SOLUTION, by Mr. P. BARLOW, the Proposer.

Let м, M, m, u represent the masses of any four planets,
M,

R, R, r, p their mean distances, and

D, D, d, & their computed densities.

Let also the required law be

RD : MP ::

R"D" : MP

ρο

making now, for the sake of abbreviation, each of these quanti ties to represent their respective logarithms, we have

Which must necessarily obtain if any such law as that we have supposed have place between those quantities; and this equality, it is also obvious, is totally independent of the law itself; and since we are at liberty to assume the mass, distance, and density of any particular planet equal to unity, the logarithm of which is o, we may simplify the above by making м, R and D each equal to unity, or their logarithms o; for then the above becomes simply

{ru —pm} × {ḍM— Dm}·

Observing only, that these symbols are the representatives of the logarithms of the quantities they were first taken to denote; that is

In order now to submit this equality to a proper trial, let us take four planets of which those quantities are supposed to be the best known, which are Venus, the Earth, Jupiter, and Saturn, assuming also the mass, distance, and density of the earth as unity. Then adopting La Place's results, we shall have

Distances Jupiter
Saturn

5202792; log.

71624 = r.

Densities

9540724; log. = +97958 = g.

Venus 1024; log. 0103
= +

Jupiter
Saturn

= D.

0 20093; 0 10349;

log. =

•69696 = d.

log. =

*98510 = d.

Now applying these numbers to the preceding formula, it will be found, that the required equality does not obtain ; whence we may conclude with certainty, that no such law as that we have supposed, has place with regard to the powers, or roots, of the masses, distances, and densities of the planetary bodies, at least if their present computed masses and densities be accurately determined.

XVII. QUESTION 347, by Mr. S. H. CHRISTIE, Royal Military Academy.

when n is an even number; and that when n is an odd number,

n (n2 — 1 ) (n2 — 9) (sin3 1 a + n sina + n' sina

2.3.4.5

n

1

13

+ &c.) + &c.

a = sin a + 1 (sin3 = a + n sin3
a+n sin3 11⁄2 a + n2 sin3

&c.

&c.

1

3

a + &c.)

n

when n is an odd number: and when n is an even number

a

3 1

2

sin a + A(P sin3 -a + Pn sin3

1

3

1

— A(P sina + Pnsin a

which are equivalent to the forms given in the question. When n is an odd number,

sin n = n sin x A sin3 x + A sin3 x —

If now m be indefinitely increased, the limit of sina is

a = sin a + a (sin31a + n sin3a+&c.)

A

n

3

n