Mathematical Physics Vol 1

4.6 Examples

155

Exercise 95 Let us observe the motion of a material point (particle) M under the action of a central force 22 . The differential equation of motion of a particle M , ofmass m , can then be represented in the form m d 2 r d t 2 = f ( r ) r 0 , where r is the position vector of the particle M measured in relation to the coordinate origin O , r 0 is the unit vector of vector r , and f ( r ) is a function of the distance of point M from O . a) Give an interpretation of the physical meaning of the expressions f ( r ) < 0 and f ( r ) > 0. b) Prove that r × d r d t = c , where c is a constant vector. c) Give a geometric an kinematic interpretation of the result under b). d) Relate the previous result to the motion of the planets in our solar system. d 2 r d t 2 is a vector opposite to vector r , and thus the force has direction from M to O , which means that an attractive force acts on the particle from point O . If f ( r ) > 0, then the force has direction from O to M , and a repulsive force acts on the particle from point O . b) Let us form a vector product of both sides of expression m d 2 r d t 2 = f ( r ) r 0 with vector r , from the left. We thus obtain m r × d 2 r d t 2 = f ( r ) r × r 0 = 0 , as the vectors r and r 0 are collinear, that is, r × r 0 = 0 . From here it follows that m r × d 2 r d t 2 = 0 , that is d d t r × d r d t = 0 . By integration of both sides we obtain r × d r d t = c , (4.181) where c is a constant vector. Solution a) For f ( r ) < 0, the acceleration

S

∂ P ∂ y

∂ Q ∂ x −

21 s

P d x + Q d y . In the special case, when P = 0 and Q = x , we obtain s S

d x d y = H C

d x d y = H C

x d y .

In the case when P = − y and Q = 0, we obtain s S y d x . 22 A force whose direction passes through a fixed point O of space (center of force) is called a central force. A particularly important category of central forces are those whose intensity depends only on the distance from the center. d x d y = − H C