use ϕ = –GM / r for the gravitational potential in the field due to a point mass

Gravitational Potential – Cambridge IGCSE / A‑Level (9702)

Learning Outcomes

• State the definition of gravitational potential (ϕ) and its units (J kg⁻¹).
• Derive and use the formula  ϕ = ‑GM/r for a point mass (or outside a spherically symmetric body).
• Show that the gravitational field  \(\mathbf{g}=‑\nabla ϕ\) and convert between field and potential.
• Apply ϕ to calculate gravitational potential energy (U = mϕ) and to solve orbital‑mechanics problems (escape speed, total specific mechanical energy).
• Analyse simple experimental data related to gravitational potential (e.g. satellite altitude‑speed data).
• Manipulate the formula algebraically, use scientific notation correctly and apply appropriate significant figures.

Key Equations

Definition of Gravitational Potential

Derivation of ϕ = ‑GM/r

1. Potential difference between two radii:

   \[

   \Delta ϕ = -\int{r1}^{r_2}\mathbf{g}\cdot d\mathbf{r}.

   \]

2. Insert the radial field of a point mass,

   \[

   \mathbf{g}= -\frac{GM}{r^{2}}\hat{r},\qquad d\mathbf{r}=dr\,\hat{r}.

   \]

3. Because the vectors are collinear,

   \[

   \Delta ϕ = -\int{r1}^{r_2}\!\!\left(-\frac{GM}{r^{2}}\right)dr

   = GM\int{r1}^{r_2}\frac{dr}{r^{2}}.

   \]

   Integrating,

   \[

   GM\int{r1}^{r2}\frac{dr}{r^{2}} = GM\Bigl[-\frac{1}{r}\Bigr]{r1}^{r2}

   = GM\left(\frac{1}{r1}-\frac{1}{r2}\right).

   \]

4. Choose the reference point at infinity: ϕ(∞)=0, i.e. let \(r_2\to\infty\). Then

   \[

   ϕ(r)-0 = -GM\left(\frac{1}{r}\right)

   \]

   giving the required result

   \[

   \boxed{ϕ(r)= -\frac{GM}{r}}.

   \]

Assumptions & Applicability (syllabus point 13‑4)

• The formula assumes a point mass or a spherically symmetric body. By the shell theorem, the same expression holds for any point outside a uniform sphere.
• Inside a uniform solid sphere the potential is ϕ = ‑\(GM/2R\)\( (3 - r^{2}/R^{2})\); this is not required for the 9702 exam but is useful for deeper understanding.

From Potential to Field (and Vice‑versa)

The gravitational field is the spatial gradient of the potential:

\[

\mathbf{g}= -\nabla ϕ.

\]

For a spherically symmetric field only the radial component matters:

\[

g_{r}= -\frac{dϕ}{dr}= -\frac{d}{dr}\!\left(-\frac{GM}{r}\right)= -\frac{GM}{r^{2}}.

\]

Conversely, if the field is known, the potential can be obtained by integration:

\[

ϕ(r)= -\int_{\infty}^{r} \mathbf{g}\cdot d\mathbf{r}

= -\int_{\infty}^{r}\!\! \left(-\frac{GM}{r^{2}}\right)dr

= -\frac{GM}{r}.

\]

Algebraic Rearrangements (AO2 practice)

• Find the distance r at which ϕ = ‑3.0 × 10⁷ J kg⁻¹ for Earth (M = 5.97 × 10²⁴ kg).

  \[

  r = -\frac{GM}{ϕ}.

  \]

• Express the mass of a planet in terms of a measured potential and distance:

  \[

  M = -\frac{ϕ\,r}{G}.

  \]

• When solving, keep extra digits through the calculation, then round the final answer to the appropriate number of significant figures (usually 2–3 for A‑Level).

