Download Chapter 5: Circular Motion and Gravitation

Chapter 5
LECTURE NOTES
CIRCULAR MOTION AND GRAVITATION
Uniform circular motion is when a body moves in a circle at constant speed. The velocity,
however, is not constant as its direction keeps changing.
A changing velocity implies an acceleration. The radial acceleration aR (sometimes called
centripetal acceleration) keeps the body moving in a circle of radius r with velocity v: aR = v2/r,
and is directed toward the center of the circle. Thus we take the radial direction toward the
center to be positive in our equations of motion.
If the object takes time T (in seconds) to complete 1 revolution, we call T the period. The
frequency (number of revolution per second) f = 1/T. Since 1 trip around the circumference of a
circle is a distance of 2  r we find v = 2  r/T.
A force is required to provide the radial acceleration. This is not a new force but one of our old
forces (tension, gravity, normal force) directed to the center and causing circular motion. This
radial (or centripetal) force FR = maR = mv2/r.
Example: Motion in a vertical circle. Consider a Ferris wheel revolving at constant speed. A
person on a seat at the top of the circular path feels gravity pulling down toward the center of the
wheel and the seat pushing up on them, away from he center. Thus, mg - N =  FR = mv2/r.
How fast must the wheel go to make N = 0 at the top of the ride (person 'floats' off seat)? We
find v  rg .
Example: Rounding a curve. A car rounding an un-banked curve is kept on its circular path
by the combined frictional forces between its tires and the road. In the vertical direction, N - mg
= may = 0 if the car stays on the road or N = mg. In the radial direction fs ≤  sN provides the
centripetal force (why static friction?). Thus fs = mv2r ≤  sN or mv2/r ≤  s mg and vmax =
 s rg . Any faster, and the car cannot make the turn.
UNIVERSAL GRAVITATION
Newton found the same force causing apples to fall to the earth could explain the motion of the
moon about the earth. He wrote the gravitational force F between bodies m1 and m2 whose
2
Gm 1m 2
-11 Nm
centers are a distance r apart as F =
,
where
G
=
6.67
x
10
is the constant of
r2
kg 2
universal gravitation.
Note: This is an inverse square law; at twice the distance, the force is four times less.
Note: The force is always attractive and acts along the line joining the center of the bodies.
Chapter 5
LECTURE NOTES
Example: Mr. and Mrs. Watson (100.kg and 50.0kg, respectively) are attracted to each other. Is
the attraction gravitational? When they are 1.00m apart, the gravitational force he exerts upon
her and she exerts upon him is 3.34 x 10-7N; this cannot be felt.
Example: Calculate the acceleration due to gravity of a body falling without air resistance near
the surface of the earth.
F = ma with +y down toward the center of the earth.
G m bodym earth
Given:
= F = mbody a = mbody g
2
rearth
G m earth
g=
= 9.80 m/s2 where we have used
2
rearth
mearth = 5.90 x 1024kg and rearth = 6.38 x 106m
SATELLITES
If an object is given just the right tangential velocity it can move in a circular path about the
earth. It is in free-fall, forever falling toward the center of the earth like a broken elevator. It
never gets any closer to the surface because at its great tangential velocity, it moves just far
enough sideways as it falls, with the earth curving away from it far below, that its altitude
remains constant. The earth still exerts gravitational forces on orbiting astronauts (in fact,
gravity provides the centripetal force to keep satellites in circular orbits). But the spacecraft and
its occupants all fall at the same rate. Weightlessness in earth orbit is just like falling freely in a
broken elevator – it just lasts longer.
F = maR for uniform circular motion of a satellite orbiting the earth, a distance r from the earth’s
center.
G m sat m earth m earth v sat

r
r2
G m earth
vsat =
.
r
2
For a circular orbit, the period T =
2r
4 2 r 2  4 2
or T2 =
 
v
v2
 G m earth
 3
r .

Example: A geosynchronous satellite seems to hover above one place on earth; its orbital
period is 24 hours. Find the altitude of such a satellite above the surface of the earth.
Using the equation above to find r we have 4.23 x 107m or about 36,000 km above the surface of
the earth. Do you know who first suggested this could be applied to a TV broadcast satellite?
Chapter 5
LECTURE NOTES
Recall we found g =
G m earth
rearth
2
near the earth’s surface; a similar derivation show g at a distance R
from the center of the earth is g =
G m earth
R2
Consider the following questions.
1.
2.
3.
4.
What is g at 300 km above the earth’s surface?
What velocity v is required for a shuttle to circularly orbit 300 km above the earth’s
surface?
What time is required for a shuttle to complete one circular orbit 300 km above the
earth's surface?
Why are astronauts "weightless" if g ≠ 0?
Kepler's Laws
By studying Tycho's observations, Kepler found Copernicus was correct. The planets (including
earth) orbit about the sun. But unlike Copernicus and the ancients, who favored circular motion,
Kepler found the planets travel in elliptical orbits with the sun at one focus (the other focus is
empty). Summarizing, we have Kepler's three laws of planetary motion.
1.
2.
3.
The path of each planet about the sun is an ellipse with the sun at one focus.
Each planet moves so a radius vector connecting the sun and planet sweeps out
equal areas in equal times (the planets move faster when closer to the sun).
The square of the period of the planetary obit is proportional to the cube of the
planet distance from the sun.
Newton showed his law of universal gravitation led to the three laws above (we have seen T2 
r3). The laws apply equally well to a moon orbiting a planet or a satellite the earth.