Define Gravitational Potential
- Define gravitational potential as $V = -GM/r$
- Relate potential to GPE per unit mass ($V = U/m$)
- State the units (J/kg or m$^2$/s$^2$) and scalar nature
Relate Potential Gradient to Field Strength
- Use $g = -dV/dr$ to connect field and potential
- Interpret graphs of $V$ vs $r$ and $g$ vs $r$
- Estimate $g$ from potential differences
Interpret Equipotential Surfaces
- Describe equipotential surfaces around spherical masses
- Explain why work done along an equipotential is zero
- Relate spacing of equipotentials to field strength
Does gravitational potential increase or decrease as you move closer to a mass? Sketch how you think $V$ varies with distance $r$.
Gravitational Potential
Define gravitational potential $V = -GM/r$, understand potential gradient, and interpret equipotential surfaces around spherical masses.
Gravitational Potential
GPE per unit mass — a scalar field quantity
Gravitational Potential
Gravitational Potential Detailed
The gravitational potential $V$ at a point in a gravitational field is defined as the gravitational potential energy per unit mass placed at that point:
Gravitational Potential
Key properties of gravitational potential:
- Scalar quantity — potential has magnitude only, no direction (unlike field strength $\vec{g}$, which is a vector)
- Negative for all finite $r$ — the negative sign reflects that gravity is attractive; work must be done to move a mass to infinity
- Zero at infinity — by convention, $V = 0$ at $r = \infty$, where there is no gravitational interaction
- Units — J/kg (joules per kilogram), equivalent to m$^2$/s$^2$
Gravitational potential $V$ describes the energy landscape of a gravitational field. Just as altitude describes a height landscape, $V$ tells us how much energy a unit mass would have at each point. Moving "uphill" in potential (toward less negative values) requires work to be done against gravity.
Worked Example — Potential at Earth's Surface
Calculate the gravitational potential at Earth's surface. $M_E = 5.97 \times 10^{24} \text{ kg}$, $R_E = 6.37 \times 10^6 \text{ m}$.
$G = 6.67 \times 10^{-11} \text{ N m}^2\text{/kg}^2$, $M_E = 5.97 \times 10^{24} \text{ kg}$, $R_E = 6.37 \times 10^6 \text{ m}$
Gravitational potential $V$ at Earth's surface
Use $V = -GM/r$ with $r = R_E$ (centre-to-centre distance at the surface).
$$V = -\frac{(6.67 \times 10^{-11} \text{ N m}^2\text{/kg}^2)(5.97 \times 10^{24} \text{ kg})}{6.37 \times 10^6 \text{ m}}$$
$$V = -\frac{3.98 \times 10^{14} \text{ N m}^2\text{/kg}}{6.37 \times 10^6 \text{ m}}$$
$$V = -6.25 \times 10^7 \text{ J/kg} = -62.5 \text{ MJ/kg}$$
$V = -62.5 \text{ MJ/kg}$ at Earth's surface. This means $62.5 \text{ MJ}$ of work is needed per kilogram to move a mass from Earth's surface to infinity.
Potential Gradient and Field Strength
Connecting the scalar potential to the vector field
The gravitational field strength $g$ is related to the gravitational potential $V$ through the potential gradient:
Field Strength from Potential Gradient
For $V = -GM/r$, we can derive the familiar expression for $g$:
$$\frac{dV}{dr} = \frac{d}{dr}\left(-\frac{GM}{r}\right) = \frac{GM}{r^2}$$
Therefore:
$$g = -\frac{dV}{dr} = -\frac{GM}{r^2}$$
The negative sign indicates that the field points radially inward (toward decreasing $r$, which is the direction of decreasing potential — more negative). When we use the magnitude only:
$$|g| = \frac{GM}{r^2}$$
Graphs of $V$ vs $r$ and $g$ vs $r$
- $V$ vs $r$: $V = -GM/r$ is always negative, approaching zero from below as $r \to \infty$. The curve is steeper near the mass (rapid change in potential) and flatter far away.
