Concept Card: Gravitational Potential Energy – Derivations, Systems & Self-Energy
1. Conservative Nature of Gravity and Definition of Potential Energy:
Newton’s law of universal gravitation describes a conservative central force field:
$$\vec{F}_g = -\frac{G M m}{r^2} \hat{r}$$
Because the curl of this central force is identically zero ($\vec{\nabla} \times \vec{F}_g = \vec{0}$), the work done by gravity along any path between two points depends solely on the initial and final endpoints, completely independent of the path taken.
By definition, the change in gravitational potential energy $\Delta U$ of a two-body system as the separation changes from $r_1$ to $r_2$ is the negative of the work done by the internal conservative gravitational force:
$$\Delta U = U(r_2) – U(r_1) = -W_{\text{grav}} = -\int_{r_1}^{r_2} \vec{F}_g \cdot d\vec{r}$$
Since $\vec{F}_g = -\frac{G M m}{r^2} \hat{r}$ and $d\vec{r} = dr \hat{r}$:
$$W_{\text{grav}} = \int_{r_1}^{r_2} \left(-\frac{G M m}{r^2}\right) dr = \left[ \frac{G M m}{r} \right]_{r_1}^{r_2} = G M m \left( \frac{1}{r_2} – \frac{1}{r_1} \right)$$
$$\Delta U = -W_{\text{grav}} = -G M m \left( \frac{1}{r_2} – \frac{1}{r_1} \right) = -\frac{G M m}{r_2} – \left(-\frac{G M m}{r_1}\right)$$
2. Rigorous Derivation of Absolute Potential Energy ($U(r)$)
Potential energy is fundamentally defined only up to an arbitrary additive reference constant. In celestial mechanics and universal gravitation, the universal standard convention sets the reference point of zero potential energy at infinite separation:
$$U(\infty) = 0$$
Let an external agent bring a test mass $m$ from infinity ($r_1 = \infty$) to a distance $r$ ($r_2 = r$) from mass $M$ without acceleration ($v = 0$ at all times):
$$U(r) – U(\infty) = -\int_\infty^r \vec{F}_g \cdot d\vec{r} = -\int_\infty^r \left(-\frac{G M m}{r’^2}\right) dr’ = -\left[ \frac{G M m}{r’} \right]_\infty^r = -\left( \frac{G M m}{r} – 0 \right)$$
Governing Formula for Gravitational Potential Energy:
$$\mathbf{U(r) = -\frac{G M m}{r}}$$
Physical Significance of the Negative Sign:
- Attractive Force: Because gravity is purely attractive, the gravitational field performs positive work ($W_{\text{grav}} > 0$) as two masses approach each other from infinity. Consequently, the system’s potential energy drops below its zero reference value at infinity, becoming negative.
- Bound State (Binding Energy): A negative total potential energy indicates a bound physical system. To disassemble the masses to infinity, an external agent must supply positive work:
$$W_{\text{ext}} = U(\infty) – U(r) = 0 – \left(-\frac{G M m}{r}\right) = +\mathbf{\frac{G M m}{r}}$$
This quantity is the binding energy of the two-particle system.
3. Potential Energy of a Multi-Particle System ($N$ Discrete Masses)
Gravitation obeys the principle of linear superposition. For an isolated system consisting of $N$ point masses $m_1, m_2, \dots, m_N$, the total gravitational interaction potential energy is the algebraic sum of the potential energies of all unique interacting pairs:
Total Multi-Particle Potential Energy:
$$\mathbf{U_{\text{total}} = \sum_{1 \le i < j \le N} -\frac{G m_i m_j}{r_{ij}} = -\frac{1}{2} \sum_{i=1}^N \sum_{j \ne i}^N \frac{G m_i m_j}{r_{ij}}}$$
The total number of interacting pairwise combinations in an $N$-particle system is:
$$N_{\text{pairs}} = \binom{N}{2} = \frac{N(N – 1)}{2}$$
- For 3 particles (e.g., triangle): $\binom{3}{2} = 3$ interaction pairs.
