Re: Moon effecting external ballistics
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: GuinnessNM</div><div class="ubbcode-body"><div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Scuzy</div><div class="ubbcode-body">Given the following:
Gravitational force
Fg=(G)((M)(m)/(R^2))
Gravitational constant G=6.67*10^-11 M^3/Kg*s^2
Mass of the moon M=7.63*10^11 Kg
Mass of 175gr bullet m=.011 Kg
Distance to the moon in its elliptical orbit R= 363,104 km to 405,696 km
The force exerted on the round by the moon depending on its location is:
.338N and .377N
So the difference of force on the round is .039N
Which translates to approximately 3% of the bullets weight.
Big deal huh?
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Mass of the moon is 7.63*10^22 kg. Using the formula I get about 4x10^-8N force on the bullet. Much less than 3% of the weight. I think you forgot to convert km to meters, which after squaring is a factor of 1 million.
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I thought a bit about how the world would look if the gravity force of the moon on a bullet really were 3% of the earth's. The effect on bullet drop at 1000 yards would be about 1 foot. The drop is given by d = .5*a*t^2, where a is the net gravitational force and t is the time to impact. d is is about 400 inches for 308Win. If the moon were straight overhead it would reduce a by 3% and thus reduce d by 3%, or about 12 inches (the time to impact is little affected). If the moon is below the horizon the drop increases 12". So over the course of a day as the earth revolves the 1000 yd point of impact would move up or down 12" from the point of aim. So it would be a big deal. Our Kestrel units would have to include a calculation of the moon's position. We would have to make range charts for different times of the day.
Luckily the effect is only 1 millionth of a foot, too small for most shooters to notice.
