Why Distance from the Centre Matters
Near Earth’s surface, gravity is often treated as having a constant strength, so the work needed to lift an object through a height \(h\) is approximated by \(mgh\). That model is useful when the change in height is small compared with Earth’s radius. Over larger distances, however, the object moves far enough from Earth’s centre that the gravitational force changes appreciably.
To calculate work over such a distance, we use a force law that depends on the object’s distance from the centre. The resulting integral also clarifies exactly when the familiar constant-\(g\) formula is a good approximation.
The Inverse-Square Gravitational Force
Assume a body of mass \(M\) is spherical, or can be modelled as a point mass, and that the object of mass \(m\) is outside it. Let \(r\) denote the object’s distance from the body’s centre. Newton’s law of gravitation gives the force magnitude \(GMm/r^2\), where \(G\) is the gravitational constant. Taking outward as the positive direction, gravity itself points inward.
The model assumes the central body’s mass is fixed and that the object moves radially. For a slow, controlled outward move, the applied force balances gravity at each position. The work required against gravity from radius \(r_1\) to radius \(r_2\), where \(0<r_1<r_2\), is therefore the integral of the outward applied force over the distance travelled.
Deriving the Work Formula
Since \(G\), \(M\), and \(m\) are constant during the motion, they can be taken outside the integral. The power rule for integration gives an antiderivative of \(r^{-2}\) as \(-r^{-1}\). Applying the bounds produces the formula for the required work.
The result is positive because \(r_1<r_2\), so \(1/r_1>1/r_2\). This sign matches the physical description: an external agent must supply energy to move the object outward against gravity. The work done by gravity over the same outward displacement has the opposite sign.
The formula also passes a units check. The product \(GMm\) has units \(\mathrm{kg\,m^3/s^2}\), and each reciprocal radius has units \(\mathrm{m^{-1}}\). Their product has units \(\mathrm{kg\,m^2/s^2}\), which are joules.
Worked Examples
Worked Example: Moving Outward from a Small Idealized Body
An object of mass \(2\ \mathrm{kg}\) is moved slowly from \(r_1=2\ \mathrm{m}\) to \(r_2=6\ \mathrm{m}\) from the centre of an idealized body with \(GM=12\ \mathrm{m^3/s^2}\). Find the work required against gravity.
First, the force that must be applied outward at distance \(r\) is
when \(r\) is measured in metres. Integrate this force over the stated interval:
Check the evaluation by differentiating the antiderivative: \(\frac{d}{dr}(-24/r)=24/r^2\), the integrand. Substituting the bounds into that antiderivative gives \(-24/6-(-24/2)=-4+12=8\ \mathrm{J}\), as calculated. The positive answer is consistent with motion outward against gravity.
Worked Example: A Large Change in Distance from a Central Body
Suppose an object of mass \(3\ \mathrm{kg}\) is moved slowly from \(6.0\times10^6\ \mathrm{m}\) to \(1.2\times10^7\ \mathrm{m}\) from a body’s centre. Let the body’s gravitational parameter be \(GM=4.0\times10^{14}\ \mathrm{m^3/s^2}\). Find the work required.
Use the two radii measured from the centre—not the height moved measured from the starting point. The work formula gives
The product \(GMm\) is \(1.2\times10^{15}\ \mathrm{kg\,m^3/s^2}\). The reciprocal-radius difference is
Therefore,
For a check, an antiderivative of \(1.2\times10^{15}/r^2\) is \(-1.2\times10^{15}/r\). Its value at the upper radius minus its value at the lower radius is \(-1.0\times10^8-(-2.0\times10^8)=1.0\times10^8\ \mathrm{J}\). The result is positive, as required for outward work against gravity.
How the Near-Surface Approximation Appears
Let an object start at the surface of a spherical body with radius \(R\), and move outward a height \(h\). Its final distance from the centre is \(R+h\). The exact inverse-square result is
Define the gravitational acceleration at the surface by \(g_0=GM/R^2\). Replacing \(GM\) with \(g_0R^2\) in the exact result gives
When \(h\) is small compared with \(R\), the factor \(R/(R+h)\) is close to \(1\). The exact work is then close to \(mg_0h\). For \(h>0\), the factor is less than \(1\), so the constant-\(g_0\) estimate is slightly larger: it treats gravity as remaining at its surface strength throughout the lift.
Worked Example: Comparing Exact Work with \(mgh\)
An object of mass \(2\ \mathrm{kg}\) is lifted \(1\ \mathrm{m}\) from the surface of an idealized body with radius \(R=10\ \mathrm{m}\) and \(GM=100\ \mathrm{m^3/s^2}\). Compare the exact work with the constant-gravity estimate.
The surface acceleration is
The initial and final distances from the centre are \(10\ \mathrm{m}\) and \(11\ \mathrm{m}\). Thus, the exact work is
The constant-gravity estimate is
To check the exact calculation, the antiderivative of \(200/r^2\) is \(-200/r\), whose endpoint difference is \(-200/11-(-200/10)=20/11\). The exact work is also \(mg_0h\cdot R/(R+h)=2\cdot10/11=20/11\ \mathrm{J}\), confirming the result. The constant-gravity estimate is close, but it is larger because the force decreases during the lift.
Work to Reach an Arbitrarily Large Distance
The formula also describes the limiting work as the object moves farther and farther away. Taking the limit as \(r_2\) increases without bound gives \(1/r_2\to0\), and hence
Thus, under the stated idealized model, the work to move an object from a finite radius \(r_1\) all the way to arbitrarily large distance is finite. This follows because the gravitational force weakens with distance quickly enough that its integral over this outward path converges.
A Common Error: Using the Starting Gravity for the Whole Trip
A tempting calculation is to find the force at the starting point and multiply it by the entire distance moved. For the final example, this gives \(F(R)h=mg_0h=2\ \mathrm{J}\). That calculation assumes the force stays constant at its initial value. But at distance \(r\), the actual force is \(GMm/r^2\), which is smaller than its starting value whenever \(r>R\). Therefore, the product \(F(R)h\) is not the exact work over a finite outward interval.
The correction is to integrate the varying force, using distance from the centre in the denominator. In particular, for a lift through height \(h\) from a surface of radius \(R\), the bounds are \(R\) and \(R+h\), not \(0\) and \(h\).
Check Your Understanding
Use the inverse-square model and show the integral or reasoning behind each answer.
- For an outward move from \(r_1\) to \(r_2\), with \(0<r_1<r_2\), why is the work required against gravity positive?
- An object moves outward from radius \(a\) to radius \(2a\). Express the work in terms of \(G\), \(M\), \(m\), and \(a\).
- For a move from the surface of radius \(R\) through height \(h\), what are the correct integration bounds when the variable is distance from the centre?
- Under what condition is \(mg_0h\) a useful approximation to the exact work, and why does it slightly overestimate that work for an outward lift?
- What is the limiting work as the final radius tends to infinity, starting from radius \(r_1\)?
