Gravitational field strength \(g\) describes how much gravitational force acts on each kilogram of mass placed at a point.
It is measured in \(\mathrm{N/kg}\), which is numerically equivalent to \(\mathrm{m/s^2}\).
1. Field strength outside a spherical body
Outside a spherical planet, moon, or star, the gravitational field is
\[
\boxed{
g=\frac{GM}{r^2}
}.
\]
Here \(M\) is the mass of the body and \(r\) is the distance from its center. At the surface, \(r=R\), so
\[
\boxed{
g_0=\frac{GM}{R^2}
}.
\]
2. Gravity at height above the surface
At height \(h\) above the surface,
\[
r=R+h.
\]
Therefore,
\[
\boxed{
g(h)=\frac{GM}{(R+h)^2}
}.
\]
Compared with surface gravity,
\[
\frac{g(h)}{g_0}
=
\frac{R^2}{(R+h)^2}
=
\left(\frac{R}{R+h}\right)^2.
\]
3. Gravity at depth below the surface
A simple ideal model assumes the planet has uniform density. In that case, only the mass inside radius \(r\) contributes to the gravitational field at that point.
\[
r=R-d.
\]
The uniform-density result is
\[
\boxed{
g(d)=g_0\left(1-\frac{d}{R}\right)
}.
\]
This gives \(g=0\) at the center in the ideal model. Real planets are not uniform-density, so real values inside Earth are more complicated.
4. Effective gravity and latitude
On a rotating planet, the apparent or effective gravity is slightly reduced by centrifugal acceleration.
A simple spherical approximation is
\[
\boxed{
g_{\mathrm{eff}}
\approx
g-\omega^2 r\cos^2\phi
}.
\]
Here \(\phi\) is latitude and
\[
\omega=\frac{2\pi}{T}.
\]
The centrifugal reduction is largest at the equator, where \(\phi=0^\circ\), and smallest at the poles, where \(\phi=90^\circ\).
5. Force on a test mass
Field strength is force per unit mass:
\[
g=\frac{F}{m}.
\]
Therefore, the weight of a mass \(m\) is
\[
\boxed{
W=mg
}.
\]
If using effective gravity,
\[
\boxed{
W_{\mathrm{eff}}=mg_{\mathrm{eff}}
}.
\]
6. Worked example: 500 km above Earth
Use
\[
M_{\oplus}=5.9722\times10^{24}\ \mathrm{kg},
\qquad
R_{\oplus}=6.371\times10^6\ \mathrm{m},
\qquad
h=5.00\times10^5\ \mathrm{m}.
\]
The center distance is
\[
r=R+h
=
6.371\times10^6+5.00\times10^5
=
6.871\times10^6\ \mathrm{m}.
\]
Then
\[
g
=
\frac{GM}{r^2}.
\]
\[
g
=
\frac{(6.67430\times10^{-11})(5.9722\times10^{24})}
{(6.871\times10^6)^2}.
\]
\[
\boxed{
g\approx8.44\ \mathrm{m/s^2}
}.
\]
This is less than the surface value because the object is farther from Earth’s center.
7. Common surface gravity values
| Body |
Approximate surface \(g\) |
Comment |
| Earth |
\(\approx9.82\ \mathrm{m/s^2}\) |
Standard everyday reference |
| Moon |
\(\approx1.62\ \mathrm{m/s^2}\) |
About one-sixth of Earth’s surface gravity |
| Mars |
\(\approx3.73\ \mathrm{m/s^2}\) |
About 38% of Earth’s surface gravity |
| Jupiter |
\(\approx25.9\ \mathrm{m/s^2}\) |
Very strong surface-level field estimate |
| Sun |
\(\approx274\ \mathrm{m/s^2}\) |
Very strong due to enormous mass |
8. Important assumptions
- The formula \(g=GM/r^2\) assumes a spherical body or a point mass.
- Height is measured from the surface, but \(r\) is measured from the center.
- Depth mode uses a simplified uniform-density model.
- Latitude mode uses a simplified rotation correction and does not include oblateness.
Key idea: gravitational field strength decreases with height, changes with depth, and is slightly modified by planetary rotation.
