Concept Card: Universal Law of Gravitation & The Gravitational Constant
1. Historical Context and Fundamental Nature of Gravity:
Sir Isaac Newton published the Law of Universal Gravitation in 1687 in his monumental work Philosophiae Naturalis Principia Mathematica. Gravitation is the weakest of the four fundamental interactions in nature, yet it dictates the large-scale structure and evolution of the cosmos. It possesses four distinctive physical properties:
- Universality: It acts universally between all pairs of objects possessing mass or energy, from fundamental particles to colossal galactic clusters.
- Always Attractive: Unlike the electromagnetic force, which can attract or repel due to the existence of positive and negative charges, gravitational mass exists in only one sign; hence, gravitational interaction is universally attractive.
- Long-Range / Infinite Action: The force follows an exact inverse-square law ($F \propto 1/r^2$), possessing infinite range without an intrinsic spatial cutoff.
- Medium Independence: The gravitational attraction between two bodies is completely unaffected by the presence, nature, or state of any intervening matter or medium. There is no gravitational counterpart to electrostatic dielectric polarization or electromagnetic shielding.
2. Mathematical Formulation: Newton’s Universal Law of Gravitation
Every point mass in the universe attracts every other point mass with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
Scalar Magnitude Form:
$$F = G \frac{m_1 m_2}{r^2}$$
where:
- $m_1, m_2$ are the gravitational masses of the interacting point particles.
- $r$ is the spatial Euclidean distance separating the two particles.
- $G$ is the Universal Gravitational Constant.
Rigorous Vector Formulation:
Let $\vec{r}_1$ and $\vec{r}_2$ denote the position vectors of particles of masses $m_1$ and $m_2$ relative to an arbitrary inertial origin $O$.
The vector displacement from particle 1 to particle 2 is:
$$\vec{r}_{12} = \vec{r}_2 – \vec{r}_1$$
The unit vector directed from $1$ toward $2$ is:
$$\hat{r}_{12} = \frac{\vec{r}_{12}}{|\vec{r}_{12}|} = \frac{\vec{r}_2 – \vec{r}_1}{|\vec{r}_2 – \vec{r}_1|}$$
The gravitational force $\vec{F}_{21}$ exerted on particle 2 by particle 1 is directed toward particle 1 (opposite to $\hat{r}_{12}$):
$$\mathbf{\vec{F}_{21} = -\frac{G m_1 m_2}{r_{12}^2} \hat{r}_{12} = -\frac{G m_1 m_2}{|\vec{r}_2 – \vec{r}_1|^3} (\vec{r}_2 – \vec{r}_1)}$$
Similarly, the gravitational force $\vec{F}_{12}$ exerted on particle 1 by particle 2 is:
$$\mathbf{\vec{F}_{12} = -\frac{G m_1 m_2}{r_{21}^2} \hat{r}_{21} = +\frac{G m_1 m_2}{|\vec{r}_2 – \vec{r}_1|^3} (\vec{r}_2 – \vec{r}_1)}$$
By direct inspection:
$$\mathbf{\vec{F}_{12} = -\vec{F}_{21}}$$
This proves that gravitational interactions strictly satisfy Newton’s Third Law of Motion in strong form (the action-reaction forces are equal in magnitude, opposite in direction, and act collinearly along the line joining the centers of the two interacting masses – a central force).
3. The Universal Gravitational Constant ($G$) & The Cavendish Experiment
The constant of proportionality $G$ is an empirical universal fundamental constant of nature.
- CODATA Accepted Value:
$$G = 6.67430(15) \times 10^{-11} \text{ N}\cdot\text{m}^2/\text{kg}^2 \quad \text{or} \quad \text{m}^3\cdot\text{kg}^{-1}\cdot\text{s}^{-2}$$ - CGS Value:
$$G = 6.674 \times 10^{-8} \text{ dyne}\cdot\text{cm}^2/\\text{g}^2 \quad \text{or} \quad \text{cm}^3\cdot\text{g}^{-1}\cdot\text{s}^{-2}$$ - Dimensional Analysis:
$$[G] = \frac{[F][r^2]}{[m_1][m_2]} = \frac{[M L T^{-2}][L^2]}{[M^2]} = \mathbf{[M^{-1} L^3 T^{-2}]}$$
The Cavendish Torsion Balance Experiment (1798):
Henry Cavendish was the first scientist to experimentally measure $G$ in a laboratory setting (historically referred to as “weighing the Earth”):
- Experimental Setup: A light, rigid horizontal rod of length $L$ carries two identical small lead spheres of mass $m$ at its opposite ends. The rod is suspended horizontally from its center by a thin, highly sensitive quartz torsion wire. Two large, heavy spherical lead balls of mass $M$ are placed near the small balls on opposite sides at a center-to-center distance $d$.