Applications in Orbital Mechanics (13‑5)

Escape Speed

The work required to move a unit mass from r to infinity is –ϕ(r). Setting this equal to the kinetic energy \(\tfrac12 v^{2}\) gives

\[

\frac{1}{2}v_{\text{esc}}^{2}= \frac{GM}{r}

\;\;\Longrightarrow\;\;

v_{\text{esc}}= \sqrt{\frac{2GM}{r}}.

\]

Total Specific Mechanical Energy of a Satellite

For a satellite of mass m in a circular orbit of radius r:

• Kinetic energy per unit mass: \(\tfrac12 v^{2}= \dfrac{GM}{2r}\) (from \(mv^{2}/r = GMm/r^{2}\)).
• Potential energy per unit mass: ϕ = ‑\(GM/r\).

Hence the total specific mechanical energy

\[

E_{\text{tot}} = \frac{v^{2}}{2}+ϕ = -\frac{GM}{2r}.

\]

For an elliptical orbit the same expression holds with r replaced by the semi‑major axis a:

\[

E_{\text{tot}} = -\frac{GM}{2a}.

\]

Energy Diagram (qualitative)

A bound orbit (negative total energy) lies below the ϕ‑curve; an unbound trajectory (escape) corresponds to \(E\ge0\).

Worked Example – Escape Speed from the Moon

Given \(M_{\text{moon}} = 7.35\times10^{22}\,\text{kg}\), \(r = 1.74\times10^{6}\,\text{m}\).

1. Calculate \(GM\):

   \[

   GM = (6.674\times10^{-11})(7.35\times10^{22}) = 4.90\times10^{12}\,\text{m}^{3}\,\text{s}^{-2}.

   \]

2. Insert into the escape‑speed formula:

   \[

   v_{\text{esc}} = \sqrt{\frac{2GM}{r}}

   = \sqrt{\frac{2(4.90\times10^{12})}{1.74\times10^{6}}}

   = \sqrt{5.63\times10^{6}} \approx 2.37\times10^{3}\,\text{m s}^{-1}.

   \]

3. Result (3 sf): \(v_{\text{esc}} \approx 2.37\;\text{km s}^{-1}\).

Practical / Experimental Extension (AO3)

Activity – Using satellite data to verify ϕ = ‑GM/r

1. Obtain the orbital altitude (h) and orbital speed (v) of a low‑Earth satellite from a reliable source (e.g., NASA TLE data).
2. Compute the orbital radius \(r = R{\oplus}+h\) (with \(R{\oplus}=6.371\times10^{6}\,\text{m}\)).
3. Calculate the measured specific mechanical energy:

   \[

   E{\text{meas}} = \frac{v^{2}}{2} - \frac{GM{\oplus}}{r}.

   \]

4. Compare with the theoretical value \(E{\text{theo}} = -\dfrac{GM{\oplus}}{2r}\). Discuss any discrepancy (measurement error, atmospheric drag, Earth’s oblateness, non‑circularity).
5. Conclude whether the data support the potential formula.

Common Mistakes (and How to Avoid Them)

• Sign error: ϕ is always negative for an attractive point mass. Remember the definition includes a minus sign.
• Reference point confusion: The standard zero is at infinity. If you choose a different reference (e.g., Earth's surface), state it explicitly and adjust the formula accordingly.
• Mixing up ϕ and \(\mathbf{g}\): ϕ is a scalar (J kg⁻¹); \(\mathbf{g}\) is a vector (m s⁻²). Use \(\mathbf{g}=‑\nabla ϕ\) to convert between them.
• Units: ϕ (J kg⁻¹) vs. U (J). Multiply by the test mass m when you need total energy.
• Significant figures: Carry extra digits through calculations; round only in the final answer, matching the precision of the given data.

Practice Questions

1. Conceptual (MCQ) – Which statement is true for the gravitational potential of a point mass?
   

   1. ϕ is positive everywhere.
   2. ϕ becomes zero at the centre of the mass.
   3. ϕ approaches zero as \(r\to\infty\).
   4. The magnitude of ϕ increases with increasing r.

