Since the radius of Earth is far greater(rough: 6400km) than a few 100m, you can neglect that distance and still have g=9.80 (g=GM/(R^2), with R being the radius of earth, a few 100m dont do much difference in g's equation)
very simple. gravity is acting upon these objects the very same. no extra or no less gravity is being exerted upon the objects. the only reason they fall differently on earth is because of air resistance. the air pushes them backwards.
In fact, it is not exactly the same because in a vacuum, there is no air to create friction that decreases acceleration so the rate of acceleration tends to be greater than in a non-vaccum environment.
The free fall acceleration of the earth is approximately related to the distance from the surface by the following formula.
g = GM/(R2 + h)
where G = universal gravitational constant, M = Mass of Earth, R = Radius of Earth, h = distance from the surface of the earth
g = 3.98 * 1014 / ( (6.0 * 106 )2+ h)
as h increases to a very low value near the surface of the earth the value of g can be approximately taken to be constant near the surface of the earth.
Air resistance.
because u are entering the atmosphere
In general, when near the Earth's surface, scientists use 9.8 meters per seond^2 since the slight difference in elevation along the Earth's surface doesn't affect the acceleration very much. But if you want the exact formula it is: (G(msub1)(msub2))/radius^2. msub1 and msub2 are the masses of the two objects, and G is 6.67x10^(-11), the gravitational constant.
silicon & oxygen.
i believe roughly 70% (not completely sure though) I'm not really sure, but something like 70% of the earths surface..
Uplift
surface
Yes, exactly. Free fall results in constant acceleration.
Constant acceleration
Force (newtons) = mass (kg) * acceleration (m/s/s) > Acceleration at earths surface radius = 9.82 m/s/s
the water cycle
Yes, relative to any observer not attached to the earth's surface.
The acceleration of gravity at its surface is currently estimated as 0.4 m/s2 .That's about 4% of the acceleration of gravity on the Earth's surface.
water cycle
The force of gravity pulls it down to the earth.
Strictly speaking its not the same . This equation calculates the acceleration: acceleration = ( G * ( m1 + m2 ) ) / d2 where: G = newtons gravity constant m1 = earths mass (kg) m2 = objects mass (kg) d = distance between centres of gravity (metres) The earths mass is so large however, only a significantly large object mass would make a real difference to the acceleration.
Because the acceleration of gravity on the surface of any given body depends on the mass of the body and its radius ... the distance of the surface from the center. Mars' mass ... about 11% of Earth's ... and Mars' radius ... about 53% of Earth's ... combine to produce about 38% of Earth's gravitational acceleration at the surface of Mars.
No, your weight is just the acceleration due to the Earth's gravity,
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