- Deflecting Gravitational Couple: Each large sphere attracts the adjacent small sphere with a horizontal gravitational force $F = \frac{G M m}{d^2}$. The two equal and opposite forces form a deflecting torque (couple) about the vertical suspension axis:
$$\tau_{\text{grav}} = F \cdot L = \left(\frac{G M m}{d^2}\right) L$$ - Restoring Elastic Couple: As the rod rotates through an angle $\theta$, the suspension wire twists, developing an elastic restoring torque:
$$\tau_{\text{restoring}} = C \theta$$
where $C$ is the torsional rigidity (restoring couple per unit twist) of the fiber. - Equilibrium Condition: In rotational static equilibrium:
$$\tau_{\text{grav}} = \tau_{\text{restoring}} \implies \frac{G M m L}{d^2} = C \theta \implies \mathbf{G = \frac{C \theta d^2}{M m L}}$$ - Measurement of Deflection & Calibration:
- The minute angle of twist $\theta$ is measured with extreme precision using a lamp-and-scale optical lever arrangement (a small mirror attached to the fiber deflects a light beam by an angle $2\theta$ across a scale at distance $D$).
- The torsional constant $C$ is calibrated dynamically by measuring the time period of torsional oscillations $T$ of the suspended dumbbell:
$$T = 2\pi \sqrt{\frac{I}{C}} \implies C = \frac{4\pi^2 I}{T^2}$$
where $I = 2 m (L/2)^2 = \frac{1}{2} m L^2$ is the moment of inertia of the dumbbell about the suspension axis.
4. The Principle of Gravitational Superposition
The total gravitational force exerted on any individual particle by a collection of multiple interacting masses is the vector sum of the individual pairwise gravitational forces exerted on it by each mass, calculated independently as if no other masses were present:
$$\mathbf{\vec{F}_{\text{net}} = \sum_{i=1}^N \vec{F}_i = \sum_{i=1}^N -\frac{G m m_i}{r_i^2} \hat{r}_i}$$
For a continuous mass distribution possessing volume mass density $\rho$, surface mass density $\sigma$, or linear mass density $\lambda$, the summation generalizes to a volume, surface, or line integral:
$$\mathbf{\vec{F}_{\text{net}} = \int d\vec{F} = -G m \int \frac{dm}{r^2} \hat{r}}$$
5. Newton’s Shell Theorems for Spherically Symmetric Bodies
Point-mass formulas cannot be applied naively to extended bodies unless justified by Newton’s Shell Theorems:
- Shell Theorem 1 (External Field): A uniform spherical shell (or any spherically symmetric solid mass distribution) of total mass $M$ attracts an external point mass $m$ located at distance $r \ge R$ from its center exactly as if its entire mass were concentrated at its geometric center:
$$F(r) = \frac{G M m}{r^2} \quad (r \ge R)$$ - Shell Theorem 2 (Internal Field): A uniform spherical shell of mass $M$ exerts identically zero net gravitational force on any particle placed at any arbitrary position inside the interior cavity of the shell:
$$F(r) = 0 \quad (r < R)$$
6. Critical Insights & High-Yield JEE Traps
- Trap 1 (Force between Extended Touching Bodies): For two identical uniform spheres of radius $R$ and density $\rho$ touching each other, the distance between their centers is $2R$. Their mass is $M = \frac{4}{3}\pi R^3 \rho$. The gravitational attraction is:
$$F = \frac{G M^2}{(2R)^2} = \frac{G \left(\frac{4}{3}\pi R^3 \rho\right)^2}{4R^2} = \frac{4}{9} \pi^2 G \rho^2 \mathbf{R^4}$$
Exam Alert: The force scales with $R^4$, not $1/R^2$! - Trap 2 (Uniform Field Inside an Eccentric Cavity): When a spherical cavity is carved out of a uniform solid sphere, the gravitational field vector everywhere inside the cavity is spatially uniform and directed parallel to the displacement vector $\vec{d}$ connecting the center of the sphere to the center of the cavity:
$$\vec{g}_{\text{cavity}} = \frac{4}{3} \pi G \rho \vec{d}$$ - Trap 3 (Gravitational vs. Inertial Mass): Inertial mass $m_i$ measures resistance to acceleration ($F = m_i a$), while gravitational mass $m_g$ measures gravitational coupling ($F_g = \frac{G M m_g}{r^2}$). Einstein’s Weak Equivalence Principle establishes $m_i \equiv m_g$ to within $10^{-15}$, explaining why all objects fall with identical acceleration in a gravitational field in vacuo.
