1
$\begingroup$

I learnt that newton's law of universal gravitation F = G(m1m2)/R^2, and thought if the R is distance and determined gravitational strength, why do we use 9.81 as default acceleration of earth's gravity when it is not even constant at different heights?

$\endgroup$

3 Answers 3

2
$\begingroup$

Who is "we"? If we're students solving physics problem, $9.81 \,\rm{m/s^2}$ (at the surface) is close enough to the nominal average of $9.80665 \,\rm{m/s^2}$.

If we're a hydrologist, cartographer, or geodesy-interested person we may use a geoid, such as EGM96 (or EGM08, or the new one), e.g.:

https://cddis.nasa.gov/926/egm96/egm96.html

Here you get a potential surface represented tide-free mean sea level (MSL) for the Earth's surface represented as degree 2500+ spherical harmonic expansion. That gives the height of MSL relative to some ellipsoid (usually WGS84), not the strength of gravity.

The local value of $g$ can be computed from the gravitational anomaly (https://www.ngdc.noaa.gov/mgg/gravity/), which is the difference from the global mean value.

Gravity doesn't always point straight down, the deviation is called the vertical deflection, and is available publicly (e.g. https://beta.ngs.noaa.gov/GEOID/xDEFLEC18/index.shtml).

There are both global models and local models to the Earth's shape and/or gravity. For instance, some well known local ellipsoids are "Everest 1830 India", "Indonesian 1984", South American 1969".

If we're submariners or we're targeting ICBMs, we use classified data sets.

Finally, if we're Earth scientists using gravity in LEO to measure dynamic processes down below, we compute the changing gravity from the differential orbits of two linked spacecraft (https://www2.csr.utexas.edu/grace/gravity/).

$\endgroup$
0
$\begingroup$

We use 9.81 m/s$^2$ only when dealing with objects close enough to the surface of the earth (heights negligible compared to the radius of the Earth) that the acceleration due to gravity can be considered constant.

The value of $g$ near the surface of the Earth can be computed when you substitute the radius of the earth for $R$ and calculate the acceleration $g$ using the universal law of gravitation equation, where $M$ is the mass of the earth,

$$g=\frac{GM}{R^2}$$

Hope this helps.

$\endgroup$
0
$\begingroup$

The local gravitational acceleration at a height $h$ above the surface (of water) is given by $$g = \frac{G\,M_{\rm earth}}{(R_{\rm earth}+h)^2}$$

with $g_0 = 9.80665\;{\rm m/s^2} \approx 9.81$ when $h=0$ on average.

Your question is why use a constant value when there is a dependency on $h$?

Well, check how much does $g$ varies given reasonable values for $h$.

Heght, $h$ [m] Gravitational Acceleration, $g$ [m/s^2] Change
0 9.80665 0
100 9.80634 -0.003%
1000 9.80357 -0.03%
2000 9.80050 -0.06%
5000 9.79130 -0.16%
10000 9.77597 -0.32%

As you can see, most high school lever experiments take place under 10,000 meters of height, where 0.3% error due to gravity is perfectly acceptable.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.