\$U = -\frac{GMm}{r}\$
\$V = \frac{U}{m}= -\frac{GM}{r}\qquad\text{Units: J kg}^{-1}\$
\$\mathbf g(r)= -\frac{GM}{r^{2}}\hat{\mathbf r}\qquad\text{Units: m s}^{-2}\$
\$E_{\text{tot}} = K + U = \frac12 mv^{2} -\frac{GMm}{r}\$
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.
| 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\$ | \$-\nabla V\$ (in 1‑D: \$g=-\dfrac{dV}{dr}\$) | m s⁻² |
| Escape energy (per unit mass) | \$|V(R)|\$ | \$\dfrac{GM}{R}\$ | J kg⁻¹ |
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\$.
For a satellite of mass \$m\$ in a circular orbit of radius \$r\$ about a planet of mass \$M\$:
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\$.
When gravitational potential energy is transformed into internal energy, the temperature of the material changes. Two syllabus‑relevant contexts are:
\$M_{\earth}=5.97\times10^{24}\;\text{kg}\$, \$G=6.674\times10^{-11}\;\text{N m}^{2}\text{kg}^{-2}\$.
\$\$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.
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.
= \frac{9.81\times10^{3}}{(2)(900)}
\approx 5.45\;\text{K}$
The block’s temperature rises by about 5.5 °C.
| 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 |
| 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). |
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