Work Against Inverse-Square Gravity

Why Gravity Changes the Work

For a short lift near Earth’s surface, it is often reasonable to treat gravitational force as constant. Over a large change in height, however, an object moves noticeably farther from Earth’s center, and gravity becomes weaker. The work required to lift the object is then not simply a constant force multiplied by distance: the force must be integrated over the path.

This tutorial models Earth as a sphere and uses the inverse-square law for gravity. We will measure position by the distance from Earth’s center, distinguish that distance from altitude above the surface, and compare the exact result with the constant-gravity approximation.

Force as a Function of Distance from Earth’s Center

Let \(M\) be Earth’s mass, \(m\) the mass of the object, and \(r\) its distance from Earth’s center. Newton’s law of universal gravitation gives the magnitude of the gravitational force as \(GMm/r^2\), where \(G\) is the universal gravitational constant. The force points toward Earth’s center.

In this model, Earth’s radius is approximately \(R=6.371\times10^6\) m, and we use the gravitational parameter \(GM=\mu\approx3.986\times10^{14}\) m\(^3\)/s\(^2\). Combining \(G\) and \(M\) into \(\mu\) makes the calculations shorter. The gravitational force magnitude is \(F_g(r)=\mu m/r^2\), measured in newtons when \(m\) is in kilograms and \(r\) is in metres.

For a slow outward lift, the external lifting force must point outward and balance gravity at each position. Its magnitude is therefore \(\mu m/r^2\). The work supplied by the lifting force is positive. Gravity itself does negative work during the same outward motion.

Setting Up and Evaluating the Work Integral

For a small outward displacement \(dr\), the work supplied against gravity is the lifting-force magnitude multiplied by that displacement. Adding these small contributions from \(r_1\) to \(r_2\) gives the integral. The endpoints must be radial distances from Earth’s center, and for a lift \(r_2\ge r_1\ge R\).

The mass and \(\mu\) are constant during the lift, so they can be taken outside the integral. Since an antiderivative of \(r^{-2}\) is \(-r^{-1}\), the Fundamental Theorem of Calculus gives the exact result.

Here is the complete evaluation. The antiderivative has units consistent with the final result: \(\mu m/r\) has units \((\text{m}^3/\text{s}^2)(\text{kg})/\text{m}=\text{kg}\,\text{m}^2/\text{s}^2=\text{J}\).

This derivation also explains why the work depends on both endpoints rather than just the distance \(r_2-r_1\). Equal-length lifts at different distances from Earth’s center do not require equal work: the gravitational force is weaker farther away.

Worked Examples

When the Constant-Gravity Approximation Is Appropriate

For a lift beginning at the surface and rising by height \(h\), the exact inverse-square result can be written as \(W=\mu m h/[R(R+h)]\). The constant-gravity estimate using the starting value \(g_0=\mu/R^2\) is \(W_{\text{constant}}=mg_0h=\mu m h/R^2\). Dividing the exact expression by the estimate gives a useful comparison.

When \(h\) is small compared with \(R\), this ratio is close to \(1\), so treating gravity as constant can be a reasonable approximation. As \(h\) becomes a substantial fraction of Earth’s radius, the difference matters more. For example, a lift from \(R\) to \(2R\) has \(h=R\), so the ratio is \(1/2\): using the surface gravitational force throughout would give twice the exact work.

The approximation is a modeling choice, not a different exact formula. Use it when the height change is small enough for the desired accuracy, and use the inverse-square integral when the change in distance from Earth’s center is significant.

Common Pitfall: Using Altitude as the Radius

The inverse-square law uses \(r\), the distance from Earth’s center. Altitude \(h\) measures distance above the surface, so \(r=R+h\). Confusing the two gives an incorrect setup. For instance, if an object rises from the surface to altitude \(h\), the incorrect integral \(\int_0^h \mu m/r^2\,dr\) uses altitude as the radius and even has a singularity at its lower endpoint.

The correction is to express every position in the integral as a distance from Earth’s center. For that same lift, integrate from \(R\) to \(R+h\):

Also keep the units consistent: if \(r\) is in metres, convert heights given in kilometres to metres before inserting them. The resulting work is in joules.