CIE AS/A Level Physics 13.4 Gravitational potential Study Notes- 2025-2027 Syllabus
CIE AS/A Level Physics 13.4 Gravitational potential Study Notes – New Syllabus
CIE AS/A Level Physics 13.4 Gravitational potential Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on AS/A Level Physics latest syllabus with Candidates should be able to:
- define gravitational potential at a point as the work done per unit mass in bringing a small test mass from infinity to the point
- use ϕ = –GM / r for the gravitational potential in the field due to a point mass
- understand how the concept of gravitational potential leads to the gravitational potential energy of two point masses and use EP = –GMm / r
Definition of Gravitational Potential
The gravitational potential at a point in a gravitational field is the work done per unit mass in bringing a small test mass from infinity (where potential is defined as zero) to that point.
\( \mathrm{\phi = \dfrac{W}{m}} \)
- \( \mathrm{\phi} \): gravitational potential (J/kg)
- \( \mathrm{W} \): work done by an external agent in moving the mass (J)
- \( \mathrm{m} \): test mass (kg)
Important Note:
- At infinity, gravitational potential is defined as zero ( \( \mathrm{\phi_\infty = 0.} \) )
- Work must be done against the gravitational attraction to bring a mass closer to the source mass.
- Hence, gravitational potential near a mass is negative — the field is attractive, and potential decreases as one approaches the mass.
Example
Explain why the gravitational potential near Earth’s surface is negative and what its physical meaning is.
▶️ Answer / Explanation
At infinity, \( \mathrm{\phi_\infty = 0.} \) To bring a unit mass from infinity to the surface of Earth, work must be done against the gravitational attraction.
This means the potential energy decreases as we move towards Earth — hence the potential is negative.
Physical Meaning: A negative potential means that an external agent would need to do positive work to move a mass from that point back to infinity — escaping the gravitational attraction.
Gravitational Potential Due to a Point Mass
Derivation:
Consider a point mass \( \mathrm{M.} \) The gravitational field strength at a distance \( \mathrm{r} \) from it is:
\( \mathrm{g = \dfrac{GM}{r^2}} \)
The potential difference between infinity and a point at distance r r is obtained by integrating the field strength:
\( \mathrm{\phi = -\displaystyle \int_\infty^r g\,dr = -\int_\infty^r \dfrac{GM}{r^2}dr.} \)
Evaluating the integral:
\( \mathrm{\phi = -GM \left[-\dfrac{1}{r}\right]_\infty^r = -\dfrac{GM}{r}.} \)
Hence:
\( \mathrm{\phi = -\dfrac{GM}{r}} \)
- \( \mathrm{\phi} \): gravitational potential at distance \( \mathrm{r} \) (J/kg)
- \( \mathrm{G} \): gravitational constant (\( \mathrm{6.67\times10^{-11}\ N\,m^2/kg^2} \))
- \( \mathrm{M} \): mass of the attracting body (kg)
- \( \mathrm{r} \): distance from the mass centre (m)
Interpretation:
- \( \mathrm{\phi} \) is negative, since the gravitational force is attractive.
- Magnitude of \( \mathrm{\phi} \) increases (becomes more negative) as one moves closer to the mass.
- At infinity, \( \mathrm{\phi = 0.} \)
Relationship Between \( \mathrm{\phi} \) and \( \mathrm{g} \):
\( \mathrm{g = -\dfrac{d\phi}{dr}} \)
This shows that gravitational field strength is the negative gradient of the potential–distance graph.
Example
Calculate the gravitational potential due to Earth at its surface. Take \( \mathrm{G = 6.67\times10^{-11}\ N\,m^2/kg^2, \ M_E = 5.97\times10^{24}\ kg, \ R_E = 6.37\times10^6\ m.} \)
▶️ Answer / Explanation
Using \( \mathrm{\phi = -\dfrac{GM}{r}}: \)
\( \mathrm{\phi = -\dfrac{6.67\times10^{-11}\times5.97\times10^{24}}{6.37\times10^6} = -6.25\times10^7\ J/kg.} \)
Hence: The gravitational potential at Earth’s surface is \( \mathrm{-6.25\times10^7\ J/kg.} \)
Interpretation: This means that \( \mathrm{6.25\times10^7\ J} \) of work must be done to move 1 kg of mass from Earth’s surface to infinity.
Example
The Moon has a mass of \( \mathrm{7.35\times10^{22}\ kg} \) and a radius of \( \mathrm{1.74\times10^6\ m.} \) Calculate the gravitational potential at its surface and compare it to that on Earth.
