Gravitational potential energy and fields - TCE Physics (Tasmania)

Syllabus dot point

“Describe gravitational potential energy in a radial field and relate work done to changes in field energy.”

TCEPhysicsUnit 3: Gravity and Electromagnetism9 min readUpdated 2 Jun 2026

Quick answer

Far from a planet the simple $mgh$ rule fails, so gravitational potential energy is given by $E_p = -\dfrac{GMm}{r}$ , which is negative and rises toward zero as you move away. The work to move a mass between two points is the change in this energy, and the area under a field-distance graph gives the change in potential.

What this dot point is asking

This dot point extends gravity from the uniform field near the surface to the full radial field around a planet or star, where field strength and potential energy depend on distance.

Field lines and the radial field

A gravitational field is represented by field lines that point toward the mass creating the field. Around a point mass or sphere the lines are radial, spreading out evenly in all directions. Where the lines are closer together the field is stronger, so the field is strongest near the surface and weakens with distance following the inverse-square law $g = \dfrac{GM}{r^2}$ .

Near the surface, over distances small compared with the planet's radius, the lines are almost parallel and evenly spaced, which is why we treat the surface field as uniform and use $g = 9.8\ \text{N kg}^{-1}$ .

Gravitational potential energy in a radial field

Close to the surface we use $E_p = mgh$ , but this assumes constant $g$ . Over large distances $g$ changes, so the correct expression for the gravitational potential energy of a mass $m$ at distance $r$ from the centre of mass $M$ is:

E_p = -\frac{GMm}{r}

The energy is negative because the zero of potential energy is defined at infinite separation, where the masses no longer interact. As you move a mass closer, the energy becomes more negative; as you move it away, it rises toward zero. Bringing the masses together releases energy, which is why falling objects gain kinetic energy.

Gravitational potential energy

The gravitational potential energy of a mass in a field is the work done to bring it from infinity to its current position. It is negative because gravity is attractive, so work is released as the mass approaches.

Work done moving through the field

The work needed to move a mass between two points equals the change in its gravitational potential energy:

W = \Delta E_p = -GMm\left(\frac{1}{r_2} - \frac{1}{r_1}\right)

Moving away from the planet ( $r_2 > r_1$ ) makes $E_p$ less negative, so $\Delta E_p$ is positive and you must do work. Moving inward releases energy. This is the energy a rocket must supply to climb to a higher orbit.

Escape velocity

A particularly important application is the escape velocity, the minimum launch speed for an object to reach infinity with no kinetic energy left. Setting the kinetic energy at the surface equal to the energy needed to lift the object to zero potential energy gives $\tfrac12 m v_\text{esc}^2 = \dfrac{GMm}{r}$ , so $v_\text{esc} = \sqrt{\dfrac{2GM}{r}}$ . For Earth this is about $11\ \text{km s}^{-1}$ . Notice the mass of the projectile cancels, so the escape speed depends only on the planet, exactly as for orbital speed.

Field-distance graphs

If you plot gravitational field strength $g$ against distance $r$ , the area under the graph between two radii gives the change in gravitational potential per unit mass. Because the field follows an inverse-square curve, this area is found by integration or by counting squares, and it confirms that most of the energy to escape a planet is spent close to the surface where the field is strongest.

Worked example: energy to raise a satellite

A $500\ \text{kg}$ satellite is raised from an initial orbital radius of $r_1 = 6.4 \times 10^6\ \text{m}$ to a final orbital radius of $r_2 = 7.4 \times 10^6\ \text{m}$ above Earth's centre ( $M = 6.0 \times 10^{24}\ \text{kg}$ ). Calculate the work required.

Step 1: Evaluate the common factor $GMm$

The work formula $W = -GMm\!\left(\dfrac{1}{r_2} - \dfrac{1}{r_1}\right)$ contains the product $GMm$ , which appears in both terms. Computing it once avoids repeating large numbers and reduces rounding errors in the final subtraction:

GMm = (6.67\times10^{-11})(6.0\times10^{24})(500) = 2.0\times10^{17}\ \text{J\,m}.