Solved Examples
Example 1 (Direct Conceptual Application – Vector Superposition on a Symmetric Polygon):
Three identical point particles, each of mass $m = 2.0\text{ kg}$, are fixed at the vertices of an equilateral triangle $ABC$ of side length $a = 1.0\text{ m}$ in a horizontal plane.
(a) Determine the net gravitational force $\vec{F}_0$ exerted on a particle of mass $m_0 = 0.50\text{ kg}$ placed at the centroid $G_0$ of the triangle.
(b) If the particle at vertex $A$ is removed, find the magnitude and direction of the net gravitational force acting on mass $m_0$ at the centroid.
(c) If instead of removing vertex $A$, the mass at vertex $A$ is doubled to $2m$, determine the new net force on mass $m_0$.
Solution:
(a) Equilibrium at the Centroid by Symmetry:
The distance from any vertex of an equilateral triangle of side $a$ to its centroid $G_0$ is:
$$r = \frac{a}{\sqrt{3}} = \frac{1.0}{\sqrt{3}}\text{ m}$$
Each particle at vertices $A, B, C$ attracts mass $m_0$ with a force of magnitude:
$$F = \frac{G m m_0}{r^2} = \frac{G m m_0}{(a/\sqrt{3})^2} = \frac{3 G m m_0}{a^2}$$
The three force vectors $\vec{F}_A, \vec{F}_B, \vec{F}_C$ point from the centroid toward vertices $A, B,$ and $C$ respectively. The angle between any two adjacent vectors is $\frac{360^\circ}{3} = 120^\circ$.
By vector symmetry:
$$\vec{F}_B + \vec{F}_C = -\vec{F}_A \implies \mathbf{\vec{F}_{\text{net}} = \vec{F}_A + \vec{F}_B + \vec{F}_C = \vec{0}}$$
(b) Net Force When Vertex $A$ is Removed:
Prior to removal: $\vec{F}_A + (\vec{F}_B + \vec{F}_C) = \vec{0} \implies \vec{F}_B + \vec{F}_C = -\vec{F}_A$.
When $A$ is removed, the remaining force is simply the resultant of $B$ and $C$:
$$\vec{F}_{\text{rem}} = \vec{F}_B + \vec{F}_C = -\vec{F}_A$$
The magnitude is:
$$|\vec{F}_{\text{rem}}| = F_A = \frac{3 G m m_0}{a^2}$$
Substituting numerical values:
$$|\vec{F}_{\text{rem}}| = \frac{3 (6.674 \times 10^{-11})(2.0)(0.50)}{(1.0)^2} = \mathbf{2.00 \times 10^{-10}\text{ N}}$$
Direction: Directly away from vertex $A$ (along the altitude from $A$, bisecting the side $BC$).
(c) Net Force When Mass at Vertex $A$ is Doubled to $2m$:
The force from vertex $A$ becomes $2\vec{F}_A$.
$$\vec{F}_{\text{new}} = 2\vec{F}_A + \vec{F}_B + \vec{F}_C = \vec{F}_A + (\vec{F}_A + \vec{F}_B + \vec{F}_C) = \vec{F}_A + \vec{0} = \vec{F}_A$$
The magnitude is:
$$|\vec{F}_{\text{new}}| = \frac{3 G m m_0}{a^2} = \mathbf{2.00 \times 10^{-10}\text{ N}}$$
Direction: Directly toward vertex $A$.
