Gravitational potential

1. Overview

Gravitational potential ($\phi$) is a fundamental scalar quantity that describes the energy per unit mass at any given point within a gravitational field. While gravitational field strength ($g$) provides a vector-based description of the force acting on a mass, gravitational potential offers a scalar framework based on energy and work. This approach is significantly more efficient for solving complex problems involving orbital mechanics, planetary motion, and energy conservation, as it avoids the complexities of vector decomposition.

The concept is built upon the principle of work done. Because gravity is a conservative and attractive force, moving a mass within a field involves a transfer of energy. By establishing a universal reference point—infinity—where the gravitational influence is zero, we can quantify the "potential" of any point in space relative to that zero-point. This leads to the characteristic negative values associated with gravitational potential, representing a "potential well" from which energy must be supplied to escape.

Key Definitions

To secure full marks in Cambridge 9702 exams, definitions must be precise and include all underlined keywords from the mark schemes.

• Gravitational Potential ($\phi$): The work done per unit mass in bringing a small test mass from infinity to a specific point in a gravitational field.
• Gravitational Potential Energy ($E_P$): The work done in bringing a mass $m$ from infinity to a specific point in a gravitational field.
• Infinity: The theoretical point at a distance so great from a mass that the gravitational force exerted by that mass is zero. It is the standard reference point where gravitational potential is defined as zero.
• Equipotential Surface: A three-dimensional surface joining points of equal gravitational potential. No work is done when a mass moves along an equipotential surface because the change in potential ($\Delta\phi$) is zero.
• Potential Gradient: The change in gravitational potential per unit change in distance from the source mass. The negative of this gradient is equal to the gravitational field strength ($g$).

Content

3.1 The Concept of Negative Potential

A common point of confusion for students is why gravitational potential is always negative. This arises from the choice of the reference point and the nature of the gravitational force:

1. Reference Point: We define $\phi = 0$ at $r = \infty$.
2. Attractive Force: Gravity is always attractive. To move a mass from infinity toward a planet, the gravitational field does the work. An external agent would have to apply a force away from the planet to prevent the mass from accelerating, meaning the external agent does negative work.
3. Energy State: As a mass moves from infinity (zero energy) toward a source mass, it loses potential energy (it becomes more negative). Therefore, all points closer than infinity have a potential less than zero.

Think of a "potential well." The closer you are to the mass $M$, the deeper you are in the well, and the more energy you must "climb out" to reach the surface (infinity).

3.2 The Gravitational Potential Formula

For a point mass $M$, or outside a uniform sphere of mass $M$, the potential $\phi$ at a distance $r$ from the center of the mass is:

$$\mathbf{\phi = -\frac{GM}{r}}$$

Status: This equation is provided on the Formulae Sheet.

Variables:

• $\phi$: Gravitational potential ($\text{J kg}^{-1}$)
• $G$: Newton’s Gravitational Constant ($\approx 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$)
• $M$: The mass creating the gravitational field ($\text{kg}$)
• $r$: The distance from the center of mass $M$ to the point in question ($\text{m}$)

Key Mathematical Relationships:

• $\phi \propto M$: Doubling the source mass doubles the potential (makes it twice as negative).
• $\phi \propto -\frac{1}{r}$: Potential is inversely proportional to the distance. As $r$ increases, $\phi$ becomes less negative (increases toward zero).

3.3 Graphical Representation of Potential

The graph of $\phi$ against $r$ is a rectangular hyperbola located in the fourth quadrant.

• At $r = R$ (Surface): The potential is at its minimum (most negative) value for a solid object.
• As $r \to \infty$: The curve approaches the x-axis asymptotically ($\phi \to 0$).
• Gradient: The gradient of the $\phi-r$ graph is $\frac{d\phi}{dr}$. Since $g = -\frac{d\phi}{dr}$, the negative of the gradient at any point gives the gravitational field strength $g$ at that distance.

3.4 Gravitational Potential Energy ($E_P$)

While potential ($\phi$) is a property of the field itself (energy per unit mass), Gravitational Potential Energy ($E_P$) is a property of a specific mass $m$ placed within that field.

$$\mathbf{E_P = m\phi = -\frac{GMm}{r}}$$

Status: This equation is NOT explicitly on the formulae sheet in this form; you must derive it from $E_P = m\phi$.

Work Done and Changes in Energy: When a mass moves between two points in a field, the work done ($W$) by an external agent is equal to the change in potential energy: $$\Delta W = \Delta E_P = m(\phi_{\text{final}} - \phi_{\text{initial}})$$ $$\Delta W = GMm \left( \frac{1}{r_{\text{initial}}} - \frac{1}{r_{\text{final}}} \right)$$

• If $\Delta W$ is positive, work is done on the mass (it moves further away).
• If $\Delta W$ is negative, work is done by the field (it moves closer).