Step 2: Calculate the change in reciprocal radius

The work depends on how much $\dfrac{1}{r}$ changes, not on $\Delta r$ directly, because the gravitational potential energy varies as $-\dfrac{GMm}{r}$ . Moving to a larger radius makes $\dfrac{1}{r}$ smaller, so this difference is negative, reflecting the fact that potential energy increases (becomes less negative) as the satellite climbs:

\frac{1}{r_2} - \frac{1}{r_1} = \frac{1}{7.4\times10^6} - \frac{1}{6.4\times10^6} = -2.11\times10^{-8}\ \text{m}^{-1}.

Step 3: Find the work done

Substituting into the work formula, the two negatives multiply to give a positive result, confirming that energy must be added to the satellite to move it to the higher orbit:

W = -(2.0\times10^{17})(-2.11\times10^{-8}) = 4.2\times10^{9}\ \text{J}.

Final answer: About $4.2 \times 10^9\ \text{J}$ of work is required to raise the satellite.

Potential energy is most negative at the surface

A mass has its most negative gravitational potential energy when closest to the planet and approaches zero at infinity. This is why an object needs energy added to escape, and why the change in energy, not the absolute value, is what matters in calculations.

Using mgh for large height changes

A frequent error is using $E_p = mgh$ for satellites or space probes. That formula only holds when $g$ is effectively constant, near the surface. For large distances you must use $E_p = -\dfrac{GMm}{r}$ , otherwise the changing field strength is ignored and the answer is badly wrong.

In the exam, check the scale of the height change. If it is small compared with the planet's radius, $mgh$ is fine. If the object moves a large distance or out to space, switch to $E_p = -\dfrac{GMm}{r}$ and work with the change in energy.

Exam-style questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice4 marks

Comet Shoemaker-Levy 9 broke apart before crashing into Jupiter in 1994. Three parts A, B and C lie along the comet at different distances from Jupiter. Explain the relative sizes of Jupiter's gravitational field at A, B and C, and explain why the comet broke apart as it approached Jupiter.

Show worked answer →

Jupiter's gravitational field is radial and follows an inverse-square law, $g = \dfrac{GM}{r^2}$ , so the field strength is larger the closer a point is to Jupiter.

The part nearest Jupiter experiences the strongest field (longest vector), the middle part an intermediate field, and the far part the weakest (shortest vector). The vectors also point slightly inward toward Jupiter's centre, not all parallel.

Because the near side is pulled much harder than the far side, there is a difference in pull across the comet's length (a tidal force) that stretches it along the line to Jupiter. When this differential pull exceeds the comet's own weak self-gravity and structural strength, the comet is torn apart into a string of fragments.

Markers reward linking the inverse-square field to each vector's size and identifying the differential (tidal) force as the cause of break-up.

Practice3 marks

Cavendish measured Earth's mass using the gravitational force

F

between lead masses of

20\ \text{kg}

and

50\ \text{kg}

at various separations

r

. State what should be plotted for a straight-line graph and outline how the universal gravitational constant is found from the gradient.

Show worked answer →

Newton's law of gravitation is $F = \dfrac{Gm_1 m_2}{r^2}$ . This is not linear in $r$ , but it is linear in $\dfrac{1}{r^2}$ .

Plotting $F$ (vertical) against $\dfrac{1}{r^2}$ (horizontal) gives a straight line through the origin with gradient $Gm_1 m_2$ .

Reading the gradient and dividing by the product of the masses gives $G$ :

G = \frac{\text{gradient}}{m_1 m_2} = \frac{\text{gradient}}{(20)(50)} = \frac{\text{gradient}}{1000}.

For example a gradient of about $6.7 \times 10^{-8}\ \text{N m}^2$ gives $G = 6.7 \times 10^{-11}\ \text{N m}^2\,\text{kg}^{-2}$ . Markers want the linearising variable $\dfrac{1}{r^2}$ and $G = \dfrac{\text{gradient}}{m_1 m_2}$ .

Practise this

Exam-style questions 2 on this page
Flashcards Physics deck

Sources & how we know this

Physics Level 4 (PHY415115) course document (2015 onwards) — TASC