Why is that the gravitational force acting on a book (of mass 3 kg) which is raised 1m from the surface of the earth is the same as that of another of same but which is raised 10m from the surface of the earth? Isn't the force inversely proportional to the distance between the two objects? Or is 10m too small a difference to be noticeable in calculations?
2006-08-08 01:06:40 · 6 answers · asked by Stavi 2 in Science & Mathematics ➔ Physics
Force is inversely proportional to the square of the distance. Above is typo!!
2006-08-08 01:07:36 · update #1
It's not the same.
Gravitational force:
F = GMm/r^2
G is the universal gravitational constant, M is the mass of the Earth, m is the mass of the object. 'r' is the distance from the center of the Earth.
The equatorial radius of the Earth is 6,378,137 meters. If an object is 1 meter above the surface of the Earth, the object is 6,378,138 meters away from the center of the Earth. If an object is 10 m from the surface, the object is 6,378,147 meters from the center of the Earth.
The force on the nearer object is 0.000282% stronger. The difference is so small it would be very hard to measure the difference physically.
Technically, each unit of mass in the Earth attracts the two objects, but if you assume a uniform density (which isn't quite true) and assume a perfectly spherical shape (which also isn't quite true), the combined force from all of the Earth's mass is the same as if all of the mass existed in the center of the Earth. The imperfections in the model are greater than the difference in force caused by a difference of only 9 meters. (In fact the Earth's equator has a radius about 20,000 meters greater than the polar radius and even mean sea level around the equator varies by up to 65 meters.
For very precise measurements, both a more accurate gravitational model of the Earth would have to be used (EGM-96, for example) plus the difference between the two objects would have to be taken into account.
2006-08-08 03:09:57 · answer #1 · answered by Bob G 6 · 0⤊ 0⤋
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2016-11-23 15:40:07 · answer #2 · answered by halyard 4 · 0⤊ 0⤋
The weight is calculated using the formula F = mg.
The value of g is given by the formula
g = 9.80616 - 0.025928 cos 2A + 0.000069 cos^2 (2A) - 0.000003 h,
where A is latitude of the place and h ( in meter) is height above sea level.
From the formula you can see that as h increases g decreases.
Since the difference is very small usually we neglect them.
2006-08-08 03:04:19 · answer #3 · answered by Pearlsawme 7 · 0⤊ 0⤋
It isn't quite the same, but it's so incredibly close that 9.8 m/s^2 is a very good approximation. Try it if you don't believe me. Plug your numbers in to F=GMm/R^2 (remembering to use the distance from the center of one object to the center of the other).
2006-08-08 01:14:26 · answer #4 · answered by AK 1 · 0⤊ 0⤋
because when you do the math, the mass of the earth is a hell of alot larger than the book, the distance tot the center of mass of the earth far outweighs the measly 100m or so as well. so Gm1m2/r^2 essentially becomes constant * m2
2006-08-08 01:11:28 · answer #5 · answered by Auggie 3 · 0⤊ 0⤋
yes u r right
there is a difference but so small to percieve
2006-08-08 01:24:41 · answer #6 · answered by Anonymous · 0⤊ 0⤋