252 



♦ KNOWLEDGE ♦ 



[Sept. 26, 1884. 



FLIGHT OF A MISSILE. 



By Richard A. Proctor. 



I HAVE been asked whether the method by which I 

 treated the movement of a body projected vertically 

 from a point on the equator — regarding such a body, from 

 the moment when its flight began, as moving in space under 

 the attraction of the earth, and therefore in obedience to 

 Kepler's laws — can be extended to any missile projected 

 in any direction from any part of the earth's surface. I 

 see no reason why the method should fail, or even why 

 there should be any difficulty in applying it. Let us, 

 however, inquire. 



Let O, the point from which a missile is projected, lie 

 iu latitude A, and let O X be part of a latitude-parallel and 

 Y part of the meridian ; O Z vertical. 



Fig. 1. 



Suppose the missile to be projected from O in direction 

 O P with velocity r, at an elevation f ; and draw P L 

 perpendicular to the plane Y X (part of the earth's 

 surface, appreciably, near O) join L, and draw L M, L N 

 perpendicular to O X and O Y respectively. Let angle 

 LOX = a. 



Suppose now that the length P represents the velocity 

 V ; then P L represents the vertical velocity of the missile, 

 or II sin £ ; L represents the horizontal velocity, or v cos t ; 

 O M is the velocity in direction O X, or r cos e cos a ; and 

 O N is the velocity in direction O Y or i- cos c sin «. But 

 we have to consider also the velocity in direction O X on 

 account of the earth's motion of rotation, which the missile 

 shares with the earth's surface at the moment of projection, 

 and retains thereafter, subject of course to such influences 

 as may thereafter be brought to bear on it. Call this 

 velocity u for convenience (we know that it is represented 

 by the circumference of the parallel of latitude though O 

 divided by the period of rotation, or is 2 n- r cos X ^ P, 

 where r is the radius of the earth in feet and P the number 

 of seconds in a sidereal day). Then if we represent 21 by 

 M m, O m represents the true velocity of the missile in 

 direction X, or v cos e cos a -f 71. 



Complete the parallelogram O N / m, draw / j) parallel 

 and equal to L P, and join O^), then O /7 represents the 

 actual velocity of the missile in space, referred to the earth's 

 centre. (Of course the missile shares the motion of the 

 earth around the sun, of the solar system in space, &c., tkc, 

 but in considering its flight abo-ve the earth's surface we 

 need take no account of these parts of its motion). Call 

 this velocity V. 



Join p P which is parallel to U and produce to meet 

 the plane Z Y in K and join O K ; then since 7; K is parallel 

 to / N, it is at right angles to the plane Z Y and O K ;> is a 

 right angle. Hence 0/)''' = OP- + P;/- + 2 P ;?. PK or 

 Y-=zv~ + If- + 2itv cos e cos a. 



This gives us the actual velocity of the missile. We 

 know its actual inclination to the earth's surface, viz. the 

 angle ;; I, from the relation 



sin pO I =. 



Op 



V sin e 



V sm £ 



■/v- + V- + 



V. I- COS I COS a 



Knowing V we can determine the major axis of the orbit 



on which the missile begins to travel, as follows : — 



Put M= moon's mean velocity in her orbit, in feet per 



second. 



'"= number of seconds in a sidereal lunar montL 



D=the moon's mean distance | from earth's centre, 



d =missile's mean distance j in feet 



Then by a well-known corollary from Kepler's third law, 



we have 



rf : D :: V-" : M« 



2rD 

 and M = 



m 



Thus 



DV- (v'-+u-+'2uv cos iCoaa)m- 



d=^—-= 



M-' "" 4--D 



Hence we have all the details necessary for the precise 

 determination of the path of our projectile, neglecting the 

 effects of atmospheric resistance. 



Fig. 2. 



Thus, suppose we have in Fig. 2 a section through the 

 earth's centre and the line ^ of Fig. 1. Let S be the 

 earth's centre, 0;) the true course of the missile (0^ of 

 Fig. 1 ). Join O S, produce p O to p', and draw O K per- 

 pendicular to pp'. Then as S is one focus of the elliptical 

 orbit A A' of the missile, the other focus lies in direction 

 O H, so drawn that angle p O H= angle SO;/. But angle 

 SOp' is equal to the angle K M since each is the com- 

 plement of S K. Hence 



angle p H= angle K O M=comp. of angle ]> O M. 

 And since S O-fO H=A A'=:2rf 



0H=2rf— S 0=2 rf— earth's radiu8=2c?— r 

 _{v- + u~-\-2uv cos E cos a)in-_ , 

 2^2 D '' 



This assigns the position of H ; and H S being thus 

 given, while A A' is known, we have the position of A, and 

 the whole of the path A a. 