- For 4 particles (e.g., square): $\binom{4}{2} = 6$ interaction pairs (4 along the edges, 2 along the diagonals).
- For 8 particles (e.g., cube vertices): $\binom{8}{2} = 28$ interaction pairs.
4. Gravitational Self-Energy of Continuous Mass Distributions
The gravitational self-energy ($U_{\text{self}}$) of an extended celestial body is the total work done by gravity (or the negative of work done by an external agent) in assembling the entire body, mass element by mass element, from an initially dispersed state at infinity.
(A) Uniform Thin Spherical Shell of Mass $M$ and Radius $R$:
$$\mathbf{U_{\text{self, shell}} = -\frac{G M^2}{2R}}$$
(B) Rigorous Derivation for a Uniform Solid Sphere of Mass $M$ and Radius $R$:
Consider the sphere being built from within, layer by concentric spherical layer.
Suppose a solid spherical core of radius $r$ ($0 \le r \le R$) and uniform density $\rho = \frac{M}{\frac{4}{3}\pi R^3}$ has already been assembled.
The mass of this core is:
$$M(r) = \frac{4}{3}\pi r^3 \rho = M \left(\frac{r}{R}\right)^3$$
Now, bring an infinitesimal mass shell of thickness $dr$ from infinity and deposit it onto the core.
The volume of this shell is $dV = 4\pi r^2 dr$, and its mass is:
$$dm = \rho dV = \left(\frac{M}{\frac{4}{3}\pi R^3}\right) 4\pi r^2 dr = \frac{3 M}{R^3} r^2 dr$$
Since all parts of this thin shell settle at distance $r$ from the center of the core of mass $M(r)$, the infinitesimal change in potential energy is:
$$dU = -\frac{G M(r) dm}{r} = -\frac{G \left[M \left(\frac{r}{R}\right)^3\right] \left[\frac{3M}{R^3} r^2 dr\right]}{r} = -\frac{3 G M^2}{R^6} r^4 dr$$
Integrating over the assembly of the entire sphere from $r = 0$ to $r = R$:
$$U_{\text{self, sphere}} = -\frac{3 G M^2}{R^6} \int_0^R r^4 dr = -\frac{3 G M^2}{R^6} \left[ \frac{R^5}{5} \right]$$
Gravitational Self-Energy of a Uniform Solid Sphere:
$$\mathbf{U_{\text{self, sphere}} = -\frac{3}{5} \frac{G M^2}{R}}$$
Astrophysical Significance: If a uniform proto-stellar cloud of mass $M$ collapses gravitationally from an infinite cloud to a radius $R$, it releases a gravitational binding energy equal to $\Delta E = \frac{3}{5}\frac{G M^2}{R}$, heating the stellar core (the Kelvin-Helmholtz mechanism).
5. Potential Energy Gain Near Earth’s Surface vs. Exact Formula
When an object of mass $m$ is elevated from the Earth’s surface ($r_1 = R$) to an altitude $h$ ($r_2 = R + h$):
$$\Delta U = U(R + h) – U(R) = -\frac{G M m}{R + h} – \left(-\frac{G M m}{R}\right) = G M m \left( \frac{1}{R} – \frac{1}{R + h} \right) = \frac{G M m h}{R(R + h)}$$
Using $G M = g_0 R^2$:
$$\mathbf{\Delta U = \frac{m g_0 h}{1 + \frac{h}{R}}}$$
- When $h \ll R$ (near the surface): $\frac{h}{R} \to 0 \implies \mathbf{\Delta U \approx m g_0 h}$.
- When $h = R$: $\Delta U = \frac{m g_0 R}{1 + 1} = \mathbf{\frac{1}{2} m g_0 R}$.
- When $h \to \infty$ (complete escape to infinity): $\Delta U = m g_0 R$.
