MHB Gravity on Earth: Does Height Change It?

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Gravity on Earth decreases slightly with height, meaning a person at a mountain's peak experiences less gravity than someone on the ground. For example, at 9 km high, gravity is about 9.78 m/s² compared to the standard 9.81 m/s² at sea level. Additionally, gravity varies slightly due to the Earth's shape, causing people at the poles to weigh more than those at the equator. Factors such as the Earth's rotation and mass distribution also influence gravitational effects. Overall, while gravity does weaken with altitude, the difference is minimal until reaching significant heights.
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Does gravity on Earth stays constant regardless of height?. For instance, if a person standing at a mountain would he experiences the same gravity as the person standing on the ground?
 
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yakin said:
Does gravity on Earth stays constant regardless of height?. For instance, if a person standing at a mountain would he experiences the same gravity as the person standing on the ground?

Not quite, but almost. For instance, astronauts on the ISS still feel gravity (otherwise they would be flung into outer space) but at only about $8.76 \mathrm{m}/\mathrm{s}^2$, compared to the $9.81 \mathrm{m}/\mathrm{s}^2$ surface gravity. The reason they are weightless is because the rest of the station is also moving with them, so they have (near) zero acceleration in the station's reference frame, and the reason they don't crash into the Earth is because they are in orbit.

The formula to calculate gravity at an arbitrary height $h > 0$ can be obtained from the Earth's surface gravity ($g_0 = 9.81 \mathrm{m}/\mathrm{s}^2$) and its radius $r = 6371 \mathrm{km}$, and is as follows:
$$g_h = g_0 \left ( \frac{r}{r + h} \right )^2$$
This can be obtained from Newton's law of gravitation, basically since gravity is inversely proportional to distance squared, you can take the squared ratio of the distances to get a ratio of gravity at different heights. This is assuming the Earth is a perfect sphere, which is not quite true, but anyway, plugging in some numbers, say $h = 9 \mathrm{km}$ (Mount Everest):
$$g_{9 \mathrm{km}} = 9.78$$
So, even standing on the highest mountain, the Earth's gravity is still very strong.

However, because the Earth is actually not a perfect sphere, but is elongated at the equator, people living at the poles weigh more than people on the equator (by a small but measurable amount). As a less extreme example, take your average person who weighs $60 \mathrm{kg}$ and lives in, say, southern Australia, where gravity is approximately $9.8 \mathrm{m}/\mathrm{s}^2$. If he were to travel to central Africa (the equator), he would feel slightly lighter, as if he only weighed $59.88 \mathrm{kg}$. Of course his mass didn't change, but gravity has, and so his scale (unless calibrated to the new gravity somehow) will show that he weighs a couple hundred grams less :D

Of course this isn't quite true because there's more than gravity at work here, there are also centrifugal effects from the Earth's rotation, and the fact that gravity does not depend strictly on distance from the Earth's center (it also depends on the mass distribution of the Earth, which is actually quite complicated), etc... but to answer your question, yes, gravity does get weaker as you go further away from the Earth, but it isn't noticeable unless you go a few hundred kilometres high.
 
yakin said:
Does gravity on Earth stays constant regardless of height?. For instance, if a person standing at a mountain would he experiences the same gravity as the person standing on the ground?

I presume that the question is really more intending to ask if a person suspended from a balloon at a given height above a smoothly round Earth would experience the same or less gravity as a person standing on the surface of said earth. The answer then is that the person suspended above the surface would experience less gravity.

OTOH, a person standing on a mountain at the given height would experience a greater gravity than when suspended from a balloon but probably less than when standing on the surface.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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