Is Gravity 9.8 m/s² at Every Height?

The value $g \approx 9.8\,\text{m/s}^2$ is used in many mechanics problems because it is a good approximation near Earth’s surface. It is not exactly the same at every height or location. Whether you should treat it as constant depends on how much the object’s distance from Earth’s center changes during the motion and on the accuracy the problem requires.

Why gravity changes with height

For an approximately spherical Earth, Newton’s law of gravitation gives the gravitational acceleration at distance $r$ from Earth’s center:

$$g(r)=\frac{GM}{r^2},$$

where $G$ is the gravitational constant and $M$ is Earth’s mass. At the surface, $r$ is approximately Earth’s radius $R$. At altitude $h$, the distance is $R+h$, so

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

where $g_0$ is the surface value for the chosen model. The key point is that gravity depends on distance from Earth’s center, not simply on height above the ground.

As $h$ increases, the denominator grows and $g$ decreases. The change is gradual: climbing a few metres barely changes gravity, while travelling hundreds or thousands of kilometres makes the difference more significant.

When is 9.8 m/s² a good approximation?

Use $9.8\,\text{m/s}^2$ for typical school-level problems involving objects near the ground, such as a ball tossed upward or a short fall. Over a small vertical distance compared with Earth’s radius, the change in $g$ is tiny. The constant-acceleration equations are then accurate enough for most purposes.

For example, Earth’s mean radius is about $6371\,\text{km}$. At an altitude of $1\,\text{km}$, the ratio of gravity to its surface value is approximately

$$\frac{g(h)}{g_0}=\left(\frac{6371}{6372}\right)^2\approx 0.9997.$$

That is a decrease of about $0.03\%$. Taking $g_0=9.8\,\text{m/s}^2$, gravity there is about $9.797\,\text{m/s}^2$. This difference is negligible in most ordinary mechanics calculations.

At $100\,\text{km}$ altitude, however,

$$\frac{g(h)}{g_0}=\left(\frac{6371}{6471}\right)^2\approx 0.969,$$

so gravity is about $0.969(9.8)\approx 9.5\,\text{m/s}^2$. That is roughly a $3\%$ decrease. A constant value may still be acceptable for a rough estimate, but a more accurate calculation should account for the change.

A practical way to choose a model

First identify the height range over which the object moves. If that range is very small compared with Earth’s radius and the requested precision is modest, use a constant $g$. If the motion covers a large range in altitude, calculate $g$ as a function of distance or use an energy or orbital model that already accounts for changing gravity.

A useful measure is the fractional change:

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

When $h$ is much smaller than $R$, this ratio is close to 1. For small $h/R$, the fractional decrease is approximately $2h/R$. This estimate helps explain why a kilometre above the surface makes little difference, while an altitude of hundreds of kilometres can matter.

For a radial fall from rest at distance $r_0$ to a smaller distance $r$, ignoring air resistance, conservation of energy gives

$$v^2=2GM\left(\frac{1}{r}-\frac{1}{r_0}\right).$$

This replaces the constant-acceleration formula when gravity’s variation over the fall cannot be ignored. Near the surface, the constant-$g$ model is a simpler approximation to this more general description.

9.8 is not exact everywhere on Earth

Even at the same altitude, measured gravity varies slightly from place to place. Earth is not a perfect sphere, its mass is not distributed uniformly, and Earth’s rotation changes the effective acceleration measured at the surface. Thus, $9.8\,\text{m/s}^2$ is a rounded, broadly useful value rather than a precise local measurement.

You may also see the standard acceleration of gravity, defined as exactly $9.80665\,\text{m/s}^2$. This is a conventional reference value, not a claim that gravity has that exact value at every point on Earth. Textbooks often use $9.8$ or $9.81$ for calculations.

Common mistakes

• Assuming gravity is exactly constant: It is nearly constant over small height changes, but not over arbitrary distances.
• Using height instead of distance from Earth’s center: In the inverse-square law, the relevant distance is $R+h$, not just $h$.
• Assuming astronauts are weightless because gravity is zero: Gravity at orbital altitudes is still substantial. Astronauts feel weightless because they and their spacecraft are in continuous free fall together.
• Mixing gravitational acceleration with apparent weight: A scale reading can also be affected by acceleration, Earth’s rotation, and local conditions; it is not always a direct measurement of the simple inverse-square value.

For a problem near the ground, follow the stated convention—usually $g=9.8\,\text{m/s}^2$—unless the problem asks for greater precision. For a large altitude change, use the inverse-square dependence or a model based on conservation of energy.