6. High-Yield Exam Pitfalls & Analytical Traps in JEE:
- Trap 1 (Pairwise Counting Overlap): When computing the potential energy of a multi-particle system, never count any interaction pair twice! The formula is $\sum_{i < j} -\frac{G m_i m_j}{r_{ij}}$, NOT $\sum_{i} \sum_j$.
- Trap 2 (Potential vs. Potential Energy):
- Gravitational Potential ($V$): Potential energy per unit mass ($V = U/m = -\frac{G M}{r}$), unit: $\text{J/kg}$.
- Gravitational Potential Energy ($U$): Total energy of interaction ($U = m V = -\frac{G M m}{r}$), unit: $\text{Joules}$.
- Trap 3 (Sign of Work in Potential Energy Problems):
$$W_{\text{ext}} = +\Delta U = U_{\text{final}} – U_{\text{initial}}$$
$$W_{\text{gravity}} = -\Delta U = U_{\text{initial}} – U_{\text{final}}$$
If a particle falls closer to Earth ($r_f < r_i$), gravity does positive work, and potential energy decreases (becomes more negative).
Solved Examples
Example 1 (Direct Multi-Particle System – Square Geometry & Disassembly Work):
Four identical point masses, each of mass $m = 2.0\text{ kg}$, are held at the four vertices of a square of side length $a = 0.50\text{ m}$.
(a) Calculate the total gravitational potential energy $U_{\text{total}}$ of the four-mass configuration.
(b) How much external work $W_{\text{disassemble}}$ is required to expand the system so that all four masses are separated to infinity?
(c) How much work must be done by an external agent to remove just one mass to infinity while the remaining three masses remain fixed in place?
Solution:
(a) Total Gravitational Potential Energy:
For $N = 4$ masses, there are $\binom{4}{2} = 6$ unique pairwise interactions:
- 4 pairs along the edges of the square, each separated by distance $r = a$:
$$U_{\text{edges}} = 4 \times \left(-\frac{G m^2}{a}\right) = -\frac{4 G m^2}{a}$$ - 2 pairs along the diagonals of the square, each separated by distance $r = a\sqrt{2}$:
$$U_{\text{diagonals}} = 2 \times \left(-\frac{G m^2}{a\sqrt{2}}\right) = -\frac{\sqrt{2} G m^2}{a}$$
Total potential energy:
$$U_{\text{total}} = U_{\text{edges}} + U_{\text{diagonals}} = -\frac{G m^2}{a} \left(4 + \sqrt{2}\right)$$
Substitute numerical values ($m = 2.0\text{ kg}, a = 0.50\text{ m}$):
$$\frac{G m^2}{a} = \frac{G (2.0)^2}{0.50} = \frac{4.0 G}{0.50} = 8.0 G$$
$$U_{\text{total}} = -(8.0 G) (4 + 1.4142) = -(8.0)(5.4142) G = -43.314 G$$
With $G = 6.674 \times 10^{-11}\text{ N}\cdot\text{m}^2/\text{kg}^2$:
$$U_{\text{total}} = -43.314 \times (6.674 \times 10^{-11}) = -\mathbf{2.89 \times 10^{-9}\text{ Joules}}$$
(b) Work to Disassemble the Entire System:
$$W_{\text{disassemble}} = U(\infty) – U_{\text{total}} = 0 – (-2.89 \times 10^{-9}\text{ J}) = +\mathbf{2.89 \times 10^{-9}\text{ Joules}}$$
(c) Work to Remove One Particle:
The removed particle interacted with 3 other particles:
– 2 adjacent particles at distance $a$.
– 1 diagonal particle at distance $a\sqrt{2}$.