Takeaway: In symmetrical configurations, removing a mass $\Delta m$ is equivalent to superimposing a negative mass $-\Delta m$, yielding a net force directed away from the removed vertex.
Example 2 (Calculus & Continuous Mass Distribution – Thin Rod Attracting an Axial Point Mass):
A uniform thin rod of mass $M$ and length $L$ lies horizontally along the $x$-axis between $x = d$ and $x = d + L$. A point mass $m$ is positioned at the coordinate origin $x = 0$.
(a) Derive the exact gravitational attraction force $F$ exerted by the rod on the point mass $m$.
(b) Analyze the limiting behavior of the force for $d \gg L$ and interpret the result physically.
(c) Compare this result with the incorrect formula obtained by concentrating the rod’s mass at its center of mass ($x_{cm} = d + L/2$).
Solution:
(a) Integration via Differential Mass Elements:
The linear mass density of the uniform rod is:
$$\lambda = \frac{M}{L}$$
Consider an infinitesimal element of the rod of length $dx$ located at distance $x$ from the origin ($x = 0$).
The mass of this element is:
$$dm = \lambda dx = \frac{M}{L} dx$$
Since $dm$ can be treated as a point mass at distance $x$ from $m$, the differential attractive force directed toward $+x$ is:
$$dF = \frac{G m dm}{x^2} = \frac{G m M}{L} \frac{dx}{x^2}$$
Integrating over the entire length of the rod from $x = d$ to $x = d + L$:
$$F = \int_d^{d+L} \frac{G m M}{L} \frac{dx}{x^2} = \frac{G m M}{L} \left[ -\frac{1}{x} \right]_d^{d+L}$$
$$F = \frac{G m M}{L} \left( \frac{1}{d} – \frac{1}{d + L} \right) = \frac{G m M}{L} \left( \frac{d + L – d}{d(d + L)} \right)$$
$$\mathbf{F = \frac{G M m}{d (d + L)}}$$
(b) Limiting Case for Distant Particles ($d \gg L$):
When $d \gg L$, $d + L \approx d$. Hence:
$$F \approx \frac{G M m}{d^2}$$
At great distances, the spatial extension of the rod becomes negligible, and its gravitational field asymptotically converges to that of a single point mass $M$ located at the origin of the rod.
(c) The Center-of-Mass Pitfall:
If one mistakenly concentrated the rod’s mass at its center of mass $x_{cm} = d + L/2$, the predicted force would be:
$$F_{\text{cm}} = \frac{G M m}{(d + L/2)^2} = \frac{G M m}{d^2 + dL + L^2/4}$$
Notice that $d(d + L) = d^2 + dL < d^2 + dL + L^2/4$.
Therefore:
$$d(d+L) < \left(d + \frac{L}{2}\right)^2 \implies \mathbf{F_{\text{exact}} > F_{\text{cm}}}$$
Crucial Conclusion: Newton’s law can never be applied by substituting the center of mass for extended bodies unless the body is spherically symmetric (Shell Theorem). For non-spherical bodies, $F_{\text{exact}} \ne F_{\text{cm}}$.
Example 3 (Advanced JEE Multi-Concept Scenario – Uniform Field Inside an Off-Center Spherical Cavity):
A uniform solid sphere of mass density $\rho$ and radius $R$ is centered at origin $O$. A spherical cavity of radius $R_c$ ($R_c < R/2$) is excavated inside the sphere such that its center $O'$ is located at displacement vector $\vec{d}$ from $O$ ($|\vec{d}| + R_c < R$).
(a) Prove analytically that the gravitational field vector $\vec{g}$ at every point inside the cavity is strictly uniform (independent of position).
(b) Determine the magnitude and direction of the gravitational force exerted on a small particle of mass $m$ placed anywhere inside this cavity.
Solution:
(a) Superposition Principle for Negative Mass:
By the principle of superposition, the sphere with the cavity is mathematically equivalent to:
$$\text{Sphere with Cavity} = \text{Complete Solid Sphere of density } +\rho \text{ centered at } O + \text{Smaller Solid Sphere of density } -\rho \text{ centered at } O’$$
Let $P$ be an arbitrary point inside the cavity. Its position vector relative to $O$ is $\vec{r}$, and its position vector relative to the cavity center $O’$ is $\vec{r}’$.