▶️ Answer / Explanation
For the Moon:
\( \mathrm{\phi_M = -\dfrac{GM_M}{R_M} = -\dfrac{6.67\times10^{-11}\times7.35\times10^{22}}{1.74\times10^6} = -2.82\times10^6\ J/kg.} \)
For Earth (from Example 2): \( \mathrm{\phi_E = -6.25\times10^7\ J/kg.} \)
Comparison: \( \mathrm{|\phi_E| / |\phi_M| = 6.25\times10^7 / 2.82\times10^6 \approx 22.2.} \)
Hence: The potential well on Earth is over 22 times deeper than on the Moon.
Interpretation: This explains why the escape velocity on Earth is much greater — a stronger gravitational potential means more energy is needed to escape Earth’s gravity.
Gravitational Potential Energy of Two Point Masses
The concept of gravitational potential directly leads to the idea of gravitational potential energy (GPE) between two masses. The gravitational potential ( \mathrm{\phi} ) at a point is the potential energy per unit mass, i.e., the energy a unit mass would have due to its position in a gravitational field. Therefore, for any mass ( \mathrm{m} ) placed at a point where the potential is ( \mathrm{\phi}, ) the gravitational potential energy is given by:
\( \mathrm{E_P = m\phi.} \)
Substituting ( \mathrm{\phi = -\dfrac{GM}{r}} ), we get:
\( \mathrm{E_P = -\dfrac{GMm}{r}} \)
Derivation from the Definition of Work Done
Consider two point masses \( \mathrm{M} \) and \( \mathrm{m} \) separated by a distance \( \mathrm{r.} \) The gravitational force between them is:
\( \mathrm{F = \dfrac{GMm}{r^2}} \)
To move the smaller mass from infinity to a distance \( \mathrm{r,} \) the work done by an external agent against this attraction is:
\( \mathrm{W = \int_\infty^r F\,dr = \int_\infty^r \dfrac{GMm}{r^2}\,dr = GMm\left[-\dfrac{1}{r}\right]_\infty^r = -\dfrac{GMm}{r}.} \)
This work is stored as the gravitational potential energy of the system.
Hence,
\( \mathrm{E_P = -\dfrac{GMm}{r}} \)
Understanding the Negative Sign:
- The negative sign shows that the potential energy is zero at infinity and negative for all finite separations.
- This reflects the fact that work must be done against gravity to separate the masses completely to infinity (escape the field).
- The system is bound — energy must be supplied to overcome the gravitational attraction.
At Infinity: \( \mathrm{E_P = 0} \)
At Finite Distance: \( \mathrm{E_P = -\dfrac{GMm}{r}} \)
Example
Calculate the gravitational potential energy between Earth (\( \mathrm{M_E = 5.97\times10^{24}\ kg} \)) and a satellite of mass \( \mathrm{m = 1200\ kg} \) orbiting at a distance of \( \mathrm{r = 7.0\times10^6\ m.} \)
▶️ Answer / Explanation
Using \( \mathrm{E_P = -\dfrac{GMm}{r}}: \)
\( \mathrm{E_P = -\dfrac{6.67\times10^{-11} \times 5.97\times10^{24} \times 1200}{7.0\times10^6}} \)
\( \mathrm{E_P = -6.83\times10^{10}\ J.} \)
Hence: The gravitational potential energy of the satellite–Earth system is \( \mathrm{-6.83\times10^{10}\ J.} \)
Interpretation: This energy must be supplied to completely remove the satellite from Earth’s gravitational influence (to infinity).
Example
Two identical asteroids each of mass \( \mathrm{8.0\times10^{12}\ kg} \) are separated by a distance of \( \mathrm{5.0\times10^5\ m.} \) Calculate the gravitational potential energy of the system and find the work required to double their separation.
▶️ Answer / Explanation
Step 1: Initial potential energy
\( \mathrm{E_{P1} = -\dfrac{Gm^2}{r_1} = -\dfrac{6.67\times10^{-11}\times(8.0\times10^{12})^2}{5.0\times10^5}} \)
\( \mathrm{E_{P1} = -8.53\times10^{10}\ J.} \)
Step 2: Final potential energy (when \( \mathrm{r_2 = 1.0\times10^6\ m} \))
\( \mathrm{E_{P2} = -\dfrac{Gm^2}{r_2} = -\dfrac{6.67\times10^{-11}\times(8.0\times10^{12})^2}{1.0\times10^6} = -4.27\times10^{10}\ J.} \)
Step 3: Work done to double the separation
\( \mathrm{W = E_{P2} – E_{P1} = (-4.27\times10^{10}) – (-8.53\times10^{10}) = 4.26\times10^{10}\ J.} \)
Hence: \( \mathrm{4.26\times10^{10}\ J} \) of work is required to increase the separation from \( \mathrm{5.0\times10^5\ m} \) to \( \mathrm{1.0\times10^6\ m.} \)
Interpretation: This positive work corresponds to energy needed to overcome the gravitational attraction between the two asteroids.