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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 '15 at 10:06
up vote 17 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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The approximation of 9.81 m/s^2 is a generalisation. The exact value is most likely different at a specific location, due to the distance from the centre of the earth to the point being evaluated.

The reference to "surface of the earth" is also a relative since the earth is known not to be perfectly round due to centrifugal forces making the radius greater at the equator.

Also, since the earth is spinning the same centrifugal forces have a slight influence on object mass at the evaluation point.

In metrology laboratories, the exact value for g is displayed for that exact location.

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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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protected by Qmechanic Aug 19 '15 at 10:08

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