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    We can solve this integral in a straightforward manner. First we factor the reciprocal of x and the constant out of the radical and then invert the reciprocal to put the square root of x into the denominator of the integrand. We thus get

(Eq’n 1)

We change the form of the integrand without changing its value by multiplying it by one and adding zero to it. The multiplication by one consists of multiplying and dividing the integrand by the square root of x and the addition of zero consists of adding and subtracting the same term, so we have

(Eq’n 2)

Finally we factor one half out of the numerator, split the integral into two integrals, and write out the solution,

(Eq’n 3)

We have already solved the second integral on the right side of that equation (with 1/c=2a), so I simply made the direct substitution.

Example Application:

Timing a Free Falling Body

    We want to determine how long it takes a small body, starting with zero velocity relative to Earth, to fall from the moon’s orbit to the top of Earth’s atmosphere (200 kilometers above equatorial sea level). We assume that the moon lies far enough along its orbit from the drop point that its gravity will not significantly affect the fall of the body.

    Although we can’t see right away how to calculate the time the body takes to fall to a given point, we do know how to calculate the velocity with which the body passes through that point: we need only equate the kinetic energy that the body has gained with the gravitational potential energy that it has lost (and divide out the body’s mass because it appears in both formulae as a simple multiplier), so we have

(Eq’n A-1)

in which r0 represents the distance between the drop point and Earth’s center. Thus we have

(Eq’n A-2)

Multiplying that equation by the differential increment of elapsed time and dividing it by the square root (while factoring the constants out of the radical) gives us

(Eq’n A-3)

Recognizing that c=1/r0, we integrate that equation and get

(Eq’n A-4)

    For the actual calculation of the time of fall we have Earth’s equatorial radius (rE=6378 kilometers) and the acceleration of gravity at the surface (g=0.00981 kilometers per second squared), so we calculate MG=grE2=399,060 kilometers cubed per second squared, which gives us

(Eq’n A-5)

We also have r=6578 kilometers and three possibilities for the radial position of the drop point: on the moon’s orbit we have the apogee radius (r0=405,500 kilometers), the mean radius (r0=384,400 kilometers), and the perigee radius (r0=363,300 kilometers). For the time of fall we have;

ta = 226607 sec = 62.95 hrs = 2 days, 14.95 hrs

tm = 209120 sec = 58.09 hrs = 2 days, 10.09 hrs

tp = 192115 sec = 53.37 hrs = 2 days, 5.37 hrs.

Those times came out less than the three days that the Apollo ships took to return to Earth from the moon. But the Apollo ships were flying directly away from the moon, so the moon’s gravity hindered their movement, thereby accounting for the additional time.

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