Temperature

Gravitational Potential

Learning Objectives

• Define gravitational potential energy (U) and gravitational potential (V).
• Derive the expression for gravitational potential from the gravitational field.
• Apply the formulas for:
  • Point‑mass fields
  • Uniform‑field (near‑Earth) approximation
  • Systems of two masses
• Explain the relationship between potential, field strength, and equipotential surfaces.
• Calculate escape velocity, orbital speed, and total mechanical energy of a satellite.
• Relate changes in gravitational potential energy to temperature changes (adiabatic heating, accretion heating).
• Identify common pitfalls and use the summary table for quick revision.

Key Definitions

• Gravitational potential energy (U): Work required to bring a mass $m$ from infinity to a point in the field of a mass $M$.
  $$U = -\frac{GMm}{r}$$
• Gravitational potential (V): Potential energy per unit mass.
  $$V = \frac{U}{m}= -\frac{GM}{r}\qquad\text{Units: J kg}^{-1}$$
• Gravitational field strength (g): Force per unit mass.
  $$\mathbf g(r)= -\frac{GM}{r^{2}}\hat{\mathbf r}\qquad\text{Units: m s}^{-2}$$
• Equipotential surface: A surface on which $V$ is constant; the field $\mathbf g$ is everywhere perpendicular to it.
• Reference point / sign convention: $V(\infty)=0$. Because the force is attractive, $V$ is negative at any finite $r$; a less‑negative value corresponds to a higher potential.
• Total mechanical energy of a two‑body system:
  $$E_{\text{tot}} = K + U = \frac12 mv^{2} -\frac{GMm}{r}$$

Deriving Gravitational Potential from the Field

The potential is defined by the line integral of the field:

$$V(r)= -\int_{\infty}^{r}\mathbf g\cdot d\mathbf r = -\int_{\infty}^{r}\frac{GM}{r^{2}}\,dr = -\frac{GM}{r}$$

This derivation is the basis of many AO2 “show that …” questions.

Fundamental Relationships (Point‑Mass Field)

Quantity	Symbol	Formula	Units
Gravitational field	$\mathbf g$	$-\dfrac{GM}{r^{2}}\hat{\mathbf r}$	m s⁻²
Gravitational potential energy	$U$	$-\dfrac{GMm}{r}$	J
Gravitational potential	$V$	$-\dfrac{GM}{r}$	J kg⁻¹
Field from potential	$\mathbf g$	$-abla V$ (in 1‑D: $g=-\dfrac{dV}{dr}$)	m s⁻²
Escape energy (per unit mass)	$|V(R)|$	$\dfrac{GM}{R}$	J kg⁻¹

Uniform‑Field Approximation (Near Earth)

When the height $h$ above the surface satisfies $h\ll R_{\earth}$, the field may be treated as constant ($g\approx9.81\;\text{m s}^{-2}$). Integrating $g$ gives

$$V = V_{0} - g h$$

If the reference level is sea‑level ($V_{0}=0$), then $V=-gh$.

Equipotential Surfaces & Field Direction

• $\mathbf g$ is always perpendicular to an equipotential surface.
• For a spherical mass the surfaces are concentric spheres; for an oblate or irregular body they are distorted.

Orbital Motion – Circular Orbits

For a satellite of mass $m$ in a circular orbit of radius $r$ about a planet of mass $M$:

• Centripetal condition: $\displaystyle\frac{mv^{2}}{r}= \frac{GMm}{r^{2}} \;\Rightarrow\; v^{2}= \frac{GM}{r}$
• Kinetic energy: $K = \frac12 mv^{2}= \frac{GMm}{2r}$
• Potential energy: $U = -\frac{GMm}{r}$
• Total mechanical energy: $E_{\text{tot}} = K+U = -\frac{GMm}{2r}$ (negative → bound orbit)

Escape Velocity & Escape Energy

To reach infinity with zero final speed, kinetic energy must equal the magnitude of the GPE at the launch radius $R$:

$$\frac12 mv_{\text{esc}}^{2}= \frac{GMm}{R}\quad\Rightarrow\quad v_{\text{esc}} = \sqrt{\frac{2GM}{R}}$$

Escape energy per unit mass is $|V(R)| = GM/R$.

