Transcript Gravitation - Physics Rocks

Gravitation
Reading: pp. 124-129; 133-137
Syll. State.: 9.2.1-9.2.9—due
Friday, October 3
Newton’s Law of Universal
Gravitation

“Every material particle in the Universe
attracts every other material particle
with a force that is directly proportional
to the product of the masses of the
particles and that is inversely
proportional to the square of the
distance between them.”

Principia, Sir Isaac Newton
Gravitation
The gravitational forces obey Newton’s 3rd
law of motion—they are a pair of
action/reaction forces
 EVERY particle in the universe obeys this
law

Gravitation
Gm1m2
F12 
  F21
2
r





F12 = strength of gravitational force body 1
exerts on body 2
F21 = strength of gravitational force body 2
exerts on body 1
m1 and m2 = masses of the two bodies
r = separation distance between the center of
mass for each of the bodies
G = Universal Gravitation constant
= 6.67 x 10-11 N m2 kg-2
Sample Problem #1

Find the force between the sun and the
earth.
R  1.5 x 1011 m
m = 5.98 x 10 24 kg
GmM
F
R2
F
(6.67 x10
11
M = 1.99 x 1030 kg
24
30
N)(5.98 x 10 kg)(1.99 x 10 kg)
(1.5 x 1011 m)2
F  3.5 x 10 N
22


If a spherical object is
uniform in mass, we can safely
Re
M
assume that the gravitational
e
force due to this mass is exactly
the same as if all the mass were
concentrated at the very center.
Therefore, for something at the surface of this mass, we
can deduce that the following is true:
( M e m)
G
 Fg  mg
2
( Re )

Thus we can conclude that the acceleration due to the
gravity of a uniform mass is related to the radius of the
large mass (in this example, Earth…)
(M e )
g G
( Re ) 2
m
Sample problem #2

Determine the acceleration due to gravity
at the surface of a planet that is 10 times
more massive than the earth and has a
radius 20 times larger.
(10M E ) 10  M E 
g G

G 2 
2
(20 Re )
400  Re 
(M )
g G
2
( R)
1  ME
g  G 2
40  Re
 1
  ( ge )
 40
 1 
g    9.80  0.25 m  s-2
 40 
Gravitational Field Strength




Universal gravitation is VERY similar to Coulomb’s Law
(review electrostatics!), except there is no such thing as
negative mass…
Therefore, Gravitational forces will always be attractive
forces.
The gravitational field is a “field of influence radiating
outward from a particle of mass m”
Gravitational Field Strength at a given point is the
force exerted per unit mass on a particle of small mass
placed at that point.


The strength of the gravitational field is essentially equal to the
acceleration of the particle (g)
Units for the field strength, although they are equal to m s-2, are
typically listed as N kg-1
Gravitational Field (diagram)

Gravitational field lines are vectors that always
are directed radially toward the center of the
mass.

(remember…the field lines (vectors) will indicate the
direction a smaller mass will be pulled as a result of
the gravitational force caused by the field…)
Gravitational Potential Energy
Energy that is “stored” in an object’s
gravitational field.
 Work must be done in order to change the
position of a second mass from an
infinitely far away point in space to a
position relatively near the first mass.
 Remember: Work-Energy Theorem

The amount of work done on an object is
equal to the change in energy of that object.
 In this case, the energy change is all
potential.

Gravitational Potential Energy


Work is done when moving the mass from a
distance infinitely far from the source of a
gravitational field to some distance R from the
source
This work is stored in the gravitational field
between the two masses: Gravitational
Potential Energy
GMm
W  Ep  
R
Gravitational Potential (V)


A field, defined at every
point in space, but a
scalar quantity
Defined as:

The work done per unit
mass to bring a small
point mass (m) from
infinity to a point P.
W Ep
V 
m m
GMm

r
V
m
GM
V 
r
Total Energy of an Object in Orbit

Total energy, as usual, is the sum of the
orbiting object’s gravitational potential and
kinetic energies:
GMm
1
E  2 mv 
r
2

However, we can simplify this…
Total Energy…
simplified!


Mm mv 2
G 2 
r
r
GMmr
2

v
r 2m
GM
2
v 
r
Solve for v2 using
Newton’s laws:
Substitute into
energy equation:
 GM  GMm
E  m

r
 r 
GMm
2
1
E
  2 mv
2r
1
2
Graphing Mechanical Energy:

Graph of gravitational potential energy as a
function of distance from a planet’s surface
Orbital Motion
What kind of force causes objects to move
along a circular path?
Centripetal Force!
 To determine the orbital speed of a mass
that is orbiting around a larger central
mass, we must first recognize that the
centripetal force keeping it in orbit is
actually the gravitational force, so…

Orbital Motion
Mm
F G 2
r

And…
mv 2
Fc 
r
Mm mv 2
G 2 
r
r

So it follows that:

The orbital speed of a satellite, therefore,
can be determined by using:
M
2
G
v
r
Orbital Motion
The previous example shows us how the
orbital speed of a satellite depends on its
orbital radius.
 The farther out a satellite is from Earth’s
surface, the slower it can go and still
remain in orbit
 How far out (what orbital radius) must a
satellite have to remain in
geosynchronous orbit?

Kepler’s laws of Planetary Motion
1st Law: Planets move in ellipses around
the sun, the sun being at one of the foci of
each ellipse.
 2nd Law: The line from the sun to he
planet sweeps out equal areas in equal
times.
 3rd Law: The time taken to complete one
full revolution is proportional to the 3/2
power of the mean distance from the sun.


Or: the orbital period squared is directly
proportional to the cube of the mean distance
from the sun.
Orbital Period, related to Kepler’s
third law of planetary motion…


Assume a circular orbit…the average orbital
speed is equal to the distance traveled by the
planet or satellite in one orbital period:
Which Means…
d
2 r
v
v
t
T
Using the relationship we previously deduced:
M
G
 v2
r

We can relate Newton’s law of gravitation to
Kepler’s 3rd law of planetary motion (see
problem #7 from your homework assignment)
Weightlessness

The forces that act on an astronaut (or
anything else) in orbit in a spacecraft,
some distance r from the center of the
earth are:
His weight (gravitational force from Earth)
 The normal force (N) from the “floor” of the
spacecraft.


What is the net force acting on the
astronaut?

Fnet = (GMm) – N = mv2
r2

r
(GMm) – mv2 = N
r2
r
N = (m/r){(GM/r) – (v2)}
When we looked at Orbital motion, we showed that
(GM/r) = v2 …so our equation above mathematically
shows that the Normal force that the astronaut feels is
equal to 0 N...
Therefore: “Weightlessness” implies not that there is no
gravity, but rather that there is no reaction force exerting
the upward force that we sense as our weight.