# Do we take gravity = 9.8 m/s² for all heights when solving problems? Why or why not?

37,491

## Solution 1

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.

## Solution 2

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.

## Solution 3

$g$ 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.

## Solution 4

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 $g$.

## Solution 5

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.

Share:
37,491 Author by

### Qmechanic

Updated on August 01, 2022

• Qmechanic 5 months

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

• I take it as 10 - it makes doing the approximation head-doable
• Tobias Kienzler almost 12 years
that's not an approximation, it's exact (as long as you assume earth as a point mass...)
• Eelvex almost 12 years
@Tobias: It's an approximation in the sense that it treats earth 1) as a point or a perfect sphere 2) not rotating, etc...
• Ted Bunn almost 12 years
In fact, the measured variations in $g$ are very useful to geophysicists, oil prospectors, etc.
• It's strange that you would measure the height in miles when the value for gravity is in meters per second squared.
• Not really; miles are a common unit for height. And it's trivial to convert when needed.
• 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
• • @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.
• • 