3.5 Relationship between $g$ and $\phi$

The relationship between field strength and potential is analogous to the relationship between force and potential energy.

$$\mathbf{g = -\frac{d\phi}{dr}}$$

Status: This equation is provided on the Formulae Sheet.

In practical exam terms:

1. Magnitude: The magnitude of $g$ is the steepness of the $\phi-r$ graph.
2. Direction: $g$ always points in the direction of decreasing potential (down the "potential hill").
3. Uniform Fields: In a uniform field (like very close to the Earth's surface), the relationship simplifies to $g = -\frac{\Delta \phi}{\Delta r}$. This is why $\Delta E_P = mgh$ works for small heights; $g$ is effectively constant, so the potential changes linearly with height.

3.6 Superposition of Gravitational Potential

Because potential is a scalar, calculating the total potential at a point due to multiple masses (e.g., a point between the Earth and the Moon) is straightforward. You simply calculate the potential from each mass individually and add them together.

$$\phi_{\text{total}} = \phi_1 + \phi_2 + \phi_3 + \dots$$ $$\phi_{\text{total}} = -\frac{GM_1}{r_1} - \frac{GM_2}{r_2} - \dots$$

Note: Since all individual potentials are negative, the total potential will be more negative than any single component.

3.7 Comparison: $g$ vs $\phi$

Feature	Gravitational Field Strength ($g$)	Gravitational Potential ($\phi$)
Type of Quantity	Vector (has direction)	Scalar (magnitude only)
Definition	Force per unit mass	Work done per unit mass
Formula	$g = \frac{GM}{r^2}$	$\phi = -\frac{GM}{r}$
SI Units	$\text{N kg}^{-1}$ or $\text{m s}^{-2}$	$\text{J kg}^{-1}$
Relationship	$g$ is the negative gradient of $\phi$	$\phi$ is the area under the $g-r$ graph
At Infinity	$g = 0$	$\phi = 0$

4. Worked Examples

Worked example 1 — Potential at a Point Between Two Masses

A point $P$ lies on the line joining the centers of the Earth and the Moon. The Earth has mass $M_E = 5.97 \times 10^{24} \text{ kg}$ and the Moon has mass $M_M = 7.35 \times 10^{22} \text{ kg}$. The distance between their centers is $3.84 \times 10^8 \text{ m}$. Calculate the gravitational potential at point $P$, located $3.00 \times 10^8 \text{ m}$ from the center of the Earth.

Step 1: Identify the distances.

• Distance from Earth center to $P$: $r_E = 3.00 \times 10^8 \text{ m}$
• Distance from Moon center to $P$: $r_M = (3.84 \times 10^8) - (3.00 \times 10^8) = 0.84 \times 10^8 \text{ m}$

Step 2: Calculate potential due to Earth ($\phi_E$). $$\phi_E = -\frac{GM_E}{r_E}$$ $$\phi_E = -\frac{(6.67 \times 10^{-11}) \times (5.97 \times 10^{24})}{3.00 \times 10^8}$$ $$\phi_E = -1.327 \times 10^6 \text{ J kg}^{-1}$$

Step 3: Calculate potential due to Moon ($\phi_M$). $$\phi_M = -\frac{GM_M}{r_M}$$ $$\phi_M = -\frac{(6.67 \times 10^{-11}) \times (7.35 \times 10^{22})}{0.84 \times 10^8}$$ $$\phi_M = -5.836 \times 10^4 \text{ J kg}^{-1}$$

Step 4: Sum the potentials (Scalar addition). $$\phi_{\text{total}} = \phi_E + \phi_M$$ $$\phi_{\text{total}} = (-1.327 \times 10^6) + (-0.058 \times 10^6)$$ $$\phi_{\text{total}} = -1.385 \times 10^6 \text{ J kg}^{-1}$$

Answer: $\phi = -1.39 \times 10^6 \text{ J kg}^{-1}$ (to 3 s.f.)

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Worked example 2 — Energy Required for Orbital Transfer

A satellite of mass $800 \text{ kg}$ is in a circular orbit around the Earth at an altitude of $500 \text{ km}$. It is moved to a new higher altitude of $2000 \text{ km}$. Calculate the work done to increase its gravitational potential energy. (Earth Mass $M = 5.97 \times 10^{24} \text{ kg}$, Earth Radius $R = 6.37 \times 10^6 \text{ m}$)

Step 1: Calculate the radial distances from the center.

