2o6

A

S T R 0 N 0 M r.

its Appearance is but a very small Part of the Athe Centre of such' an Ellipsis being removed t° trjjfl

Plate

LX 1 I.Fig. I.

iplis being removed t Jii®a Distance must occasion the Curvature at ea cto be vastly near that of a Parabola having st ie 0 { *focal Distance ; and consequently the M ot ' |0 !'qth sllltComet may be calculated in a Parabolic Orbitany sensible Error. , p e ri^ l>

11; Therefore the Velocity of a Comet in E(viz. in the Vertex of the Parabola P) is toVelocity of a Planet describing a Circle about t ^,,il

at the same focal Distance 8 P, as \/ z to ,csupposing the Earth to be that Planet, Jet us r ftRadius of its Orbit SP = 100000, and the 11 . tKthe whole Periodical Time of the Earth 365' ^whole Periphery 62^318, so is 1 Day to 1 7 . 2 ?^ e scf*^described in one Day; and in one Hour it wist ^71,67 Parts. But as 1 : %/ % :: 1720,2 1the Parts described by the Comet in one Day> ‘jjl “the Parts described by the Comet in one Hour101,364. . , he P<

12. Whence if the Lutui Refhirn L R of 1 g^tl 1bola be equal to 4 times the Radius S P ot tb eOrbit, and we put S P 2 = 100000000; the Ar e ^the Comet will describe each Day, by a EV 3to the Sun, will be 1216373s of those Parts,^ "jgjTsstHour an Area of 50602 -J. of those Parts. ^ st £ j,e V .strate this we must consider, that the Square® +ameter of any Circle is to its Area as 1 : ®’ 7 p3,14159 ; therefore the Square of Radius ° r stjWhence the Area of the Circle is to the

PM as 3,14159 to 1. And the RectangleBut the Parabolic Area P L S —tienc® this Area P L S is to ti

3

X

k

s,

T

Area

of rliL

c> sC 1

to

3,14159. And if the Velocity

of the

c°r

th?

3 . ~

and Planet at P were the fame, the Time in v^i^pPComet would describe the Arch of the i*

woinii bo to the Time in which the Planet dejer> .

Orbit in the same Ratio of

-4

to 3,14^59-

pu

Vc.

■1 <P>

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