Transcript m - Humble ISD

Newton’s Law of
Universal Gravitation
by
Daniel Silver
AP Physics C
Chapter 11
Physics for Scientists and Engineers, 4th ed
Modified by Stephanie Ingle, January 2009
Universal Law of Gravitation
Every object attracts every other object
with a force that is directly
proportional to the product of the
masses of the objects and inversely
proportional to the square of the
distance between them.
Calculating Gravitational Force
Fg = gravitational force
w = weight
m1m2
Fg  w  G 2
d
G = universal constant of
gravitation (6.67E-11)
m1, m2 = masses of objects
d = distance from center to
center between
objects
Fg
Fg
m1
d
m2
Orbital Velocity for a Satellite
moving in a Circular Orbits
Centripetal Force = Gravitational Force
Fc  Fg
v
2
mv
Mm
G
2
r
r
so...
v
GM
r
Useful Constants
• G = 6.67E-11 Nm2/kg2
• Mass of Earth = 5.98E24 kg
• Radius of Earth = 6.4E6 m
Kepler’s Laws
overview
• 1st Law: “Planets move in elliptical orbits
with the sun at one focus of the ellipse.”
• 2nd Law: “A line from the sun to a given
planet sweeps out equal areas in equal
times.”
• 3rd Law: The cube of the ratio of the orbital
radius is equal to the square of the ratio of
the period of rotation
Kepler’s 1st Law – Elliptical Orbits
• “Planets move in elliptical orbits with the sun at
one focus of the ellipse.”
Eccentricity is a property of an
ellipse that describes how
elongated the ellipse is.
Minor axis
Major axis
Planet
Empty focus
Sun
b
a
a = semi-major axis
b = semi-minor axis
Satellite Motion w/ Keplers 1st Law
• Planetary motion is just one example of
satellite motion…therefore this law can be
extended to any satellite orbiting around
any body.
• Circular orbits are a special case of an
elliptical orbit. (eccentricity =0)
• Elliptical orbits have an eccentricity
between zero and one…the larger the
number, the more elongated the ellipse.
Kepler’s 2nd Law –
Equal Areas in Equal Time
• “A line from the sun to a given planet
sweeps out equal areas in equal times.”
d
a
S
b
If a planet takes 20 days to move
from point “a” to point “b” and also
moves from point “c” to point “d” in
the same amount of time…the areas
of aSb and cSd will be equal.
c
This law helps to
illustrate that
planets (or
satellites) in
elliptical orbits
travel faster when
they are closer to
the Sun (or body
they are orbiting.
Kepler’s 3rd Law –
Orbital period and distance related
• The cube of the ratio of the orbital radius is equal to
the square of the ratio if the period of rotation
So for a satellite moving
in a circular orbit…
v
2r
v
T
and
v
GM
r
With a little substitution and algebra,
we can solve for the orbital period.
4 3
2
T 
r
GM
2r 2
T
GM
3
2
or
Kepler’s 3rd Law –
Elliptical Orbits
For elliptical orbits there is not a constant orbital
radius so we must use the semi-major axis “a” in
place of the “r” for orbital distance.
v
2
4

2
3
T 
a
GM
or
2a 2
T
GM
3
Gravitational Potential Energy
The Physics 1 way – “local Ug”
• So far, we have defined gravitational potential energy (Ug)
relative a reference point at the lowest point the object can go
(often the ground) which is given a value of zero.
• In doing this we have been able to calculate gravitational
potential energy simply in terms of height above the lowest
possible point (or surface of the earth), which is great for
conserving energy “locally” or near the surface of the earth.
In fact, the formula:
U g  mgh
Assumes the use of the local acceleration due to gravity “g”
at the surface of the earth, indicating that it is only valid for
energy conservation near the surface of the earth.
Gravitational Potential Energy
With Calculus and anywhere in the universe
Now that we are defining gravitation in a more universal sense,
we must re-define the zero reference point to be infinitely
away from the earth’s center. In otherwords…
U g  0 at
r 
Recall that force and potential energy are related …
dU
F 
dr
Where “r” is a
position vector
measured
radially rather
than along an x-,
y-, z- axis.
r
U g (r )    Fdr

If we rearrange
this equation and
do some
substitution and
algebra…
GMm
dr
2

r
r 1
U g (r )  GMm 2 dr
 r
GMm
U g (r )  
r
U g (r )   
r
Gravitational Potential Energy at the
Earth’s Surface
m
RE
Earth, ME
An object of mass “m” has a
Gravitational Potential Energy of
GM E m
Ug  
RE
Conservation of Energy in a
Gravitational Field
m
If an object (i.e. rocket) of mass “m”
were launched from the earth’s
surface at velocity “vi”, calculate the
velocity “vf” at a height “h” above
the earth’s surface.
h
m
RE
Earth, ME
1
Start by conserving energy the same
way as before… K1+U1=K2+U2 …
then substitute the equations for K &
U at each point…then solve for “vf”
GM E m

2 mv 
RE
2
i
1
GM E m
2 mv 
RE  h
2
f
Escape Velocity
m
Escape velocity is defined as the minimum velocity an
object must have in order to completely escape the planet’s
(earth’s) gravitational pull.
According to the potential energy formula…
GM E m
Ug  
r
m
RE
The planet’s gravitational potential
energy approaches zero as “r” approaches
∞, so we need to give our rocket exactly
the amount of Kinetic energy to start with
so that it will use it all up to get to ∞. In
other words, at r= ∞, Ug=0 and K=0. To
calculate the escape velocity from the
GM E m
earth’s surface…
1 mv 2 
 00
2
Earth, ME
esc
RE
So escape velocity from
2GM E
v

the earth’s surface is … esc
RE