Gravitational Potential Energy (con`t)
Download
Report
Transcript Gravitational Potential Energy (con`t)
4.3 Newton’s Law of Universal Gravitation
From Kepler to Newton
Newton used Kepler’s Laws to derive a law describing the nature of the
gravitational force that causes planets to move in their orbits.
Newton concluded that this force was a mutual force. If one objects
pulls on another, then that objects pulls back on the first object with
an equal but opposite force.
New showed that the force between two objects varied directly
with their individual masses, and inversely with the square of the
distance between them.
Fg = G
Mm
R2
The universal gravitational constant,
G = 6.67 x 10-11 Nm2/kg2 was not
determined by Newton.
p. 140
4.3 Newton’s Law of Universal Gravitation
Cavendish’s Experiment to Measure G
In 1797, Henry Cavendish performed a very sensitive experiment which both
confirmed Newton’s Gravitation Law and gave a value for the Gravitational constant.
A 2.0 m long rod was suspended from the ceiling by a
wire. Two small lead spheres were fixed to the end of
the rod. Two large masses were placed near the lead
spheres. Cavendish was able to measure the period of
the rotation of the rod due to the gravitation force
between the lead spheres and the large masses.
Through many difficult measurements Cavendish
was able to calculate the value for the universal
gravitation, G = 6.67 x 10-11 Nm2/kg2 (in modern
units).
p.
4.3 Newton’s Law of Universal Gravitation
Gravitation Field Strength of the Earth
Remember that force of gravity near the
Earth’s surface is calculated by:
Fg = m * g
Law of Universal Gravitation:
Since both give the
Force of gravity, let
them equal each other:
Fg = G
m*g
Cancel out the small, m from each side
results in an equation to determine the
acceleration due to gravity:
(where M = mass of Earth)
=
Mm
R2
G
g =
Mm
R2
G
M
R2
p. 141
4.3 Newton’s Law of Universal Gravitation
Weightlessness
According to Newton’s law of universal gravitation, the force of gravity varies
inversely as the square of the distance between the center of masses.
As this graph shows the gravitational force
never reaches zero unless R = infinity.
F 0 as R ∞
True weightlessness can only be achieved at
infinite distances between two objects.
Apparent weightless can be experienced when zero force is felt from supporting
structure s like your chair seat, the floor, or the Earth’s surface.
p. 145
4.3 Newton’s Law of Universal Gravitation
Example of Apparent Weightlessness
1. A Falling Elevator
A person standing on a scale in an elevator. When the elevator is stationary the
person will experience the true weight of the person as measure by the scale.
The normal force (which the scale actually measures) equals the force of
gravity. (Figure 4.3.3 (a))
p. 144
4.3 Newton’s Law of Universal Gravitation
Example of Apparent Weightlessness
1. A Falling Elevator (con’t)
When the elevator accelerates downwards the scale will measure a smaller
apparent weight. The elevator and the scale are falling away from the person
and hence will measure a smaller normal force. (Figure 4.3.3 (b))
If the elevator cable breaks, the elevator will accelerate downwards at 9.80
m/s2 (a = g) and be in free-fall. The elevator and the scale are falling
downwards at the same rate as the acceleration of gravity, hence no force will
be placed on the scale and the person will achieved zero apparent
weightlessness. (Figure 4.3.3 (c))
p.
4.3 Newton’s Law of Universal Gravitation
Example of Apparent Weightlessness
2. Orbiting Weightlessness
Astronauts in an orbiting space vehicle feel weightlessness for the same
reason as a person in a free-falling elevator. The astronauts are free-falling
towards the Earth but have enough horizontal velocity to actually travel in
orbit around the Earth.
3. Momentary Weightlessness
When a person jumps off the ground or dives off a diving board, or travels
over a hill in car, that person may experience apparent weightlessness for
brief periods of time while the person is no longer in contact with the Earth.
p. 144
4.3 Newton’s Law of Universal Gravitation
Satellites in Circular Orbits – Orbital Velocity
The orbital (tangential) speed that an object has to achieve to
maintain an orbit at a given altitude can be determined
through Newton’s Gravitational Law and centripetal force.
Fc = Fg
m*
v2
R
v2
Where M = mass of the Earth or other large body,
and R = distance of orbit to center of Earth.
v
= G
= G
=
√
Mm
R2
M
R
G
M
R
p. 145
4.3 Newton’s Law of Universal Gravitation
Gravitational Potential Energy
To move a spaceship to another position above a planet requires work to be done on
the spaceship against the gravitational field of the plant. Since the gravitational field
varies as the distance changes this is not a simple W = F x d situation.
The amount of work that must be done to escape
a planet’s gravitational field is equal to the area
underneath the force vs. distance graph, between
R = Re and R ∞.
By using some calculus it can be shown that the
change in gravitational potential energy is equal
to:
Mm
ΔEp = G
R
p. 146 -147
4.3 Newton’s Law of Universal Gravitation
Gravitational Potential Energy (con’t)
In situations involving space travel to distances away from a plant it is much more
convenient to make the gravitational potential energy equal to zero when R ∞.
When this happens the gravitational potential energy equations changes to:
Ep (at R = Re) = - G
Mm
R
This would make the gravitational potential energy at some point above a planet a
negative value. This makes sense if you understand that at R = ∞, there would be
no force and therefore the Ep = 0. At any distance less than R = ∞, you would have
to have a smaller Ep which would yield a negative value.
p. 147
4.3 Newton’s Law of Universal Gravitation
Escape Velocity
Escape velocity is the speed that a spaceship has to achieve to completely escape a planets
gravitational field. To determine this speed the kinetic energy of the spaceship plus the
gravitational potential energy of the spaceship at some altitude above the planet will be equal
to zero.
E + E = 0
K
½
mv2
+
p
-G
½ mv2
v2
Minimum speed for an object
to completely escape the
gravitational field of a planet.
=
= 0
R
G
2GM
=
vescape
Mm
Mm
R
R
=
√
2GM
R
p. 147
4.3 Newton’s Law of Universal Gravitation
In this section, you should understand how to solve the following key questions.
Page #142 - 143 Practice Problems 4.3.1 Newton’s Law of Universal Gravitation #1 – 3
Page# 145
Quick Check #1 – 4
Page# 146
Quick Check #1 – 2
Page# 148
Quick Check #1 - 2
Page #149 – 150
4.3 Review Questions #1 – 12
To be sure you understand the concepts presented in the entire chapter on Universal
Gravitation:
Page #151 – 154
Chapter 4 Review Questions #1 – 19