CHAPTER-13
Gravitation
Ch 13-2 Newton’s Law of Gravitation
 Newton's Law of Gravitation- a
key in understanding
gravitational force holding
Earth, moon, Sun and other
galactic bodies together
 Magnitude of gravitational force
F between two mases m1 and m2
separated by a distance r:
 F=G(m1m2/r2)
 Gravitational constant
 G= 6.67x10-11 m3/kg.s
Ch 13-2 Newton’s Law of Gravitation
 Shell theorem: A uniform
spherical shell attracts a
particle outside the shell as
if all the shell mass was
concentrated at the shell
center
 For any particle located
inside the shell, the net
gravitational force between
the particle and shell is
zero
Ch 13 Checkpoint 1
•A particle is to be placed , in
turn, outside four objects,
each of mass m: (1) large
uniform solid sphere, (2) large
uniform spherical shell, (3) a
small uniform solid sphere, (4)
a small uniform shell. In each
situation, the distance
between the particle and the
center of the object is d.
Rank the objects according to
the magnitude of the
gravitational force they exert
on the particles, greatest
first
• all tie
Ch 13 Checkpoint 2
The figure shows four
arrangements of three particles
of equal masses. (a) Rank the
arrangements according to the
magnitude of the net
gravitational force on the
particle labeled m, greatest
first. (b) In arrangement 2, is
the direction of the net force
closer to the line of length d or
to the length D.
(1)
(2)
(3)
(4)
Fnet=Gm2(1/d2+1/D2)i
Fnet=Gm2[(1/d2 )i+(1/D2) j]
Fnet=Gm2[-(1/d2 )+(1/D2)]i
Fnet=Gm2[-(1/d2 )j+(1/D2)i]
(a) 1, tie of 2 and 4, then 3;
(b) line d
Ch 13 Checkpoint 3
In the figure her, what is
the direction of the net
gravitational force on the
particle of mass m1 due to
other particles, each of mass
m and arranged symmetrically
relative to the y axis?
Net force downward along yaxis with all x-components
cancelled out
Ch 13-4 Gravitation near Earth Surface
 Principal of superposition of Gravitational force
 For n interacting particles, the net gravitational
force on particle 1 due to other is F1,net=i=2F1i
 Gravitation near Earth Surface:
 Force of attraction between the Earth and a
particle of mass m located outside Earth at a
distance of r from Earth’s center;

F=GmME/r2 but F=mag where ag is
gravitational acceleration given by:
 ag= GME/r2
Ch 13-4 Gravitation Near Earth
 Acceleration of gravity g
differs from g because:
 Earths mass is not
uniformly distributed
 Earth is not a sphere
 Earth is rotating
 Analyze forces on a crate
with mass m located at the
equator
 FN-mag=-FR=-mv2/R=-mR2
 But FN=mg then
 mg = mag-mR2
 g = ag-R2
Ch 13-5 Gravitation Inside Earth
For any particle
located inside
uniform shell of
matter , the net
gravitational force
between the
particle and shell is
zero
Ch 13-6 Gravitational Potential Energy
 Gravitational Potential Energy of
two-particles system:
 Work done on the ball when the
ball move from point P to a point
at infinity from earth center
 W= F(r).dr= F(r)dr cos
For =180 and F(r)=GMm/r2


W= R -F(r).dr=R -(GMm/r2)dr


2
=-GMmR (dr/r )= GMm[1/r]R

W=0-GMm/r=-GMm/r
Ch 13-6 Escape Speed
 Potential Energy and Force:
Force F(r)=-dU/dr=-d/dr(GMm/r)=-GMm/r2
Negative sign indicates direction of force opposite to
that increasing r
 Earth Escape Speed
Minimum initial speed required at Earth surface to
send an object to infinity with zero kinetic energy
(velocity) and zero potential energy. Then
Ki+Ui=mvesc2/2-GMm/R=0
vesc=2GM/R
Ch 13 Checkpoint 4
You move a ball of mass m
away from a sphere of
mass M.
(a) Does the gravitational
potential energy of the
ball-sphere system
increase or decrease?
(b) Is positive or negative
work done by the
gravitational force between
the ball and the sphere?
(a) U= -(GmM)/r
[ U=0 for r= and U becomes
more negative as particles
move closer].
(a) U becomes less
negative and it increases
(b) Wg=-Wa
negative
Ch 13-7 Planets and Satellites:
Kepler’s Laws
Kepler’s Law of Planetory Motion:
 Three laws namely Law of
Orbits, Law of Areas and Law of
Periods
 Law of Orbits:
All planets move in elliptical
orbits , with the sun at one
focus.
 Semi major axis a, semi minor
axis b; eccentricity e, ea
distance of one of the focal
point from the center of the
ellipse
 For a circle eccentricity e is zero
Ch 13-7 Planets and Satellites:
Kepler’s Law- Law of Areas
Law of Areas:
A line that connects a planet to
the sun sweeps out equal
areas in the plane of the
planet’s orbit in equal time
intervals; that is the rate
dA/dt at which it sweeps out
area A is constant
Area A of the wedge is the
area of the triangle i.e.
A=(r2)/2;
dA/dt = r2(d/dt)/2= r2/2
But angular momentum L=mr2
Then dA/dt= r2/2=L/2m
Ch 13 Checkpoint 5
Satellite 1 is in a
certain circular orbit
around a planet,
while satellite 2 is in
a large circular
orbit. Which satellite
has
(a) the longer period
and
(b) the greater speed
T2=(42/GM)R3
Since R1<R2 then T1<T2
Longer period for satellite 2;
(b) K=mv2/2=GmM/2R
v2=GM/R
Greater v for smaller R i.e R1
the greater speed for satellite
1
Ch 13-7 Planets and Satellites:
Kepler’s Law- Law of Periods
 Law of Periods
The square of period of any planet
is proportional to the cube of the
semi major axis of the orbit
 Considering the circular orbit with
radius R (the radius of a circle is
equivalent to the semi major axis of
an ellipse)
 Newton’s law applied to an orbiting
planet gives
GmM/R2=mv2/R=m2/R
GM/R3= 2=(2/T)2
R3/GM=T2 /42

T2 =(42/GM) R3
Ch 13-8 Satellites Orbits and Energy
 For an orbiting satellite, speed
fixes its kinetic energy K and its
distance from earth fixes its
potential energy U. Then
mechanical energy E (E = K+U) of
the Earth-satellite system
remains constant.
 K=mv2/2 but mv2/R=GmM/R2
Then K=mv2/2=GmM/2R
U=-GmM/R = -2K ; we have
K=- U/2 and
E=K+U= K+(-2K)=-K(circular
orbit)
 For an elliptical orbit R=a
Then E=-K=-GmM/2a
For same value of a, E is constant