Chapter 13
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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:

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CHAPTER-13

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Chapter 13

CHAPTER-13

Gravitation


Ch 13 2 newton s law of 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 gravitation1

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 1


Gravitation

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) Fnet=Gm2(1/d2+1/D2)i

(2)Fnet=Gm2[(1/d2 )i+(1/D2) j]

(3) Fnet=Gm2[-(1/d2 )+(1/D2)]i

(4) Fnet=Gm2[-(1/d2 )j+(1/D2)i]

(a) 1, tie of 2 and 4, then 3;

(b) line d


Gravitation

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


Ch 13 4 gravitation near earth surface

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 isF1,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

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

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

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)= GMm[1/r]R

  • W=0-GMm/r=-GMm/r


Ch 13 6 escape speed

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


Gravitation

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

    [ U=0 for r= and U becomes more negative as particles move closer].

  • U becomes less negative and it increases

    (b) Wg=-Wa

    negative


Ch 13 7 planets and satellites kepler s laws

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

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


Gravitation

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 forsatellite 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

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

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


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