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Do we take gravity = 9.8 m/s² for all heights when solving problems?

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I take it as 10 - it makes doing the approximation head-doable –  Dirk Bruere Jun 19 at 10:06

6 Answers 6

up vote 16 down vote accepted

No, the value $9.8\frac{\mathrm{m}}{\mathrm{s}^2}$ is an approximation that is only valid at or near the Earth's surface. You can go a few miles up or down and it'll still be good enough, but once you get any significant distance away from the surface of Earth, you would need to use a different value for gravitational acceleration. You can calculate the value from Newton's law of gravitation, $F = Gm_1m_2/r^2$, and you'll get

$$g = \frac{GM}{r^2} = \frac{3.99\times 10^{14}\ \mathrm{m^3/s^2}}{r^2}$$

where $M$ is the mass of the Earth and $r$ is the distance from the Earth's center to the point for which you are doing the calculation.

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It's strange that you would measure the height in miles when the value for gravity is in meters per second squared. –  LDC3 Nov 3 '14 at 2:53
Not really; miles are a common unit for height. And it's trivial to convert when needed. –  David Z Nov 3 '14 at 2:57
Maybe you could cite that $g = \frac{GM}{r^2} \approx \frac{4 \cdot 10^{14}}{r^2}$ (error: 0.3%). r in meters, g in m/s² –  André Neves Nov 3 '14 at 3:00
Maybe if you are a geologist, but for most scientist, we rarely use anything but SI units. –  LDC3 Nov 3 '14 at 3:03
@LDC3 some scientists rarely use anything but SI units, I'm sure, but many branches of science have their own conventional unit systems. In particle physics we use natural units ($c$ and $\hbar$ set to 1), in condensed matter they often use some lattice spacing as a length unit, in cosmology they use megaparsecs or the Hubble radius, and so on. The point is, a qualified scientist is capable of understanding the science regardless of what units are used. –  David Z Nov 3 '14 at 4:03

To expand a little on David's point assume we move from the nominal "surface" where $g$ is $9.8\text{ m}/\text{s}^2$ to another point at radius $r + \Delta r$. How much does the acceleration of gravity change? $$ g = \frac{GM}{(r+\Delta r)^2} = \frac{GM}{r^2(1 + \Delta r/r)^2} $$ and as long as $\Delta r$ is small compared to $r$ we can reasonably approximate this as $$ g \approx \frac{GM}{r^2}\left(1 - 2\frac{\Delta r}{r}\right) . $$ Well, the radius of the Earth is about $6000 \text{ km}$ so the approximation is good at less than 1% error for around $30\text{ km}$ up or down from the nominal surface, which is all the land and sea floor, and a bit up and down from there.

It is also worth noticing that due to variations in the local mass density of the Earth the measured value of $g$ even at the surface can vary by several tenth of a percent.

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In fact, the measured variations in $g$ are very useful to geophysicists, oil prospectors, etc. –  Ted Bunn Feb 28 '11 at 14:56
I was reading an article on the use of optical lattice clocks today which explained how such clocks allow an even more precise measurement of these changes - useful to evaluate height of water table, prospect for oil and gas, etc. –  Floris Nov 3 '14 at 4:18

$g$ becomes $ g \approx 9.7 \frac{m}{s^2}$ at a height of about 35km, so it would be ok to use the value $9.81$ for "down to earth" problems.

The relevant wikipedia article has lots of useful information, like for example the following approximation formula for different heights: $$ g_h=g_0\left(\frac{r_e}{r_e+h}\right)^2 $$ Where $g_h$, is the gravity measure at height $h$ above sea level; $r_e$, is the Earth's mean radius and $g_0$, is the standard gravity.

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that's not an approximation, it's exact (as long as you assume earth as a point mass...) –  Tobias Kienzler Feb 28 '11 at 8:50
@Tobias: It's an approximation in the sense that it treats earth 1) as a point or a perfect sphere 2) not rotating, etc... –  Eelvex Feb 28 '11 at 9:51

It might also be worth mentioning that $g$ isn't even constant over the earth's surface at sea level. Depending on the mass distribution and the shape (not perfectly spherical!) of the earth, different parts of the world have different $g$.

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Acceleration due to gravity, g is not a universal constant like G. Its calculated by formula mentioned in previous answers. So, for a constant mass system, g depends only on r (distance between center of earth & object in problem). As r = R + h (R is radius of earth & h is height of object from surface) & R is constant, g depends mainly on height.
The relation: Increase the height, g will become less (as per formula)

The value 9.8 m/s² is valid for the object at the surface of earth (at sea level). When height is small (with respect to radius of earth), the value is slightly less than 9.8 m/s². So, this variation can be neglected for a high school etc problems. When accuracy is important (due to scientific reasons etc), the value of g can't be 9.8 m/s².

Once again, This consideration is valid only for constant mass system. Plus, for relativistic systems, the formula isn't valid with constant space & time scale.

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As we go above or below the surface of earth the value of g decreases since g is inversely proportional to height

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