Free-fall acceleration is the acceleration an object would have if gravity were the only force acting on it.
Near Earth’s surface it is often approximated as \(9.8\ \mathrm{m/s^2}\), but it actually changes with distance from the planet’s center.
1. Newton’s gravitational field formula
For a spherical planet or moon of mass \(M\), the gravitational acceleration at distance \(r\) from its center is
\[
\boxed{
g=\frac{GM}{r^2}
}.
\]
Here \(G\) is the universal gravitational constant:
\[
\boxed{
G=6.67430\times10^{-11}\ \mathrm{N\,m^2/kg^2}
}.
\]
2. Acceleration at altitude
If the planet radius is \(R\) and the object is at altitude \(h\) above the surface, the center distance is
\[
r=R+h.
\]
Therefore, the free-fall acceleration at altitude is
\[
\boxed{
g(h)=\frac{GM}{(R+h)^2}
}.
\]
This is why gravity decreases with altitude. The farther the object is from the center, the weaker the gravitational field becomes.
3. Surface acceleration and ratio
At the surface,
\[
\boxed{
g_0=\frac{GM}{R^2}
}.
\]
The fraction of surface gravity remaining at altitude is
\[
\boxed{
\frac{g(h)}{g_0}
=
\left(\frac{R}{R+h}\right)^2
}.
\]
This ratio is useful because it shows that gravity remains significant even hundreds of kilometers above Earth.
4. Worked example: 400 km above Earth
Use approximate Earth data:
\[
M_{\oplus}=5.9722\times10^{24}\ \mathrm{kg},
\qquad
R_{\oplus}=6.371\times10^6\ \mathrm{m},
\qquad
h=4.00\times10^5\ \mathrm{m}.
\]
First find the center distance:
\[
r=R+h
=
6.371\times10^6+4.00\times10^5
=
6.771\times10^6\ \mathrm{m}.
\]
Then calculate:
\[
g(h)
=
\frac{GM}{r^2}.
\]
\[
g(h)
=
\frac{(6.67430\times10^{-11})(5.9722\times10^{24})}
{(6.771\times10^6)^2}.
\]
\[
\boxed{
g(h)\approx 8.69\ \mathrm{m/s^2}
}.
\]
This is close to the sample result \(8.68\ \mathrm{m/s^2}\). Small differences come from rounding the Earth mass, Earth radius, and \(G\).
5. Why astronauts feel weightless in orbit
At the altitude of the International Space Station, gravity is still about \(88\%\) to \(90\%\) of surface gravity.
Astronauts feel weightless because they and the spacecraft are continuously falling together around Earth, not because gravity is absent.
6. Air resistance option
In real atmospheres, an object moving through air experiences drag. A common quadratic drag model is
\[
\boxed{
D=\frac12\rho C_dAv^2
}.
\]
Here:
- \(\rho\) is air density,
- \(C_d\) is the drag coefficient,
- \(A\) is cross-sectional area,
- \(v\) is speed relative to the air.
If the object is falling downward, drag acts upward. The instantaneous downward acceleration is approximately
\[
\boxed{
a_{\mathrm{net}}=g-\frac{D}{m}
}.
\]
This is not a complete trajectory simulation, because velocity and density change during the fall. It is an instantaneous estimate at the entered altitude and speed.
7. Exponential atmosphere model
A simple atmosphere model is
\[
\boxed{
\rho(h)=\rho_0e^{-h/H}
}.
\]
\(H\) is the atmospheric scale height. For Earth, a rough value is about \(8.5\ \mathrm{km}\).
This model is useful for educational estimates but becomes less reliable at very high altitudes.
8. Terminal speed estimate
If drag balances weight, then
\[
mg=\frac12\rho C_dAv_t^2.
\]
Solving for terminal speed gives
\[
\boxed{
v_t=\sqrt{\frac{2mg}{\rho C_dA}}
}.
\]
At high altitudes, \(\rho\) can be very small, so the terminal speed estimate becomes very large.
9. Summary table
| Quantity |
Formula |
Meaning |
| Center distance |
\(r=R+h\) |
Distance from planet center to object |
| Surface gravity |
\(g_0=GM/R^2\) |
Free-fall acceleration at the surface |
| Altitude gravity |
\(g(h)=GM/(R+h)^2\) |
Free-fall acceleration at altitude |
| Gravity ratio |
\(\left(R/(R+h)\right)^2\) |
Fraction of surface gravity remaining |
| Drag force |
\(D=\frac12\rho C_dAv^2\) |
Air resistance magnitude |
| Net downward acceleration |
\(a_{\mathrm{net}}=g-D/m\) |
Instantaneous downward acceleration with drag |
Key idea: free-fall acceleration decreases with altitude, but gravity remains strong in low Earth orbit.
