CENTRAL – FORCE MOTION
When a particle moves under the influence of a force directed toward a fixed
center of attraction, the motion is called central-force motion.
Motion of a Single Body
Consider a particle of mass m moving under the action of the central gravitational
attraction, as shown in Fig, (9.1).
Where m0 is the mass of the attracting body, and the particle of mass m could be
the moon moving about the earth, or a satellite in its orbital motion about the
earth above the atmosphere.
The most convenient coordinate system to use is polar coordinates in the plane
The force F is always in the negative r-direction and there is no force in θ-
mm 0 2
G m r r ma r
0 m r 2 r ma
0 r 2 2r r r 2
r 2 h cons tan t .......... .........( 2)
The angular momentum of m about m0 is r x mv and has the magnitude of mr 2 .
This states that the angular momentum of m about m0 remains constant (is
During time dt, the radius vector sweeps out a shaded area equal to dA as
shown in Figure (9.1).
dA r rd
The rate at which area is swept is
A r 2 cons tan t
So, areas swept out in equal times are equal “ Kepler’s second law”.
The shape of the path followed by m may be obtained by solving equation (1)
with time t, as
r 1 u 2 u h u hdu d
and r h d 2u d 2 h 2 u 2 d 2 u d 2
Substitute into equation (1)
Gm 0 u 2 h 2u 2 2 h 2u 4
The solution of the above second order equation is
C cos 2 0
C and δ are the two integration constants.
The phase angle δ can be eliminated by choosing the x- axis so that r is
minimum when θ = 0, so
C cos 2 0 .......... .....( 3)
The conic section is formed by the locus of a point which moves so that the ratio
e of its distance from a point (focus) to a line (directrix) is constant. So from Fig.
d r cos
Which may rewritten as
1 1 1
cos .......... .......( 4)
r d ed
This has the same form of equation (3). The motion of m is along a conic section
with d and ed
C Gm 0
or e .......... .......... ..5
Three cases to be investigated corresponding to the value of e. The trajectory for
each of these cases is shown in Fig. (9.2).
Case 1: ellipse (e < 1).
From equation (4), we deduce that
r is minimum when θ = 0, and
r is maximum when θ =π.
ed ed ed
2a rmin rmax or a
1 e 1 e 1 e2
1 1 e cos
a 1 e2
rmin a1 e rmax a1 e Also b a 1 e2
When e = 0, r = a, and the ellipse becomes a circle.
For the period
A ab 2ab
1 2 h
h2C 1 ed
e d a b a 1 e2 , and Gm 0 gR 2
Where R is the mean radius of central attracting body and g is the absolute value
of acceleration due to gravity at the surface of the attracting body.
Case 2: parabola (e = 1).
Equations (4) and (5) become
and h 2 C Gm0
The radius vector becomes infinite as θ approaches π, so the dimension a is
Case 3: hyperbola e > 1).
From equation (4), the radial distance r becomes infinite for the two polar angles
θ1 and – θ1 defined by cos θ1 = - 1/e.
Branch I represent physical possible motion which correspond to – θ1 < θ < θ1,
Branch II corresponds to angles in the remaining sector (with r negative)
1 1 1 1
r d ed r ed d
The energy of particle m is constant
E m r r 2
1 gR 2
at 0, r 0, C from equation (3)
r from equation (2)
2E g 2R 4
Substitute value of C from equation (5)
mg 2R 4
elliptical orbit e < 1, E is negative
parabolic orbit e =1, E is zero
hyperbolic orbit e > 1, E is positive
For elliptical orbit
Substitute in energy equation
2 2gR 2
g 1 e g rmax
Perigee R R
a 1 e a rmin
g 1 e g rmin
Apogee R R
a 1 e a rmax