The interaction potential energy of this single particle with the rest of the system is:
$$U_1 = 2 \left(-\frac{G m^2}{a}\right) + 1 \left(-\frac{G m^2}{a\sqrt{2}}\right) = -\frac{G m^2}{a} \left(2 + \frac{1}{\sqrt{2}}\right)$$
$$U_1 = -(8.0 G)(2 + 0.7071) = -(8.0)(2.7071) G = -21.657 G = -1.445 \times 10^{-9}\text{ J}$$
The work required to remove this one particle to infinity is:
$$W_{\text{ext}} = -U_1 = +\mathbf{1.45 \times 10^{-9}\text{ Joules}}$$
(Notice that $U_{\text{total}}$ is not simply $4 \times U_1$; rather, $U_{\text{total}} = \frac{1}{2} \sum_{i=1}^4 U_i = \frac{1}{2}(4 U_1) = 2 U_1$, verifying the double-sum identity!).
Example 2 (Astrophysical Self-Energy – Total Gravitational Binding Energy of Earth):
Model the Earth as a uniform solid sphere of mass $M = 6.0 \times 10^{24}\text{ kg}$ and mean radius $R = 6.4 \times 10^6\text{ m}$.
(a) Compute the gravitational self-energy $U_{\text{self}}$ of the Earth.
(b) How much energy would be required to completely disintegrate and disperse the Earth into cosmic dust at infinity?
(c) If this total gravitational binding energy were released as radiant heat over a contraction period of $30\text{ million years}$, calculate the average power output in Watts.
Solution:
(a) Gravitational Self-Energy:
$$U_{\text{self}} = -\frac{3}{5} \frac{G M^2}{R}$$
Substitute values:
$$M^2 = (6.0 \times 10^{24})^2 = 3.6 \times 10^{49}\text{ kg}^2$$
$$U_{\text{self}} = -\frac{3}{5} \frac{(6.674 \times 10^{-11})(3.6 \times 10^{49})}{6.4 \times 10^6} = -0.6 \times \frac{2.4026 \times 10^{39}}{6.4 \times 10^6} = -\mathbf{2.25 \times 10^{32}\text{ Joules}}$$
(b) Energy to Completely Disintegrate the Earth:
$$E_{\text{disintegrate}} = 0 – U_{\text{self}} = +\mathbf{2.25 \times 10^{32}\text{ Joules}}$$
(This colossal quantity is equivalent to the total energy output of the Sun over nearly a week!).
(c) Kelvin-Helmholtz Contraction Power:
Time duration:
$$t = 30 \times 10^6\text{ years} = (3.0 \times 10^7)(3.156 \times 10^7\text{ s}) \approx 9.47 \times 10^{14}\text{ s}$$
Average power:
$$P_{\text{avg}} = \frac{2.25 \times 10^{32}\text{ J}}{9.47 \times 10^{14}\text{ s}} \approx \mathbf{2.38 \times 10^{17}\text{ Watts}}$$
Example 3 (Coupled Dynamics – Mutual Gravitational Infall of Two Free Masses):
Two point particles of masses $m_1 = 3.0\text{ kg}$ and $m_2 = 6.0\text{ kg}$ are initially held stationary in empty space at a separation distance $r_0 = 12.0\text{ m}$. They are released simultaneously from rest and accelerate toward each other under mutual gravitational attraction.
(a) When their separation distance reduces to $r = 4.0\text{ m}$, calculate the linear speeds $v_1$ and $v_2$ of each particle.
(b) Determine the relative velocity of approach $v_{\text{rel}}$ at this separation.
(c) What is the position of their center of mass throughout the motion?
Solution:
(a) Conservation of Linear Momentum & Mechanical Energy:
Since no external forces act on the two-particle system, its Center of Mass remains permanently stationary ($v_{cm} = 0$).