From vector triangle rule:
$$\vec{r} = \vec{d} + \vec{r}’ \implies \vec{r}’ = \vec{r} – \vec{d}$$
Field of the complete solid sphere at internal point $P$:
By Gauss’s Law for gravity / Shell Theorem, only the concentric mass inside radius $r$ contributes:
$$M_{\text{enclosed}} = \frac{4}{3}\pi r^3 \rho$$
The gravitational field is directed radially inward toward $O$:
$$\vec{g}_1 = -\frac{G M_{\text{enclosed}}}{r^2} \hat{r} = -\frac{G \left(\frac{4}{3}\pi r^3 \rho\right)}{r^3} \vec{r} = -\frac{4}{3}\pi G \rho \vec{r}$$
Field of the removed sphere of density $-\rho$ at internal point $P$:
Point $P$ is also inside the removed sphere at distance $\vec{r}’$ from $O’$:
$$\vec{g}_2 = -\frac{4}{3}\pi G (-\rho) \vec{r}’ = +\frac{4}{3}\pi G \rho \vec{r}’ = +\frac{4}{3}\pi G \rho (\vec{r} – \vec{d})$$
Total gravitational field inside the cavity:
$$\vec{g}_{\text{cavity}} = \vec{g}_1 + \vec{g}_2 = -\frac{4}{3}\pi G \rho \vec{r} + \frac{4}{3}\pi G \rho \vec{r} – \frac{4}{3}\pi G \rho \vec{d}$$
$$\mathbf{\vec{g}_{\text{cavity}} = -\frac{4}{3}\pi G \rho \vec{d}}$$
Remarkably, all spatial coordinates ($\vec{r}$ and $\vec{r}’$) vanish completely! The gravitational field inside the cavity is completely uniform and constant.
(b) Force on Mass $m$:
$$\mathbf{\vec{F} = m \vec{g}_{\text{cavity}} = -\frac{4}{3}\pi G \rho m \vec{d}}$$
The magnitude of the force is:
$$F = \frac{4}{3}\pi G \rho m d$$
Direction: Strictly antiparallel to $\vec{d}$ (directed from the cavity center $O’$ toward the main sphere center $O$).
Example 4 (Dynamic Equilibrium – Orbital Motion of $N$ Point Masses Under Mutual Gravity):
Four identical particles, each of mass $m$, are located at the vertices of a square of side length $a$. The particles revolve in a circular orbit under the sole influence of their mutual gravitational attraction, preserving their square formation at all times.
(a) Determine the radius $R$ of the circular orbit.
(b) Derive the net gravitational force acting on any one particle directed toward the common center of rotation.
(c) Calculate the orbital speed $v$ and orbital period $T$ of each particle.
Solution:
(a) Radius of Orbit:
The center of rotation is the geometric center of the square (the circumcenter).
