70 
nearer apsis, where r=a( 1— e) (a being half the axis major of 
the planet's orbit, and e the eccentricity of the same), we 
2 2 
have v 2 = — + C,and v' 2 — — a — -+ C, whence, subtracting, we 
r a(L—e) 
have 
v 
( 2 ) 
2 __ _ —“I 1 , 
a( 1 — e) r ' 
Since a(l — e) is the least value of r, it is evident that v 2 is 
greater than v 2 , except when the planet is at the nearer apsis, 
and then they are equal. At any other place v' 2 —v 2 is the 
kinetic energy lost since the planet was at the nearer apsis, 
and which, as it will be regained on its return thither, is the 
potential energy. Now, if v 2 be put to the right-hand side of 
equation (2) (its sign being of course changed), we learn 
o 9 
that v 2 -\- — — — — — - =v ' 2 , that is to say, that the sum of the 
a{ 1 — e) r 
kinetic and potential energies is constant, and equal to the 
maximum kinetic energy. The maximum potential energy is 
at the point wnere r is greatest, because , the quantity to be 
subtracted from the constant 
2m 
is then least. It is 
a(l — e)’ 
therefore at the point where r=a{ 1+e), i.e., at the remote 
apsis. After this point has been passed, the potential energy 
diminishes, and at any point in the return half of the orbit 
both kinds of energy are of the same amount as they were 
when the planet was equally distant from the sun in the former 
half. 
In the case of the pendulum vibrating through small arcs, 
equation (1) takes the form v 2 =—2 g sds + C, where s de- 
notes the variable distance of the pendulum at any point 
during its oscillation from the lowest point (that distance 
being measured on the arc which it describes, and g being the 
constant force of gravity). Performing the integration, we 
have v 2 =— gs 2 +C. If we denote by / the distance of the 
point where the motion ceases, / = 0,andvve have 0= — p/ 3 + C, 
whence C = gs' 2 . Substituting this in the equation v 2 = — gs 2 + C, 
and subtracting, we get the equation 
v 2 =gs' 2 ~gs 2 . ( 3 ) 
This is the actual kinetic energy at the distance s. It 
vanishes at the greatest distance /, since there gs' 2 —gs 2 = 0, 
and it increases as s decreases until s = 0 (i.e., until it reaches 
the lowest point), when it is greatest, being equal to gs' 2 . If 
now we remove gs 2 to the left-hand side of the equation, we 
have v 2 -\-gs <2 =gs' 2 ; and as gs 2 is the difference between the 