By conservation of linear momentum:
$$m_1 v_1 = m_2 v_2 \implies (3.0) v_1 = (6.0) v_2 \implies \mathbf{v_1 = 2 v_2}$$
By conservation of mechanical energy:
$$E_{\text{initial}} = E_{\text{final}}$$
$$0 – \frac{G m_1 m_2}{r_0} = \left( \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \right) – \frac{G m_1 m_2}{r}$$
The gain in kinetic energy equals the loss in gravitational potential energy:
$$\Delta K = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = G m_1 m_2 \left( \frac{1}{r} – \frac{1}{r_0} \right)$$
Substitute $v_1 = 2 v_2$:
$$\frac{1}{2}(3.0)(2 v_2)^2 + \frac{1}{2}(6.0) v_2^2 = \frac{1}{2}(12.0) v_2^2 + 3.0 v_2^2 = 6.0 v_2^2 + 3.0 v_2^2 = 9.0 v_2^2$$
Right-hand side:
$$G (3.0)(6.0) \left( \frac{1}{4.0} – \frac{1}{12.0} \right) = 18.0 G \left( \frac{3 – 1}{12.0} \right) = 18.0 G \left( \frac{2}{12.0} \right) = 3.0 G$$
Equating:
$$9.0 v_2^2 = 3.0 G \implies v_2^2 = \frac{G}{3.0} \implies \mathbf{v_2 = \sqrt{\frac{G}{3.0}}}$$
$$v_1 = 2 v_2 = \mathbf{2\sqrt{\frac{G}{3.0}} = \sqrt{\frac{4G}{3.0}}}$$
Numerically (with $G = 6.674 \times 10^{-11}$):
$$v_2 = \sqrt{\frac{6.674 \times 10^{-11}}{3}} = \mathbf{4.72 \times 10^{-6}\text{ m/s}}$$
$$v_1 = 2 v_2 = \mathbf{9.43 \times 10^{-6}\text{ m/s}}$$
(b) Relative Speed of Approach:
$$v_{\text{rel}} = v_1 + v_2 = 3 v_2 = 3 \sqrt{\frac{G}{3.0}} = \mathbf{\sqrt{3.0 G}} = \mathbf{1.415 \times 10^{-5}\text{ m/s}}$$
(c) Center of Mass Position:
The Center of Mass divides the separation distance inversely as the ratio of their masses:
$$r_1 = \frac{m_2}{m_1 + m_2} r_0 = \frac{6.0}{9.0}(12.0) = 8.0\text{ m from } m_1$$
The center of mass remains perfectly stationary at this fixed spatial location throughout the entire infall until impact.
Example 4 (Continuous Mass Integration – Potential Energy of a Rod and Point Mass):
A uniform thin rod of mass $M$ and length $L$ lies horizontally along the $x$-axis from $x = d$ to $x = d + L$. A point particle of mass $m$ is positioned at the coordinate origin $x = 0$.
(a) Derive the exact gravitational potential energy $U$ of the combined system.
(b) Differentiate your result to verify the exact gravitational force $F = -\frac{dU}{dd}$ derived in Lecture 65.
Solution:
(a) Integration Over Continuous Elements:
The linear mass density of the rod is $\lambda = \frac{M}{L}$.
An infinitesimal element of length $dx$ at position $x$ has mass $dm = \lambda dx = \frac{M}{L} dx$.
The potential energy between particle $m$ at the origin and this element $dm$ at distance $x$ is:
$$dU = -\frac{G m dm}{x} = -\frac{G M m}{L} \frac{dx}{x}$$
Integrating over the entire rod from $x = d$ to $x = d + L$:
$$U = -\frac{G M m}{L} \int_d^{d+L} \frac{dx}{x} = -\frac{G M m}{L} [\ln x]_d^{d+L}$$
$$\mathbf{U(d) = -\frac{G M m}{L} \ln\left( \frac{d + L}{d} \right) = -\frac{G M m}{L} \ln\left( 1 + \frac{L}{d} \right)}$$
(b) Verification of Gravitational Force:
The conservative force exerted on the point mass $m$ (toward $+x$, i.e., in the direction of decreasing separation $d$) is:
$$F = -\frac{dU}{dd} = -\frac{d}{dd} \left[ -\frac{G M m}{L} (\ln(d + L) – \ln d) \right]$$
$$F = \frac{G M m}{L} \left[ \frac{1}{d + L} – \frac{1}{d} \right] = \frac{G M m}{L} \left[ \frac{d – (d + L)}{d(d + L)} \right] = -\frac{G M m}{d(d + L)}$$
The negative sign indicates an attractive force directed toward the rod (along $-d$, pulling mass $m$ toward the rod).
