How Does Gravity Change With Altitude?

Gravity does not decrease linearly with altitude. Above Earth’s surface, its strength follows an inverse-square law: doubling your distance from Earth’s center reduces the gravitational acceleration to one quarter.

For ordinary heights such as a building, mountain, or aircraft, the decrease is small but measurable. At hundreds of kilometres above Earth, it becomes substantial. The calculation below gives the standard idealized result and then explains the corrections needed for real measurements.

The altitude formula

Let:

• $R$ be Earth’s radius,
• $h$ be the altitude above the surface,
• $M$ be Earth’s mass,
• $G$ be the gravitational constant.

The distance from Earth’s center is $R+h$. Newton’s law of gravitation gives

$$g(h)=\frac{GM}{(R+h)^2}.$$

At the surface,

$$g_0=\frac{GM}{R^2}.$$

Dividing the first equation by the second eliminates $G$ and $M$:

$$\boxed{g(h)=g_0\left(\frac{R}{R+h}\right)^2}.$$

Using $R\approx 6371\ \text{km}$ and $g_0\approx 9.80665\ \text{m/s}^2$ gives a good estimate near Earth’s surface. The altitude and radius must use the same units.

The change relative to the surface value is

$$\Delta g=g(h)-g_0 =g_0\left[\left(\frac{R}{R+h}\right)^2-1\right].$$

This quantity is negative because gravity becomes weaker with altitude.

Small-altitude approximation

If $h$ is much smaller than $R$, write

$$\frac{R}{R+h}=\frac{1}{1+h/R}.$$

Using the first-order approximation $(1+x)^{-2}\approx 1-2x$ for small $x$ gives

$$g(h)\approx g_0\left(1-\frac{2h}{R}\right).$$

Therefore,

$$\boxed{\frac{\Delta g}{g_0}\approx -\frac{2h}{R}}.$$

So increasing altitude by $h$ reduces the gravitational acceleration by roughly the fraction $2h/R$. This approximation is useful for low altitudes, but the exact inverse-square expression is easy to use and should be preferred when the altitude is not very small compared with Earth’s radius.

Worked example: 1 kilometre above sea level

Take

$$R=6371\ \text{km},\qquad h=1\ \text{km},\qquad g_0=9.80665\ \text{m/s}^2.$$

The exact formula gives

$$\begin{aligned} g(h)&=9.80665\left(\frac{6371}{6371+1}\right)^2\\ &=9.80665\left(\frac{6371}{6372}\right)^2\\ &\approx 9.80357\ \text{m/s}^2. \end{aligned}$$

Thus the decrease is approximately

$$\Delta g=9.80357-9.80665\approx -0.00308\ \text{m/s}^2.$$

The percentage decrease is

$$\frac{0.00308}{9.80665}\times 100\%\approx 0.0314\%.$$

For comparison, the small-altitude estimate gives

$$\frac{2h}{R}\times 100\% =\frac{2(1)}{6371}\times100\% \approx 0.0314\%,$$

which agrees closely.

Worked example: the International Space Station

The International Space Station orbits roughly $400\ \text{km}$ above Earth’s surface. Using the same ideal model,

$$\begin{aligned} g(400\ \text{km}) &=9.80665\left(\frac{6371}{6371+400}\right)^2\\ &\approx 8.69\ \text{m/s}^2. \end{aligned}$$

That is about $88.5\%$ of the surface value, not close to zero. Astronauts feel weightless mainly because the station and its occupants are continuously falling around Earth together. Their apparent weight is nearly zero even though Earth’s gravitational acceleration there remains strong.

Gravity is not exactly the same everywhere

The formula assumes a spherical, nonrotating Earth with a spherically symmetric mass distribution. Actual surface measurements also depend on several effects:

Effect	Consequence
Latitude	Effective gravity is generally lower near the equator than near the poles because Earth rotates and is slightly flattened.
Earth’s shape	The distance to Earth’s center and the local elevation vary across the surface.
Local geology	Mountains, rock formations, and density variations create small gravitational anomalies.
Definition of altitude	Geodetic height, height above mean sea level, and distance from Earth’s center are not identical.

For everyday estimates, $g_0=9.80665\ \text{m/s}^2$ and the inverse-square formula are usually sufficient. Surveying, geophysics, and precision experiments use a latitude- and height-dependent gravity model.

What if you go below the surface?

The above formula applies outside Earth, where the distance from the center is at least approximately Earth’s radius. Below the surface, you cannot generally substitute a negative $h$ into the outside formula.

For a spherically symmetric Earth, the shell theorem says that only the mass enclosed within your radius contributes to the net gravitational force; spherical shells outside your radius produce no net force inside. In a further simplified spherically symmetric, uniform-density Earth, gravity inside is proportional to distance from the center:

$$g(r)=g_0\frac{r}{R},$$

where $r$ is the distance from Earth’s center. It decreases to zero at the center. Earth is not uniform, so the actual underground variation differs from this simple linear model.

Common mistakes

Using altitude instead of distance from the center

The inverse-square law uses $R+h$, not $h$. Using $g\propto 1/h^2$ would incorrectly predict an enormous gravitational acceleration near the surface.

Assuming the decrease is exactly linear

Gravity follows an inverse-square law. The linear expression $g\approx g_0(1-2h/R)$ is only an approximation for $h\ll R$.

Confusing gravitational acceleration with weight

An object’s weight is $W=mg$. If the object’s mass is unchanged, its weight decreases in the same proportion as $g$. A scale reading can also be affected by acceleration, rotation, and buoyancy.

Thinking orbit means no gravity

At low Earth orbit, gravity is still several metres per second squared. Weightlessness results from continuous free fall, not from the absence of gravity.

Ignoring Earth’s rotation in precision work

The inverse-square result describes gravitational acceleration from Earth’s mass. A measurement made relative to the rotating surface measures effective gravity, which includes the effect of centrifugal acceleration.