The diagonal length of the square is $a\sqrt{2}$. The radius $R$ of the circumscribing circle is:
$$\mathbf{R = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}}$$
(b) Net Mutual Gravitational Force on One Particle:
Consider particle 1 at vertex $(0, a)$. Three other particles act on it:
- Two adjacent particles (at distance $a$): Each exerts an attractive force along the sides of the square:
$$F_1 = F_2 = \frac{G m^2}{a^2}$$
The angle between these two forces is $90^\circ$. Their vector resultant points along the diagonal toward the center:
$$F_{12} = \sqrt{F_1^2 + F_2^2} = \sqrt{2} \frac{G m^2}{a^2}$$ - One diagonally opposite particle (at distance $2R = a\sqrt{2}$): Exerts an attractive force pointing directly along the diagonal toward the center:
$$F_3 = \frac{G m^2}{(a\sqrt{2})^2} = \frac{G m^2}{2a^2}$$
Since both resultants are collinear and point toward the center:
$$F_{\text{net}} = F_{12} + F_3 = \sqrt{2} \frac{G m^2}{a^2} + \frac{1}{2} \frac{G m^2}{a^2}$$
$$\mathbf{F_{\text{net}} = \frac{G m^2}{a^2} \left( \sqrt{2} + \frac{1}{2} \right) = \frac{G m^2}{2a^2} (2\sqrt{2} + 1)}$$
(c) Orbital Velocity and Period:
The net gravitational force provides the necessary centripetal force for uniform circular motion:
$$F_{\text{net}} = \frac{m v^2}{R}$$
Substitute $R = \frac{a}{\sqrt{2}}$:
$$\frac{m v^2}{a / \sqrt{2}} = \frac{G m^2}{2a^2} (2\sqrt{2} + 1)$$
$$\frac{\sqrt{2} m v^2}{a} = \frac{G m^2}{2a^2} (2\sqrt{2} + 1)$$
$$v^2 = \frac{G m}{2\sqrt{2} a} (2\sqrt{2} + 1) = \frac{G m}{a} \left( 1 + \frac{1}{2\sqrt{2}} \right) = \frac{G m}{a} \left( \frac{4 + \sqrt{2}}{4\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} \right) = \frac{G m}{a} \frac{2\sqrt{2} + 1}{2\sqrt{2}}$$
$$\mathbf{v = \sqrt{\frac{G m}{a} \left( 1 + \frac{1}{2\sqrt{2}} \right)} = \sqrt{\frac{G m}{a} \left( \frac{2\sqrt{2} + 1}{2\sqrt{2}} \right)}}$$
The time period $T$ of revolution is:
$$T = \frac{2\pi R}{v} = \frac{2\pi (a/\sqrt{2})}{v} = \mathbf{2\pi \sqrt{\frac{a^3}{G m (2 + \frac{1}{\sqrt{2}})}}}$$
Practice Worksheet: Universal Gravitation & Cavendish Experiment
Section A: Single Correct Questions (Q1 to Q5)
Q1. If the dimensions of the Universal Gravitational Constant $G$ are expressed in terms of fundamental SI units as $[M^a L^b T^c]$, the value of the algebraic expression $(2a + b + c)$ is:
(A) $-1$
(B) $0$
(C) $1$
(D) $2$
Q2. Two identical uniform solid lead spheres, each of radius $R$ and uniform mass density $\rho$, are placed in direct physical contact with each other on a flat surface. The gravitational attraction force $F$ between them is proportional to:
(A) $R^2$
(B) $R^3$
(C) $R^4$
(D) $R^{-2}$
Q3. Two stationary point masses $M$ and $4M$ are held at a fixed separation distance $L$. A third point mass $m$ is placed on the line connecting them such that the net gravitational force on $m$ is zero. The distance of mass $m$ from mass $M$ is:
(A) $\frac{L}{5}$
(B) $\frac{L}{4}$
(C) $\frac{L}{3}$
(D) $\frac{L}{2}$
Q4. A thin uniform wire of total mass $M$ is bent into a semicircle of radius $R$. A point particle of mass $m$ is placed at the center of curvature of the semicircle. The magnitude of the net gravitational force exerted by the bent wire on particle $m$ is:
(A) $\frac{G M m}{\pi R^2}$
(B) $\frac{2 G M m}{\pi R^2}$
(C) $\frac{\pi G M m}{2 R^2}$
(D) $\frac{2\pi G M m}{R^2}$
Q5. A point mass $m$ lies on the symmetry axis of a uniform thin circular ring of mass $M$ and radius $R$ at a distance $x$ from the ring’s center. The distance $x$ at which the gravitational attraction force between the ring and mass $m$ achieves its maximum value is:
(A) $x = R$
(B) $x = \frac{R}{\sqrt{2}}$
(C) $x = \frac{R}{2}$
(D) $x = R\sqrt{2}$
Section B: One or More than One Correct Questions (Q6 to Q8)
Q6. Which of the following statements regarding Newton’s Universal Law of Gravitation and the Cavendish experiment is/are correct?
(A) The gravitational force between two point masses is completely independent of the chemical composition, temperature, and nature of the intervening medium.
(B) In the Cavendish torsion balance experiment, the angle of deflection of the reflected light spot on the scale is twice the mechanical angle of twist of the suspension wire ($2\theta$).
(C) The gravitational force satisfies Newton’s third law in strong form, meaning it is conservative, central, and directed along the line joining the centers of mass.
