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and the orbit is convex or concave according as R

or Rp – 7P2 P2

pl is positive or negative. But R = 93,000,000 miles

237,000 miles P - 365 days

p = 29 days is round numbers

.. rP2 = 79213000000

Rp? = 31574325000 consequently the orbit is concave at conjunction.

468. Generally, let it be required to find the change of force or quantity of matter, in order that the eccentricity of the new elliptical orbit may be the of the radius of the circular orbit.

n

Letr be the radius of the circular orbit.

Then since at the point where the change in the force takes place, the velocity and direction of motion are the same as in the circular orbit, and the elliptical orbit touches the circular in that point, we have

V2

1 PV PV

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ra

F =

2ra
.. F =

F
202
But bo = a - (eccent.)

n+1
and a =rt eccentricity = r. by the question,

n

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(a)

.

and substituting, we get

n+1

F.

n + 2
In the problem n = 2

3
:. F -:

F

4

or the quantity of matter in the Earth must be diminished by

Hence, also,

If the quantity of matter be suddenly increased or diminished m.fold, required the eccentricity of the new orbit, and the change in the Periodic Time.

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Consequently if T be the Periodic Time in the circle, and T in the new orbit, we have, since TO

(axis maj.)

Nu
T:T::

Nu (2m – 1) (mu

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m2

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and if a be the whole distance to the centre, we have

a– ga

a's de

gde : dt =

Nu

(ui - 8)

a

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a

and t =

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Then

vla* – p)

NE Now put a

p= vers. 0 the space fallen through. sin. 8 = N (2a. amp

ap – )

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a

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a

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sin. 0

vu .. tiu::

:

tan. A su

a :: ai sin. 8 : x tan. 0. which shews the enunciation to be inaccurate.

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470. The mean angular velocity being V, and those round the foci S and H being denoted by S and H, we have easily

1
V : S ::

AC. BC SP2
and S: H :: HP X SP : AC X BC

::: V : H:: HP X SP3 : AC2 X BC 2 But when the eccentricity is very small, HP = SP = AC = BC nearly, :: V = H nearly.

Q. E. D.

This is named Seth Ward's Hypothesis. By means of it he approximated to the Solution of Kepler's Problem.

471. Let Q, 9; D, d, be the quantities of matter and densities of the Earth and Moon, respectively. Then supposing

them to be perfect spheres whose magnitudes or volumes are S, s, we have

Q= SD and
But if R and r be the radii of these globes, we also have
S =

FR3, and s =

9 = sd

4

4

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3

3

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which are therefore known.

Hence, if a be the distance from the centres of the Earth and Moon, we have that between their surfaces, viz.,

a - Rtr. Now it is evident that the body must be projected with such a velocity as shall just convey it to the point of equal attraction of the Earth and Moon, in order to place it in equilibrium between them, and therefore that the least additional velocity will cause it to proceed towards the Earth. But if x be the distance of the body at any point of its progress from the centre of the Moon, then

x is its distance from the centre of the Earth, and in order to the point of equal attraction, we have 9

Q

(a - x) which gives

a

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VQ + 7 9 Hence, the velocity requisite to carry the body from the moon's surface to this point of equal attraction, is the same as would be acquired through

axa

NQ+NI by falling from rest by the force

(a - )

?

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ava

and .. a-X= But v = 0, when x =

vQ+wa

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a

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a

2Q :.0=

29 2 (Q

(Q – 9) Let x =r=

( )

39

4πd
Then
2Q

29

*(Q – 9) which gives the velocity required. Hence is deducible a refutation of that theory of meteoric stones which supposes them to be projected from volcanoes in the moon.

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472. Let 2R be the diameter of the primary planet, g the force of gravity upon its surface, and p and a the periodic time and distance of its secondary. Then, since generally (Fo

F have

pl (by 453)

we

Period = 2aš

Vu where x = M + m, M, m being the masses of the primary and secondary respectively.

But since the attraction at the surface of a sphere whose radius is R is (see Vince's Fluxions, 144.)

4R

X: density.

3

supposing density to be uniform ;

4R
.:9 = x density

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