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Gravitation

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... a, semi minor axis b; eccentricity e, ea distance of one ... For a circle eccentricity e is zero. Ch 13-7 Planets and Satellites: Kepler's Law- Law of Areas ... – PowerPoint PPT presentation

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Title: Gravitation


1
CHAPTER-13
  • Gravitation

2
Ch 13-2 Newtons 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
  • FG(m1m2/r2)
  • Gravitational constant
  • G 6.67x10-11 m3/kg.s

3
Ch 13-2 Newtons 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

4
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

5
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) FnetGm2(1/d21/D2)i
  • (2) FnetGm2(1/d2 )i(1/D2) j
  • (3) FnetGm2-(1/d2 )(1/D2)i
  • (4) FnetGm2-(1/d2 )j(1/D2)i
  • (a) 1, tie of 2 and 4, then 3
  • (b) line d

6
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 y-axis with all
x-components cancelled out
7
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?i2F1i
  • 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 Earths center
  • FGmME/r2 but Fmag where ag is
    gravitational acceleration given by
  • ag GME/r2

8
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 FNmg then
  • mg mag-mR?2
  • g ag-R?2

9
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

10
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
  • -GMm?R?(dr/r2) GMm1/rR?
  • W0-GMm/r-GMm/r

11
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
  • KiUimvesc2/2-GMm/R0
  • vesc?2GM/R

12
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?
  • U -(GmM)/r
  • U0 for r? and U becomes more negative as
    particles move closer.
  • U becomes less negative and it increases
  • (b) Wg-Wa
  • negative

13
Ch 13-7 Planets and Satellites Keplers Laws
  • Keplers 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

14
Ch 13-7 Planets and Satellites Keplers 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 planets
    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 Lmr2?
  • Then dA/dt r2?/2L/2m

15
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 R1ltR2 then T1ltT2 Longer
period for satellite 2 (b) Kmv2/2GmM/2R
v2GM/R Greater v for smaller R i.e R1 the
greater speed for satellite 1
16
Ch 13-7 Planets and Satellites Keplers 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)
  • Newtons law applied to an orbiting planet gives
  • GmM/R2mv2/Rm?2/R
  • GM/R3 ?2(2?/T)2
  • R3/GMT2 /4?2
  • T2 (4?2/GM) R3

17
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 KU) of the Earth-satellite system
    remains constant.
  • Kmv2/2 but mv2/RGmM/R2
  • Then Kmv2/2GmM/2R
  • U-GmM/R -2K we have
  • K- U/2 and
  • EKU K(-2K)-K(circular orbit)
  • For an elliptical orbit Ra
  • Then E-K-GmM/2a
  • For same value of a, E is constant
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