbit Earth.
THE GRAVITATIONAL FIELD
Read Section.
Gravity acts over a distance.
touching.
It acts on objects that are not
Michael Faraday – invented the concept of the field to explain how a
magnet attracts objects. Later the field concept was applied to
gravity.
Gravitational Field – is equal to the universal gravitational
constant times the object’s mass divided by the square of the
distance from the object’s center.
g = Gm / r2
The Gravitational Field can be measured by placing an object with a
small mass in the gravitational field and measuring the force on it.
Then the gravitational field is the force divided by a mass. It is
measured in Newtons per kilogram (N/kg) which = m/s2. Thus we have
g = F / m
The strength of the field varies inversely with the square of the
distance from the center of Earth. The gravitational field depends
on Earth’s mass but not on the mass of the object experiencing it.
TWO KINDS OF MASS
Read Section.
Inertial Mass – is equal to the net force exerted on the object
divided by the acceleration of the object. It is a measure of the
object’s resistance to any type of force.
mInertial = FNet / a
Physics Principals and Problems © 2005 Started 2006-2007 School Year
Mr. Borosky
Physics Section 7.2 Notes
Page 4 of 4
Gravitational Mass – is equal to the distance between the objects
squared times the gravitational force divided by the product of the
universal gravitational constant times the mass of the other object.
mGravitational = r2FGravitational / Gm
Newton claimed that Inertial and Gravitational Mass are equal in
magnitude. All experiments done since then show this is the case.
EINSTEIN’S THEORY OF GRAVITY
Read Section.
Einstein proposed that Gravity is not a force but rather an effect
of space itself. According to Einstein mass changes the space
around it. Mass causes space to be curved and other bodies are
accelerated because of the way they follow this curved space.
Einstein’s General Theory of Relativity – makes many predictions
about how massive objects affect one another. It predicts the
deflection or bending of light by massive objects. In 1919, an
eclipse of the sun proved Einstein’s theory.
Do 7.2 Section Review p. 185 # 15-21
Physics Principals and Problems © 2005 Started 2006-2007 School Year