Relation to Temperature

When gravitational potential energy is transformed into internal energy, the temperature of the material changes. Two syllabus‑relevant contexts are:

1. Adiabatic heating of descending air: loss of GPE → increase in kinetic energy → rise in temperature. The dry adiabatic lapse rate is ≈ 9.8 K km⁻¹.
2. Accretion heating: gas or dust falling into a deep potential well releases energy that raises the temperature of an accretion disc or protostar.

Worked Example 1 – Gravitational Potential at a Given Distance

1. Calculate the gravitational potential $V$ at $r = 2.0\times10^{7}\;\text{m}$ from the centre of the Earth.
   $M_{\earth}=5.97\times10^{24}\;\text{kg}$, $G=6.674\times10^{-11}\;\text{N m}^{2}\text{kg}^{-2}$.
2. Solution:

$$V = -\frac{GM}{r} = -\frac{(6.674\times10^{-11})(5.97\times10^{24})}{2.0\times10^{7}} \approx -1.99\times10^{7}\;\text{J kg}^{-1}$$

Interpretation: a 1 kg mass at this altitude has $1.99\times10^{7}$ J less potential energy than it would have at infinity.

Worked Example 2 – Temperature Rise from Falling Mass

A 2 kg block of metal (specific heat $c=900\;\text{J kg}^{-1}\text{K}^{-1}$) falls from $h=500\;\text{m}$ to the ground. Assume all lost GPE becomes internal energy.

1. Loss of GPE: $\displaystyle\Delta U = mgh = (2)(9.81)(500)=9.81\times10^{3}\;\text{J}$
2. Temperature increase: $\displaystyle\Delta T = \frac{\Delta U}{mc} = \frac{9.81\times10^{3}}{(2)(900)} \approx 5.45\;\text{K}$

The block’s temperature rises by about 5.5 °C.

Common Mistakes

• Confusing $V$ (J kg⁻¹) with $g$ (m s⁻²). Remember $V$ is energy per unit mass, $g$ is acceleration.
• Dropping the negative sign in $V=-GM/r$. The sign shows that work must be done *against* the field to increase $r$.
• Using $V=-gh$ when $h$ is not $\ll R_{\earth}$; the point‑mass formula must be used instead.
• Equating escape velocity with orbital speed; $v_{\text{esc}}=\sqrt{2}\,v_{\text{orb}}$.
• Assuming the total mechanical energy of a bound orbit is positive; it is always negative.

Summary Table

Quantity	Symbol	Formula	Units
Gravitational constant	$G$	$6.674\times10^{-11}$	N m² kg⁻²
Gravitational field (point mass)	$\mathbf g$	$-\dfrac{GM}{r^{2}}\hat{\mathbf r}$	m s⁻²
Gravitational potential energy	$U$	$-\dfrac{GMm}{r}$	J
Gravitational potential	$V$	$-\dfrac{GM}{r}$	J kg⁻¹
Uniform‑field potential (near Earth)	$V$	$V_{0}-gh$	J kg⁻¹
Escape velocity	$v_{\text{esc}}$	$\sqrt{\dfrac{2GM}{R}}$	m s⁻¹
Orbital speed (circular)	$v$	$\sqrt{\dfrac{GM}{r}}$	m s⁻¹
Total mechanical energy (circular orbit)	$E_{\text{tot}}$	$-\dfrac{GMm}{2r}$	J