- $g$ vs $r$: $g = -GM/r^2$ is also always negative (inward). The magnitude $|g| \propto 1/r^2$ falls off more rapidly than $|V| \propto 1/r$.
The slope of a $V$ vs $r$ graph at any point gives $-g$ at that point. A steep slope means strong field; a gentle slope means weak field. Where $V$ is flat (far from any mass), $g \approx 0$.
Worked Example — Estimating $g$ from Potential Difference
At a point in a gravitational field, $V = -50 \times 10^6 \text{ J/kg}$. At a point 1000 m closer to the central mass, $V = -55 \times 10^6 \text{ J/kg}$. Estimate the gravitational field strength in this region.
$V_1 = -50 \times 10^6 \text{ J/kg}$, $V_2 = -55 \times 10^6 \text{ J/kg}$, $\Delta r = 1000 \text{ m}$
Gravitational field strength $g$
Use $g \approx -\Delta V/\Delta r$ for a finite separation. Note: moving inward (decreasing $r$) gives a more negative $V$, so $\Delta V = V_2 - V_1 = -5 \times 10^6 \text{ J/kg}$.
$$g \approx -\frac{\Delta V}{\Delta r} = -\frac{(-55 \times 10^6) - (-50 \times 10^6)}{1000 \text{ m}}$$
$$g = -\frac{-5 \times 10^6 \text{ J/kg}}{1000 \text{ m}} = \frac{5 \times 10^6}{1000} \text{ N/kg}$$
$$g = 5000 \text{ N/kg}$$
$g = 5000 \text{ N/kg}$ (inward). This extremely strong field indicates a location very close to a massive object — about 500 times Earth's surface gravity. The large potential change over just 1000 m confirms the proximity to a massive central body.
Equipotential Surfaces
Surfaces of constant gravitational potential
An equipotential surface is a surface on which the gravitational potential $V$ has the same value at every point. Key properties:
- Spherical around a point mass — for a single spherical mass, equipotential surfaces are concentric spheres centred on the mass
- Closer together near the mass — the spacing indicates field strength: closely spaced equipotentials mean a strong field; widely spaced means a weak field
- No work done moving along an equipotential — since $\Delta V = 0$, the work done per unit mass $W/m = \Delta V = 0$
- Field lines are perpendicular to equipotentials — if there were a component of $\vec{g$} along the surface, work would be done; since $W = 0$, $\vec{g}$ must be purely normal to the surface
Equipotential surfaces are a powerful tool because they convert a vector problem (field strength with direction) into a scalar problem (energy per unit mass). Energy calculations become simple: the work done per unit mass moving between any two points is just the difference in potential, $W/m = \Delta V$.
Worked Example — Equipotential Surfaces Around Earth
Describe the equipotential surfaces around Earth at $V = -60$, $-50$, and $-40 \text{ MJ/kg}$. Calculate the radius of each surface. $M_E = 5.97 \times 10^{24} \text{ kg}$.
$V_1 = -60 \times 10^6 \text{ J/kg}$, $V_2 = -50 \times 10^6 \text{ J/kg}$, $V_3 = -40 \times 10^6 \text{ J/kg}$, $G = 6.67 \times 10^{-11} \text{ N m}^2\text{/kg}^2$, $M_E = 5.97 \times 10^{24} \text{ kg}$
Radius of each equipotential surface and description
Rearrange $V = -GM/r$ to give $r = -GM/V$. Since both $V$ and $-GM$ are negative, $r$ is positive.