Its magnitude is precisely:
$$|F| = \mathbf{\frac{G M m}{d(d + L)}}$$
This rigorously confirms the force derived via direct Newton-integral in Lecture 65!
Practice Worksheet: Gravitational Potential Energy & Self-Energy
Section A: Single Correct Questions (Q1 to Q5)
Q1. Three identical particles of mass $m$ are situated at the vertices of an equilateral triangle of side length $a$. The total gravitational potential energy of the system is:
(A) $-\frac{G m^2}{a}$
(B) $-2\frac{G m^2}{a}$
(C) $-3\frac{G m^2}{a}$
(D) $-6\frac{G m^2}{a}$
Q2. What is the minimum work required to lift a body of mass $m$ from the Earth’s surface ($r = R$) to an altitude equal to the Earth’s radius ($h = R$)?
(A) $m g_0 R$
(B) $\frac{1}{2} m g_0 R$
(C) $\frac{1}{4} m g_0 R$
(D) $2 m g_0 R$
Q3. The ratio of the gravitational self-energy of a uniform solid sphere of mass $M$ and radius $R$ to that of a thin uniform spherical shell of the same mass $M$ and radius $R$ is:
(A) $3 : 2$
(B) $5 : 3$
(C) $6 : 5$
(D) $5 : 6$
Q4. Two particles of mass $m$ each are initially separated by a distance $r$. How much work must an external agent perform to slowly increase their separation to $2r$?
(A) $\frac{G m^2}{r}$
(B) $\frac{G m^2}{2r}$
(C) $-\frac{G m^2}{2r}$
(D) $\frac{2 G m^2}{r}$
Q5. A body of mass $m$ is dropped from rest from an altitude $h = R$ above the Earth’s surface. Neglecting air resistance, the speed with which the body strikes the Earth’s surface is:
(A) $\sqrt{2 g_0 R}$
(B) $\sqrt{g_0 R}$
(C) $\sqrt{\frac{g_0 R}{2}}$
(D) $\sqrt{\frac{g_0 R}{4}}$
Section B: One or More than One Correct Questions (Q6 to Q8)
Q6. Which of the following statements regarding gravitational potential energy is/are correct?
(A) Choosing the reference point at infinity ($U(\infty) = 0$) makes the potential energy of any bound gravitational system strictly negative.
(B) The change in potential energy between two states is independent of the reference point chosen.
(C) The conservative force $\vec{F}$ and potential energy $U$ are related by $\vec{F} = -\vec{\nabla} U$.
(D) If $U(r) = -\frac{G M m}{r}$, the force is attractive because the potential energy decreases as separation decreases.
Q7. Four identical masses $m$ are located at the vertices of a square of side $a$. Which of the following statements is/are correct?
(A) The total gravitational potential energy of the system is $U = -\frac{G m^2}{a}(4 + \sqrt{2})$.
(B) If the side length of the square is doubled from $a$ to $2a$, the potential energy becomes half of its initial value ($U/2$).
(C) The work required to expand the square from side $a$ to $2a$ is $\frac{G m^2}{2a}(4 + \sqrt{2})$.
(D) The potential at the center of the square is $V = -4\sqrt{2} \frac{G m}{a}$.
Q8. A uniform solid spherical planet of mass $M$ and radius $R$ contracts quasi-statically under its own gravity to a smaller radius $R/2$ while preserving spherical symmetry and uniform density. Which of the following statements is/are correct?
(A) The gravitational self-energy of the planet decreases (becomes more negative).