Further Practice Questions

1. Derive $V = V_{0} - gh$ by integrating the uniform field $g$ from a reference height $h_{0}$ to $h$.
2. A satellite of mass $500\;\text{kg}$ orbits $300\;\text{km}$ above the Earth’s surface. Calculate:
   1. Orbital speed.
   2. Total mechanical energy.
   3. Escape velocity from that altitude.
3. A 3 kg rock falls from $800\;\text{m}$ onto snow. If $80\%$ of the lost GPE is converted to heat in $0.5\;\text{kg}$ of snow (specific heat $c=2100\;\text{J kg}^{-1}\text{K}^{-1}$), find the temperature rise of the snow.
4. Explain why equipotential surfaces around an oblate (flattened) planet are not perfect spheres and describe how this affects satellite orbits.
5. Show that the total mechanical energy of a satellite in a circular orbit is half the gravitational potential energy (i.e. $E_{\text{tot}} = \tfrac12 U$) and discuss the physical significance.

Key Take‑aways

• Gravitational potential is a scalar field: $V=-GM/r$, with $V(\infty)=0$.
• The field is the negative gradient of the potential: $\mathbf g = -abla V$.
• For a bound system $E_{\text{tot}}<0$; for escape $E_{\text{tot}}\ge0$.
• Escape velocity is $\sqrt{2}$ times the circular orbital speed at the same radius.
• Conversions between GPE and internal energy can raise temperature (adiabatic heating, accretion).
• Equipotential surfaces help visualise the direction of $\mathbf g$ and simplify work‑energy problems.

Actionable Review of the Lecture Notes (Cambridge AS & A‑Level Physics 9702)

Area of focus	What the syllabus expects	What the notes currently provide	Suggested improvement (concise, actionable)
Definition of terms	Clear, separate definitions of $U$, $V$, $\mathbf g$ and equipotential surfaces; correct units and sign conventions.	All definitions are present but $g$ is not listed as a separate term; sign‑convention explanation is brief.	Add a dedicated bullet for $g$, emphasise units (m s⁻²) and negative sign, and expand the sign‑convention note.
Derivation from the field	Show the integral $V=-\int_{\infty}^{r}\mathbf g\cdot d\mathbf r$ and explain each step.	Derivation is included but lacks a brief commentary on why the limits are $\infty$ to $r$.	Insert a one‑sentence justification for the limits and note that $V(\infty)=0$ is the reference.
Uniform‑field approximation	State when $h\ll R_{\earth}$, give $V=V_{0}-gh$, and discuss the choice of $V_{0}$.	Formula is given; the condition $h\ll R_{\earth}$ is mentioned but not highlighted.	Bold the condition, add a quick example (e.g., $h=100\,$m) to illustrate validity.
Orbital mechanics	Derive $v$, $K$, $U$, $E_{\text{tot}}$ for circular orbits and relate $E_{\text{tot}}$ to $U$.	All formulas are listed; the derivation linking $E_{\text{tot}}$ to $U$ is implicit.	Insert a short algebraic step: $E_{\text{tot}} = K+U = \frac12U$.
Temperature connection	Explain adiabatic lapse rate and accretion heating, linking $\Delta U$ to $\Delta T$ via $c$.	Two bullet points are provided; no quantitative example of the lapse rate.	Add a quick calculation: a 1 km descent → $\Delta T \approx 9.8\,$K.
Worked examples	Include at least one example that uses the uniform‑field formula and one that uses the point‑mass formula.	Both examples use point‑mass; uniform‑field example is missing.	Insert a short example: calculate $V$ for $h=50\,$m above ground using $V=-gh$.
Common mistakes	List typical errors and how to avoid them.	Good list, but could mention the pitfall of mixing up $V$ (per unit mass) with total $U$.	Add a bullet: “Do not confuse $V$ (J kg⁻¹) with $U$ (J); multiply by $m$ when required.”
Practice questions	Provide a range of AO1–AO2 questions covering derivations, calculations and conceptual explanations.	Five questions are present, covering most required skills.	Label each question with the relevant assessment objective (AO1, AO2, AO3).