• $r_1 = R + h_1 = 6.37 \times 10^6 + 0.50 \times 10^6 = 6.87 \times 10^6 \text{ m}$
• $r_2 = R + h_2 = 6.37 \times 10^6 + 2.00 \times 10^6 = 8.37 \times 10^6 \text{ m}$

Step 2: State the work done equation. $$W = \Delta E_P = GMm \left( \frac{1}{r_1} - \frac{1}{r_2} \right)$$

Step 3: Substitute values. $$W = (6.67 \times 10^{-11}) \times (5.97 \times 10^{24}) \times (800) \times \left( \frac{1}{6.87 \times 10^6} - \frac{1}{8.37 \times 10^6} \right)$$ $$W = (3.189 \times 10^{17}) \times (1.455 \times 10^{-7} - 1.195 \times 10^{-7})$$ $$W = (3.189 \times 10^{17}) \times (2.60 \times 10^{-8})$$

Step 4: Final calculation. $$W = 8.29 \times 10^9 \text{ J}$$

Answer: $8.29 \text{ GJ}$

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Worked example 3 — Determining $g$ from a $\phi-r$ Graph

A graph of gravitational potential $\phi$ against distance $r$ from the center of a planet shows that at $r = 1.5 \times 10^7 \text{ m}$, the tangent to the curve has a gradient of $4.5 \times 10^{-1} \text{ J kg}^{-1} \text{ m}^{-1}$. Determine the gravitational field strength at this point.

Step 1: Recall the relationship. $$g = -\text{gradient of } \phi-r \text{ graph}$$

Step 2: Analyze the gradient. On a standard $\phi-r$ graph, the potential is negative and increasing (getting less negative) as $r$ increases. Therefore, the gradient $\frac{d\phi}{dr}$ is positive.

Step 3: Calculate $g$. $$g = - (4.5 \times 10^{-1})$$ $$g = -0.45 \text{ N kg}^{-1}$$

Step 4: Interpret the result. The magnitude of the field strength is $0.45 \text{ N kg}^{-1}$. The negative sign indicates the field acts in the direction of decreasing $r$ (towards the planet).

Answer: $g = 0.45 \text{ N kg}^{-1}$

Key Equations

Equation	Description	Data Sheet?
$\phi = -\frac{GM}{r}$	Gravitational potential due to a point mass	Yes
$E_P = -\frac{GMm}{r}$	Gravitational potential energy of two point masses	No (Memorize)
$g = -\frac{d\phi}{dr}$	Relationship between field strength and potential	Yes
$W = m\Delta\phi$	Work done to move a mass between two points	No (Memorize)
$\Delta E_P = GMm(\frac{1}{r_1} - \frac{1}{r_2})$	Change in GPE between two radii	No (Memorize)

Common Mistakes to Avoid

• ❌ Wrong: Forgetting the negative sign in $\phi = -GM/r$.
  • ✓ Right: Potential must be negative. If your calculation results in a positive potential, you have made an error.
• ❌ Wrong: Using $r$ as the height above the surface.
  • ✓ Right: $r$ is always the distance from the center of mass. If a question gives "altitude" or "height," you must add the planet's radius: $r = R + h$.
• ❌ Wrong: Treating potential as a vector.
  • ✓ Right: Potential is a scalar. When finding the total potential from two masses, do not use Pythagoras. Simply add the two negative values together ($\phi_1 + \phi_2$).
• ❌ Wrong: Using $E_P = mgh$ for satellite orbits.
  • ✓ Right: $mgh$ only works when $g$ is constant (near the surface). For large changes in $r$, you must use $\Delta E_P = m\Delta\phi$.
• ❌ Wrong: Confusing the $1/r$ and $1/r^2$ relationships.
  • ✓ Right: Field strength $g \propto 1/r^2$ (Inverse Square Law). Potential $\phi \propto 1/r$.

Exam Tips

1. Definition Precision: When asked to define gravitational potential, ensure you include the phrase "small test mass." This ensures that the mass being moved does not significantly alter the field of the mass $M$ being measured.
2. Units Check: Always check your units. Potential is $\text{J kg}^{-1}$. If you are asked for Potential Energy, the unit is $\text{J}$. Forgetting to multiply by the mass $m$ is a common way to lose easy marks.
3. Graph Gradients: If provided with a $\phi-r$ graph, the gradient is often very small. Pay close attention to the powers of 10 on the axes (e.g., $\phi / 10^7 \text{ J kg}^{-1}$).
4. Significant Figures: Cambridge 9702 usually requires answers to 2 or 3 significant figures. Using $G = 6.67 \times 10^{-11}$ usually necessitates 3 s.f. in your final answer.
5. The "Change" in Potential: In questions asking for the "increase" in potential energy, ensure your final value is positive. Since $E_P$ becomes less negative as you move away, the change ($\text{final} - \text{initial}$) will be a positive value.
6. Equipotentials and Field Lines: Remember that equipotential surfaces are always at right angles ($90^\circ$) to gravitational field lines. If asked to draw them, ensure they are perpendicular at every intersection.