For $V = -60 \text{ MJ/kg}$:
$$r = -\frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{-60 \times 10^6} = \frac{3.98 \times 10^{14}}{60 \times 10^6} = 6.63 \times 10^6 \text{ m}$$
For $V = -50 \text{ MJ/kg}$:
$$r = \frac{3.98 \times 10^{14}}{50 \times 10^6} = 7.96 \times 10^6 \text{ m}$$
For $V = -40 \text{ MJ/kg}$:
$$r = \frac{3.98 \times 10^{14}}{40 \times 10^6} = 9.95 \times 10^6 \text{ m}$$
Description: The equipotential surfaces are concentric spheres centred on Earth. The $-60 \text{ MJ/kg}$ surface (radius $6.63 \times 10^6 \text{ m}$, only 260 km above the surface) is closest to Earth and has the strongest field. The $-40 \text{ MJ/kg}$ surface ($r = 9.95 \times 10^6 \text{ m}$, 3580 km altitude) is furthest out with the weakest field. The spacing between surfaces increases with distance, reflecting the $1/r^2$ decrease in field strength.
Essential Formulae — Gravitational Potential
Common Misconceptions
"Potential and potential energy are the same" — Gravitational potential $V$ is per unit mass (J/kg); gravitational potential energy $U$ is for a specific mass (J). The relationship is $V = U/m$ or $U = mV$. Potential is a property of the field; potential energy is a property of a mass in the field.
"Equipotential surfaces have constant field strength" — Equipotential surfaces have constant potential $V$, not constant field strength. Field strength varies as $g \propto 1/r^2$. The spacing between adjacent equipotentials indicates field strength: closely spaced means strong field, widely spaced means weak field.
Mapping Earth's Gravitational Field from Space
NASA's Gravity Recovery and Climate Experiment (GRACE, 2002-2017) used two satellites in the same orbit measuring changes in distance between them to map Earth's gravitational field with unprecedented precision.
As the lead satellite approaches a region of slightly stronger gravity (higher mass concentration), it accelerates slightly, changing the distance between the two satellites. By precisely tracking these changes, GRACE mapped variations in Earth's gravitational potential.
Key discoveries:
- Groundwater depletion in India: Over 200 km$^3$ of groundwater lost between 2002 and 2008, detected as changes in the local gravitational potential
- Ice mass loss in Greenland: Average loss of 280 Gt/yr, contributing to global sea level rise
- Post-glacial rebound: Slow rise of land masses previously compressed by ice sheets
GRACE-FO (2018 onward) uses laser ranging for even higher precision. These missions literally map equipotential surfaces and track their changes over time, turning variations in $V = -GM/r$ into measurements of mass redistribution on Earth.
Calculation Drills
1 Find the gravitational potential at $r = 2R_E$ from Earth's centre ($R_E = 6.37 \times 10^6 \text{ m}$). Express your answer in MJ/kg.
2 If the gravitational potential changes from $-40 \text{ MJ/kg}$ to $-45 \text{ MJ/kg}$ over a distance of 500 km, estimate the average gravitational field strength in this region.
3 Describe the shape of equipotential surfaces for two equal masses placed close together. How do the equipotentials differ from those around a single mass?
Connecting Potential and Field
A student looks at a graph of $V$ vs $r$ for a planet and says: "The curve is steepest near the planet, so the potential is strongest there. The curve is flattest far away, so the potential is weakest there." Explain why this reasoning is incorrect. In your answer, discuss: (a) what the steepness of the $V$ vs $r$ graph actually represents, (b) how potential itself varies with $r$, and (c) the difference between potential and field strength.
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▾Key Definitions
- Gravitational potential: $V = -GM/r$ (J/kg, scalar)
- Potential gradient: $dV/dr$ (rate of change of $V$ with distance)
- Equipotential surface: surface of constant $V$
- $V = 0$ at infinity (by convention)
Key Formulae
- $V = -GM/r$
- $g = -dV/dr = -GM/r^2$
- $W/m = \Delta V$ (work per unit mass)
- $V_{\text{surface}} = -GM/R$
Important Points
- $V$ is a scalar; $g$ is a vector
- $V$ is always negative for finite $r$
- Steeper $V$ vs $r$ slope = stronger field
- Work along equipotential = zero
- Field lines perpendicular to equipotentials
Common Errors
- Confusing potential (J/kg) with potential energy (J)
- Forgetting $V$ is a scalar
- Thinking equipotentials = constant field strength
- Dropping the negative sign in $V = -GM/r$
- Using $h$ instead of $r = R + h$
Revisit Your Answer
Now that you have studied gravitational potential, revisit your earlier prediction about how $V$ varies with distance $r$.
