a stone of mass 10gms falls from a height of 5 feet under gravitational force...the same stone is made to fall from 500 feet...according to the formula of force..F=mg....the force remains constant in both cases...yet we know the latter case will have more force...will anyone tell me the flaw??
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The force of gravity is derived from the gravity field strength, g = GM/r^2 acting on a target mass m to yield W = mg; where G is a constant, M is the source of gravity mass, and r is the distance from the center of mass to the field location.
If g ~ 1/r^2 and g' ~ 1/R^2; where r = 6400E3 + h and h is the low height (e.g., 2 m) and R = 6400E3 + H and H = 200 m is the high drop height, solve for g' = g ((6400E3 + 2)/(6400E3 + 200))^2 = 0.999938128 g.
And there you are, the field strength at your higher height H is 99.994 % the field strength at the lower height h. Which means, from W = mg and W' = mg', the two forces of gravity are almost identical at the two heights. So, for simplicity, we treat g = g' as a constant value near Earth's surface. QED
If g ~ 1/r^2 and g' ~ 1/R^2; where r = 6400E3 + h and h is the low height (e.g., 2 m) and R = 6400E3 + H and H = 200 m is the high drop height, solve for g' = g ((6400E3 + 2)/(6400E3 + 200))^2 = 0.999938128 g.
And there you are, the field strength at your higher height H is 99.994 % the field strength at the lower height h. Which means, from W = mg and W' = mg', the two forces of gravity are almost identical at the two heights. So, for simplicity, we treat g = g' as a constant value near Earth's surface. QED
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Allowing for the slight difference in gravity between 5 ft and 500 ft.
F=MA is dead on correct. If you want to intrgrate acceleration as a function of distance it is still dead on correct as to the Force acting on the rock as it falls
When it strikes the ground Impulse/Momentum enters the calculation but
F=MA is still dead on. The acceleration to bring the rock to a stop will give the exact same answer as the Impulse/momentum calculations.
Guess I don't understand your question in the macro world between quantum mechanics and very high gravity fields of General Relativity neither of which apply to a rock falling 5 to 500 ft
F=MA is dead on correct. If you want to intrgrate acceleration as a function of distance it is still dead on correct as to the Force acting on the rock as it falls
When it strikes the ground Impulse/Momentum enters the calculation but
F=MA is still dead on. The acceleration to bring the rock to a stop will give the exact same answer as the Impulse/momentum calculations.
Guess I don't understand your question in the macro world between quantum mechanics and very high gravity fields of General Relativity neither of which apply to a rock falling 5 to 500 ft
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The F in F=mg is the force of gravity on the stone, otherwise known as its weight.
It's true that the stone will weigh less at 500 ft than at 5 ft. That's because at 500 ft the stone is further from the earth's center, so the value of g is less. But not much less! In fact, you would not notice the difference at all.
It's true that the stone will weigh less at 500 ft than at 5 ft. That's because at 500 ft the stone is further from the earth's center, so the value of g is less. But not much less! In fact, you would not notice the difference at all.
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