Transcript you can`t use the symbol g for this because it is not at the earth`s

Gravitaton
Universal Gravitation
Gravity Near the Earth’s Surface
Satellites and Weightlessness
5.6 Newton’s Law of Universal Gravitation
Any two objects of mass attract each other. This force
(gravitational force) acts at a distance; no contact is needed.
The gravitational force between objects of mass M and m,
separated by a distance R, is given by
OSE :
GmM
FG =
, attractive .
2
R
G is a constant, and has a value of G=6.67x10-11 N·m2/kg2.
M
FMm
FmM
m
FMm = -FmM
Example 5-12: What is the force of gravity acting on a 2000
kg spacecraft when it orbits two earth radii from the earth’s
center (that is, a distance RE=6380 km above the earth’s
surface)? The mass of the earth is ME=5.98x1024 kg.
Solution #1: plug the numbers into the equation on the
previous slide.
Solution #2: Giancoli’s physicist-type solution—pull the answer
seemingly out of nowhere.
Solution #3: Use the equation on the previous slide to
calculate the force at RE and 2RE, and take the ratio. I’ll do
this one on the blackboard. (Note: the space shuttle orbits at
a height of a few hundred km.)
5.7 Gravity near the Earth’s Surface
For the first half of this class we have been assuming g (as in
W = mg) is a constant. It is not.
Weight (on earth) is the force on an object due to the
gravitational attraction of the earth.
GmM
 GM 
FG =
= m  2  = mg
2
R
R 
suggests that
 GM 
g =  2 .
R 
We have been treating the earth as an object with all its mass
concentrated at a point at its center. Clearly, the value of g
depends on how far you are from this center.
An object with a spherically symmetric mass distribution can
be treated, for the purposes of calculating gravitational forces,
as a point mass with all the mass concentrated at the center.
The earth is a flattened spheroid, and its mass distribution is
nonuniform.
Slightly fatter at equator.
Greatly exaggerated in drawing.
Near the surface of the earth, g is approximately 9.80 m/s2.
g varies measurably from place to place,
depending on how the nearby mass is
distributed, and the value of g can be measured
quite accurately (say, 6 decimal places) with a
pendulum (we’ll see how this works later).
See this web page for depictions of the earth’s gravitational
field, and a bigger rotating earth image.
g is different in the basement of Physics than it is on the 2nd
floor, and the difference is measurable. The difference,
however, is small enough that we can treat g as a constant for
our 3-figure calculations.
Example 5-14: Estimate the effective value of g at the top of
Mount Everest (8848 m above the earth’s surface). Use
RE=6380 km, ME=5.98x1024 kg, and assume (unrealistically)
that the earth’s mass distribution is spherically symmetric.
Of course, the existence of Mount Everest proves that the
earth’s mass distribution is not spherically symmetric.
Nevertheless, we can use
geffective
GM
= 2
R
you can’t use the symbol
g for this because it is not
at the earth’s surface
and get geffective=9.77 m/s2.
Gosh, I wonder what you get for g if you plug in RE and ME?
Alternatively, if we know G and the radius of the earth, we can
calculate the mass of the earth.
Note: mg = FG (done four slides back) is not an OSE; it only
works at surface of earth with g = 9.8 m/s2.
5.8 Satellites and “Weightlessness”
A satellite in orbit is acted on by gravity (the centripetal force).
You already know how to handle centripetal force problems:
Fr = Fgrav,r = mar = mv2/r
and the force is directed towards the center of the earth.
This is old stuff; nothing new here!
Now, if you have thought much about satellites, you probably
have thought you could put a satellite in orbit at any height
above the earth you wanted, with any speed you wanted.
Not so! I’ll show the simple calculation on the board that
proves there is a unique orbital velocity for any orbital radius
(or vice versa).
F =F
r
grav
=mar
mEmsat
v2
G
= msat
2
r
r
v2 =
GmE
r
Do you know what a
geosynchronous satellite is?
It’s a satellite that orbits with a
speed so that it always stays
above the same point above the
earth.
Such satellites are extremely
valuable for communications,
weather monitoring, and spying.
Trouble is, we’ve virtually run out of room in space for
geosynchronous satellites.
Huh? How can you run out of room in space?
The speed of the satellite is v=2R/T where T is one day.
If you use V=2a/T for our orbital speed V at height a, and
set FG=mV2/r, you get another OSE:
OSE:
T2
 4π2  3
=
a .

 GM 
I’ve used the symbol “a” for the orbital radius, for consistency
with Physics 23.
You can solve this equation for the orbital radius of a
geosynchronous satellite (T = 1 day for this satellite):
 GM  2
a3 = 
T .
2
 4π 
Everything on the right hand side is a constant! Plugging in
the values gives a=42,300 km, or a height of about 36,000 km
above the earth’s surface.
Easy calculation for you to try: what is the orbital speed of a
geosynchronous satellite? Compare your result with the
answer on page 130 of Giancoli.
Weightlessness
You have probably felt “weightless” on a carnival ride, roller
coaster, or rapidly-accelerating elevator.
Were you truly weightless? One could only wish. You simply
experienced free fall.
An astronaut in the space station is not weightless either, but
experiences the weightless sensation because he/she is
constantly falling around the earth.
The effective value of g at the orbital radius of the space
station is not zero! Nevertheless, it is probably valid to claim
that the apparent weightlessness of free fall simulates true
zero-g weightlessness in outer space.