   Answer: iii

2. Numerical – rearrangement – The measured potential at a distance of \(1.0\times10^{7}\,\text{m}\) from an unknown planet is \(-4.0\times10^{6}\,\text{J kg}^{-1}\). Find the planet’s mass. (G = 6.674×10⁻¹¹ N m² kg⁻²)
   

   Solution: \(M = -\dfrac{ϕ r}{G} = \dfrac{(4.0\times10^{6})(1.0\times10^{7})}{6.674\times10^{-11}} \approx 6.0\times10^{23}\,\text{kg}\) (3 sf).

3. Orbital energy – A satellite of mass 800 kg orbits Earth in a circular orbit of radius \(7.0\times10^{6}\,\text{m}\). Calculate:

   • (i) its speed,
   • (ii) its total mechanical energy (U + K),
   • (iii) the energy that must be supplied to move it to a geostationary orbit (\(r = 4.22\times10^{7}\,\text{m}\)).

   Solution sketch:

   1. \(v = \sqrt{GM/r} = \sqrt{(3.986\times10^{14})/7.0\times10^{6}} \approx 7.5\times10^{3}\,\text{m s}^{-1}\).
   2. Specific energy \(E = -GM/(2r) = -2.85\times10^{7}\,\text{J kg}^{-1}\). Total energy \(U+K = mE \approx -2.28\times10^{10}\,\text{J}\).
   3. New specific energy \(E' = -GM/(2r') = -4.73\times10^{6}\,\text{J kg}^{-1}\). Energy required = \(m(E' - E) \approx 1.86\times10^{10}\,\text{J}\).

4. Escape speed – Show that the escape speed from the surface of a planet of radius R and mass M is \(\sqrt{2GM/R}\). Then compute the escape speed from Mars (\(M = 6.42\times10^{23}\,\text{kg}\), \(R = 3.39\times10^{6}\,\text{m}\)).
   

   Derivation: Set kinetic energy \(\tfrac12 v^{2}\) equal to the work \(-ϕ(R)=GM/R\). → \(v_{\text{esc}}=\sqrt{2GM/R}\).
   

   Numerical: \(v_{\text{esc}} = \sqrt{2(6.674\times10^{-11})(6.42\times10^{23})/3.39\times10^{6}} \approx 5.0\times10^{3}\,\text{m s}^{-1}\) (5.0 km s⁻¹).

5. Data‑analysis (AO3) – Using the satellite activity, the measured specific mechanical energy is \(-3.0\times10^{7}\,\text{J kg}^{-1}\) while the theoretical value from \(-GM/(2r)\) is \(-3.1\times10^{7}\,\text{J kg}^{-1}\). Discuss possible reasons for the 3 % difference.
   

   Possible reasons:

   • Atmospheric drag reduces the speed slightly, lowering kinetic energy.
   • Earth is not a perfect sphere; equatorial bulge changes the effective GM at the satellite’s latitude.
   • Measurement uncertainties in altitude (r) and speed (v) propagate into the energy calculation.
   • Assumption of a perfectly circular orbit; a small eccentricity would alter the instantaneous kinetic energy.

Summary

The gravitational potential of a point mass is a fundamental scalar quantity defined as the work per unit mass required to bring a test mass from infinity to a distance r. Its compact form,

\[

ϕ = -\frac{GM}{r},

\]

follows directly from the definition of potential and the inverse‑square gravitational field. The relationship \(\mathbf{g}=‑\nabla ϕ\) links the scalar potential to the vector field, allowing easy conversion between them. ϕ underpins calculations of gravitational potential energy, escape velocity, and the total specific mechanical energy of satellites—key topics in the Cambridge 9702 syllabus. Mastery of the formula, careful handling of signs and units, and the ability to manipulate the expression algebraically and to interpret real data will give students confidence in both calculation‑focused and conceptual exam questions.