Back to Contents

    For convenience I define

(Eq'n 1)

so our integral becomes

(Eq'n 2)

We have solved the integral of the inverse square root of a quadratic formula before and gotten a natural logarithm as the solution, so we may expect the solution of this integral also to give us a natural logarithm of some function. Based on that expectation, we want to transform our integrand to conform to

(Eq'n 3)

in which k represents a constant. Thus we expect to find that

(Eq'n 4)

We know that

(Eq'n 5)

From that fact we can infer that

(Eq'n 6)

in which p, q, and r represent constants. Differentiating that equation gives us

(Eq'n 7)

If we combine that result with Equation 6 by way of Equation 4, we get

(Eq'n 8)

Gathering like terms gives us

(Eq'n 9)


(Eq'n 10)

We extract p=kq from Equation 9 and substitute it into Equation 10, which then becomes

(Eq'n 11)

Equating coefficients of x tells us that k=1/. The second term on the right side of Equation 11 consists entirely of constants, so we can choose an arbitrary solution to r=k2qb/2 and then adjust the proportions in our final solution if needed; thus, I choose r=b, which gives us q=2a. We then calculate that p=2. Thus Equation 6 becomes

(Eq'n 12)

and, consequently, we have

(Eq'n 13)

So we have at last

(Eq'n 14)

Example Application:

Angle Traversed on an Orbit

    Kepler's second law of planetary motion tells us that the radius vector connecting an object to the primary focus of the object's orbit sweeps out equal areas in equal intervals of time. Based on the law of conservation of angular momentum, Kepler's second law specifies that in a minuscule interval of time dt the orbiting object's radius vector traces out a long, narrow triangle in accordance with

(Eq'n 15)

in which L represents the orbiting body's angular momentum per unit mass and dθ represents the vertex angle of the triangle.

    Now I want to translate dt into some function involving dr, the incremental change in the radius of the orbit. I begin with the basic description of the square of the orbiting body's velocity, via the Pythagorean theorem, as

(Eq'n 16)

which I solve for dt to obtain

(Eq'n 17)

But we also know from basic orbital dynamics the law of conservation of energy in the form

(Eq'n 18)

in which v0 represents the speed with which the body passes through its orbit's peritelion, a represents the radial distance of the peritelion point from the orbit's prime focus, and E represents the body's total orbital energy per unit mass. With that equation I transform Equation 17 into

(Eq'n 19)

I then substitute that result into Equation 15 and solve for dθ to get

(Eq'n 20)

which I then integrate. So now we know that

(Eq'n 21)

Finally, for the sake of the act of calculation, I want to put that equation into an alternate form, one that eliminates the square root of minus one and leaves me with real numbers only. I rewrite Equation 21 as

(Eq'n 22)

In going from the first line to the second line in that equation I added 0=M2G2-M2G2 under the radical in both the numerator and the denominator in the argument of the natural logarithm. In going from the second line to the third line I divided both the numerator and the denominator of the logarithm's argument by , so we have in that third line

(Eq'n 23)


(Eq'n 24)

At last I apply the fact that lnA/B = lnA-lnB and the identity

(Eq'n 25)

and obtain

(Eq'n 26)


Back to Contents