(B) The change in gravitational self-energy is $\Delta U_{\text{self}} = -\frac{3}{5}\frac{G M^2}{R}$.
(C) The total work done by gravity during this contraction is $+\frac{3}{5}\frac{G M^2}{R}$.
(D) This released gravitational energy is converted into thermal radiation according to the Virial Theorem.
Section C: Numerical Value Type Questions (Q9 to Q10)
Q9. An external agent performs an amount of work $W = 0.75 m g_0 R$ to slowly lift a body of mass $m$ from the surface of Earth to an altitude $h$. Given that the radius of Earth is $R$, the altitude can be expressed as $h = n R$. Find the integer value of $n$.
Q10. Three identical particles of mass $m = 2.0\text{ kg}$ each are initially held at the vertices of an equilateral triangle of side $a = 1.0\text{ m}$. The external work required to slowly expand the system so that the side of the equilateral triangle becomes $2a = 2.0\text{ m}$ is equal to $k G$ Joules, where $G$ is the universal gravitational constant. Find the integer value of $k$.
Comprehensive Solutions and Answer Key
Answer Key
| Question | Correct Answer | Core Physics Concept |
|---|---|---|
| Q1 | (C) | Multi-particle potential energy: $\binom{3}{2} = 3$ pairs $\implies -3 \frac{G m^2}{a}$ |
| Q2 | (B) | Work near Earth: $\Delta U = \frac{m g_0 h}{1 + h/R} = \frac{1}{2} m g_0 R$ |
| Q3 | (C) | Self-energy ratio: $\frac{3/5}{1/2} = 6/5$ |
| Q4 | (B) | External work: $W = U(2r) – U(r) = \frac{G m^2}{2r}$ |
| Q5 | (B) | Energy conservation: $\frac{1}{2}mv^2 = \frac{1}{2} m g_0 R \implies v = \sqrt{g_0 R}$ |
| Q6 | (A, B, C, D) | Fundamental principles and vector gradient of potential energy |
| Q7 | (A, B, C, D) | Square geometry: 4 edge pairs, 2 diagonal pairs, potential scaling |
| Q8 | (A, B, C, D) | Gravitational self-energy contraction & Virial theorem release |
| Q9 | 3 | Altitude inversion: $\frac{h/R}{1 + h/R} = 0.75 \implies h = 3R$ |
| Q10 | 6 | Expansion work: $W = 12G – 6G = 6G \implies k = 6$ |
Detailed Explanations
Solution 1:
For three masses $m$ at the vertices of an equilateral triangle of side $a$, there are $\binom{3}{2} = 3$ pairwise interactions, each having separation $a$.
$$U_{\text{total}} = \left(-\frac{G m^2}{a}\right) + \left(-\frac{G m^2}{a}\right) + \left(-\frac{G m^2}{a}\right) = -\mathbf{3\frac{G m^2}{a}}$$
Correct Answer: (C)
Solution 2:
The work done by an external agent is:
$$W_{\text{ext}} = \Delta U = U(R + h) – U(R) = \frac{m g_0 h}{1 + \frac{h}{R}}$$
For $h = R$:
$$W_{\text{ext}} = \frac{m g_0 R}{1 + 1} = \mathbf{\frac{1}{2} m g_0 R}$$
Correct Answer: (B)
Solution 3:
– Gravitational self-energy of a uniform solid sphere: $U_{\text{solid}} = -\frac{3}{5}\frac{G M^2}{R}$.
– Gravitational self-energy of a uniform spherical shell: $U_{\text{shell}} = -\frac{1}{2}\frac{G M^2}{R}$.
$$\frac{|U_{\text{solid}}|}{|U_{\text{shell}}|} = \frac{3/5}{1/2} = \frac{6}{5}$$
Correct Answer: (C)
Solution 4:
Initial potential energy: $U_i = -\frac{G m^2}{r}$.
Final potential energy: $U_f = -\frac{G m^2}{2r}$.