Gravitational potential $V = -GM/r$ is negative for all finite distances and becomes more negative as $r$ decreases (closer to the mass). As $r \to \infty$, $V \to 0$. The graph of $V$ vs $r$ is a curve that starts at large negative values near the mass and asymptotically approaches zero from below. The steepness of the curve near the mass reflects the strong gravitational field in that region.
Has your understanding changed? Write a revised explanation:
1. Gravitational potential $V$ is defined as:
2. The relationship between gravitational field strength $g$ and potential $V$ is:
3. Equipotential surfaces around a point mass are:
4. The work done moving a mass along an equipotential surface is:
5. At a point where $V = -30 \text{ MJ/kg}$, the gravitational field strength:
Calculate the gravitational potential at the surface of the Moon ($M = 7.35 \times 10^{22} \text{ kg}$, $R = 1.74 \times 10^6 \text{ m}$). How much work is needed per kilogram to move a payload from the Moon's surface to infinity?
The gravitational potential at distance $r$ from Earth's centre is $-40 \text{ MJ/kg}$. At $r + 2000 \text{ km}$, it is $-35 \text{ MJ/kg}$. Estimate the average gravitational field strength in this region.
Evaluate the usefulness of representing gravitational fields using equipotential surfaces rather than field lines. Discuss the advantages of each representation and when one is more informative than the other.
Model Answers
▾Question 1 (3 marks)
$V = -\dfrac{GM}{R} = -\dfrac{(6.67 \times 10^{-11})(7.35 \times 10^{22})}{1.74 \times 10^6}$ (1 mark)
$V = -\dfrac{4.90 \times 10^{12}}{1.74 \times 10^6} = -2.82 \times 10^6 \text{ J/kg} = -2.82 \text{ MJ/kg}$ (1 mark)
Work to infinity per kilogram: $W/m = \Delta V = V_\infty - V_{\text{surface}} = 0 - (-2.82) = 2.82 \text{ MJ/kg}$ (1 mark)
Question 2 (3 marks)
$g \approx -\dfrac{\Delta V}{\Delta r}$ (1 mark)
$\Delta V = (-35 \times 10^6) - (-40 \times 10^6) = 5 \times 10^6 \text{ J/kg}$
$\Delta r = 2000 \text{ km} = 2 \times 10^6 \text{ m}$
$g = -\dfrac{5 \times 10^6}{2 \times 10^6} = -2.5 \text{ N/kg}$ (magnitude $2.5 \text{ N/kg}$) (2 marks)
Question 3 (4 marks)
Equipotential surfaces — advantages: They are scalar quantities (easy to calculate), work done moving between points is simply $\Delta V$ (enabling direct energy calculations), and they represent surfaces of constant energy. They are ideal for solving energy problems such as calculating escape velocity or orbital energy changes. (1 mark)
Equipotential surfaces — disadvantages: They do not show the direction of the gravitational force. A test mass at rest on an equipotential will not move, so the direction of acceleration is not immediately apparent. (1 mark)
Field lines — advantages: They show the direction of the gravitational force at every point. The density of field lines indicates field strength, and the tangent to a field line gives the direction of force on a test mass. (1 mark)
Comparison: Equipotentials are best for energy-related problems (work, potential energy, escape velocity). Field lines are best for visualising force direction and predicting motion. The two representations are complementary — field lines are always perpendicular to equipotential surfaces, and using both together gives a complete picture of the field. (1 mark)
Mark Lesson 15 Complete
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