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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all tieCh 13 Checkpoint 1
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.
(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
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
For any particle located inside uniform shell of matter , the net gravitational force between the particle and shell is zero
For =180 and F(r)=GMm/r2
Negative sign indicates direction of force opposite to that increasing r
Minimum initial speed required at Earth surface to send an object to infinity with zero kinetic energy (velocity) and zero potential energy. Then
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=0 for r= and U becomes more negative as particles move closer].
Kepler’s Law of Planetory Motion:
All planets move in elliptical orbits , with the sun at one focus.
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.
dA/dt = r2(d/dt)/2= r2/2
But angular momentum L=mr2
Then dA/dt= r2/2=L/2m
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
Since R1<R2 then T1<T2
Longer period forsatellite 2;
Greater v for smaller R i.e R1
the greater speed for satellite 1
The square of period of any planet is proportional to the cube of the semi major axis of the orbit
U=-GmM/R = -2K ; we have
K=- U/2 and
E=K+U= K+(-2K)=-K(circular orbit)
For same value of a, E is constant