The work required by an external agent is:
$$W_{\text{ext}} = U_f – U_i = -\\frac{G m^2}{2r} – \left(-\frac{G m^2}{r}\right) = \mathbf{\frac{G m^2}{2r}}$$
Correct Answer: (B)
Solution 5:
By conservation of mechanical energy:
$$E_{\text{initial}} = E_{\text{final}}$$
$$0 – \frac{G M m}{2R} = \frac{1}{2} m v^2 – \frac{G M m}{R}$$
$$\frac{1}{2} m v^2 = \frac{G M m}{R} – \frac{G M m}{2R} = \frac{G M m}{2R} = \frac{1}{2} m g_0 R$$
$$v^2 = g_0 R \implies \mathbf{v = \sqrt{g_0 R}}$$
Correct Answer: (B)
Solution 6:
– (A) True: Setting $U(\infty) = 0$ makes all bound states have $U < 0$.
- (B) True: Only differences in potential energy $\Delta U$ are physically measurable, which are invariant under gauge shift.
- (C) True: By definition of conservative forces, $\vec{F} = -\vec{\nabla} U$.
- (D) True: $F_r = -\frac{dU}{dr} = -\frac{G M m}{r^2} < 0$ (attractive force).
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 7:
– (A) 4 side pairs at distance $a$, 2 diagonal pairs at distance $a\sqrt{2}$: $U = -\frac{G m^2}{a}(4 + \sqrt{2})$. (True).
– (B) Since $U \propto 1/a$, doubling $a$ halves the magnitude: $U’ = U/2$. (True).
– (C) $W = U_f – U_i = U/2 – U = -U/2 = +\frac{G m^2}{2a}(4 + \sqrt{2})$. (True).
– (D) Distance from center to any vertex is $r = a/\sqrt{2}$. Potential $V = 4 \left(-\frac{G m}{a/\sqrt{2}}\right) = -4\sqrt{2}\frac{G m}{a}$. (True).
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 8:
– (A) $U_{\text{self}} = -\frac{3}{5}\frac{GM^2}{R}$. As $R$ decreases, $U$ becomes more negative. (True).
– (B) $\Delta U = U_f – U_i = -\frac{3}{5}\frac{GM^2}{R/2} – \left(-\frac{3}{5}\frac{GM^2}{R}\right) = -\frac{6}{5}\frac{GM^2}{R} + \frac{3}{5}\frac{GM^2}{R} = -\frac{3}{5}\frac{GM^2}{R}$. (True).
– (C) Work done by gravity: $W_g = -\Delta U = +\frac{3}{5}\frac{GM^2}{R}$. (True).
– (D) Half of this energy is radiated away as heat, and half increases the internal kinetic energy. (True).
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 9:
The work required to lift mass $m$ to height $h$ is:
$$W = \Delta U = \frac{m g_0 h}{1 + \frac{h}{R}} = 0.75 m g_0 R = \frac{3}{4} m g_0 R$$
$$\frac{h/R}{1 + h/R} = \frac{3}{4} \implies 4\left(\frac{h}{R}\right) = 3 + 3\left(\frac{h}{R}\right) \implies \frac{h}{R} = 3$$
$$h = 3R \implies \mathbf{n = 3}$$
Correct Answer: 3
Solution 10:
Initial potential energy with $m = 2.0\text{ kg}$ and $a = 1.0\text{ m}$:
$$U_i = -3 \frac{G m^2}{a} = -3 \frac{G (2.0)^2}{1.0} = -12 G\text{ Joules}$$
Final potential energy with side $2a = 2.0\text{ m}$:
$$U_f = -3 \frac{G m^2}{2a} = -3 \frac{G (4.0)}{2.0} = -6 G\text{ Joules}$$
The work required by an external agent is:
$$W_{\text{ext}} = U_f – U_i = -6 G – (-12 G) = +6 G\text{ Joules}$$
Comparing with $W = k G$:
$$\mathbf{k = 6}$$
Correct Answer: 6