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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). Figure (9.1) mm 0 FG r2 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 of motion. The force F is always in the negative r-direction and there is no force in θ- direction. mm 0 2 G m r r ma r r2 .......... .......(1) 0 m r 2 r ma r multiply by m d 0 r 2 2r r r 2 dt So 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 conserved). During time dt, the radius vector sweeps out a shaded area equal to dA as shown in Figure (9.1). 1 dA r rd 2 The rate at which area is swept is 1 A r 2 cons tan t 2 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 1 r u 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) d2u 1 Gm 0 u 2 h 2u 2 2 h 2u 4 d u or d2u Gm u 20 d 2 h The solution of the above second order equation is Gm C cos 2 0 1 u r h 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 1 Gm C cos 2 0 .......... .....( 3) r h Conic Sections 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. (9.1), r e 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 1 h2 with d and ed C Gm 0 h2C or e .......... .......... ..5 Gm 0 Three cases to be investigated corresponding to the value of e. The trajectory for each of these cases is shown in Fig. (9.2). Figure (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 r 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 or 1 2 h A r 2 For h2C 1 ed e d a b a 1 e2 , and Gm 0 gR 2 Gm 0 , C , 1 e2 , a3 2 2 R g 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 1 cos 1 1 and h 2 C Gm0 r d The radius vector becomes infinite as θ approaches π, so the dimension a is infinite. 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, Fig. (9.3). Branch II corresponds to angles in the remaining sector (with r negative) Figure (9.3) 1 cos cos 1 1 1 1 or r d ed r ed d Energy Analysis The energy of particle m is constant E=T+V 1 2 2 mgR 2 E m r r 2 2 r 1 gR 2 at 0, r 0, C from equation (3) r h h r from equation (2) r 2E g 2R 4 h2C2 2 m h Substitute value of C from equation (5) 2Eh 2 e 1 mg 2R 4 For: elliptical orbit e < 1, E is negative parabolic orbit e =1, E is zero hyperbolic orbit e > 1, E is positive For elliptical orbit gR 2m E 2a Substitute in energy equation 1 1 2 2gR 2 r 2a 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

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posted: | 12/1/2011 |

language: | English |

pages: | 6 |

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