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L' = 1, or L = 1

V for the limit of whether the eccentricity become Indefinitely

great or Indefinitely small.

501.

By (495,) the time down any space r — ç is

(

vers.-1 29

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The periodic time in a circle whose radius is r and Fc

is

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OTHERWISE
By prop. XXXVI, and iv. of the Principia,

SA
T: + (time in O rad. = :: 1 : 2

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a and b being the semiaxes, p and the I upon the tangent and
radius vector from the focus.
2pdp

be
dp
2a

2a)
2abop*
64%

P Р

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.: F - cdp

2a

p* 62

ge

2ac? 1
pådę
1

and is repulsive.

pa 504. If the distance of the centre of the O from the centre of co-ordinates be a, and the radius of the O be r, then we easily know that #p=

que – a2 + g.

2r Hence

dp

dp
and F=
cdp

Scopo X pide (r2 — a° +8°)3 according as the centre of force falls within or without the tangent.

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a being the distance at which the body begins to fall. If a = o,

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Now if the force be constant for the space g, we have

Ú= N2F2 =

P

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506. Since the force is in the focus we have

F=
THE

p2 and the velocity down any portion 3 of the radius vector p, beginning from the curve is

zu x

see 505,

px

But the velocity in the curve is

с

р

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Х

= 2

= 24 x

ple-x) Hence co

$?

c?
p? (1 + *).

2p po

be But p2 = and c = b... (453) :. after substitution and the requisite reductions we get e)

(a)

2a

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p(2a

CENTRAL FORCES.

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Hence when x - maximum or minimum, we have
dr 2a
P

+ Ç. (2a-ş) = 0
de 4a

8 (

4-8) or 2 (a - ).(42 – 8) + p. (2a – g) = 0 whence we get

pl – sag = 8a% and solving the equation

$ = 4a + a2 2 = 2a (2+ 2) which being substituted in (a) will give either maximum or minimum values of x. To ascertain whether they are maximum or minimum, we have

* – Sag + 8a de (4a —s)

8ae + 8a?)
del (4a – )

(4a - p)
16a?
(4a – 8)

dir

d2 x

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+ 262

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and ..

p= 2a (2+2) gives the maximum value of x

and s = 2a . (2-2) gives the minimum, (see Simpson's Fluxions, Appendix to the New Edition, or Lacroix.)

These values of x are
po(2a – s)
2a.(2+2) X

2a-2a.(2+2) 4a

4a-2a.(2+x2) = 2(1 IV 2) x (1 +N 2) a= 2(1 V2)ea,

507. Let F = Mig", m and y being at present unknown but determinate. Then

Đdo = - Fde = р and vp =

2.

x (amts - pati )

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v (et de portes ) m Surat het

का)

24

xv (amti - p+1)
m + 1
de

m

24
m + 1

dz
X
24

and t =

But

(20+1$+2)* = a* 1-B), smat?

being put = B,

amti

1 3 1.3.5 1. + B + Be + &c.}

2 4 2.4.6 2u

3 gan+de +

t&c} m + 1

anti

2.4

2n +
mt-2
e

3
=+
+

+ &c. + C. 2.(m+2)a ti

2 4 (2m+3)a mt? Let t = 0, when g = a and we have C

1
a (1 +

+ &c.)
2. (m + 2)
24
Xt=gt

&c m + 1

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pats 2.(m+2)amti + &c.

1 - a(1+

+ &c.) .......(a)

2.(m+2) Let now p = 0. Then the time to the centre is

1 T={1+

m + 1

taa 2(m + 2)

24 and consequently for different altitudes (a),

1-m

+ &c.}. mm

Τ α α 9
But by the question

Τα α"

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