(D) The principle of superposition implies that the presence of a third mass alters the pairwise gravitational force between two existing masses.
Q7. An infinite collection of identical point masses, each of mass $m$, are positioned along the positive $x$-axis at coordinates $x = 1\text{ m}, 2\text{ m}, 4\text{ m}, 8\text{ m}, 16\text{ m}, \dots$. A test mass $m_0$ is placed at the origin $x = 0$. Choose the correct statement(s):
(A) The magnitude of the net gravitational force acting on mass $m_0$ is $\frac{4}{3} G m m_0$.
(B) The magnitude of the net gravitational force acting on mass $m_0$ is $2 G m m_0$.
(C) If the masses at alternating positions ($x = 2, 8, 32, \dots$) have their masses replaced with $-m$ (conceptually repelling), the net force at the origin becomes $\frac{4}{5} G m m_0$.
(D) The net gravitational potential at the origin due to this infinite mass distribution is $-2 G m$.
Q8. A uniform solid sphere of mass $M$, radius $R$, and density $\rho$ has a spherical cavity of radius $R/2$ excavated such that the cavity’s surface touches both the center and the outer surface of the sphere. Which of the following statements is/are correct?
(A) The gravitational field vector everywhere inside the cavity is uniform in both magnitude and direction.
(B) The magnitude of the gravitational field inside the cavity is $g = \frac{2}{3}\pi G \rho R$.
(C) A particle placed at the center of the cavity experiences zero gravitational force.
(D) The gravitational field at the center of the original sphere is directed toward the center of the cavity.
Section C: Numerical Value Type Questions (Q9 to Q10)
Q9. Three identical stars, each of mass $m = 3.0\text{ kg}$, revolve in a common circular orbit under their mutual gravitational attraction, always maintaining an equilateral triangle formation of side length $a = 1.0\text{ m}$. The orbital speed of each star can be written as $v = \sqrt{n G}$, where $G$ is the universal gravitational constant in SI units. Find the integer value of $n$.
Q10. Two stationary point masses $M_1 = 1.0\text{ kg}$ and $M_2 = 9.0\text{ kg}$ are held at a fixed separation distance $d = 8.0\text{ m}$ in empty space. A point on the segment connecting them where the net gravitational field is zero is located at a distance $x$ (in meters) from mass $M_1$. Determine the value of $x$.
Comprehensive Solutions and Answer Key
Answer Key
| Question | Correct Answer | Core Physics Concept |
|---|---|---|
| Q1 | (A) | Dimensional Formula of $G$: $[M^{-1} L^3 T^{-2}]$ |
| Q2 | (C) | Contact Gravitational Scaling ($F \propto R^4$) |
| Q3 | (C) | Null Neutral Point for Unequal Masses ($x = L/3$) |
| Q4 | (B) | Continuous Integration over Semicircular Arc |
| Q5 | (B) | Maximum Axial Force for Ring ($x = R/\sqrt{2}$) |
| Q6 | (A, B, C) | Cavendish Balance Principles & Superposition Invariance |
| Q7 | (A, C, D) | Geometric Series Superposition for Infinite Point Masses |
| Q8 | (A, B) | Superposition in Spherical Cavities: Uniform Internal Field |
| Q9 | 3 | Orbital Speed in Symmetrical 3-Body System ($v = \sqrt{3G}$) |
| Q10 | 2 | Zero Field Null Point Location ($x = 2\text{ m}$) |
Detailed Explanations
Solution 1:
From Newton’s Law of Gravitation:
$$F = G \frac{m_1 m_2}{r^2} \implies G = \frac{F r^2}{m_1 m_2}$$
Substituting fundamental dimensions:
$$[G] = \frac{[M L T^{-2}][L^2]}{[M^2]} = [M^{-1} L^3 T^{-2}]$$
Comparing with $[M^a L^b T^c]$:
$$a = -1, \quad b = 3, \quad c = -2$$
Evaluate $(2a + b + c)$:
$$2(-1) + 3 + (-2) = -2 + 3 – 2 = \mathbf{-1}$$
Correct Answer: (A)
Solution 2:
For two spheres of radius $R$ in contact, the distance between their centers is:
$$r = R + R = 2R$$
The mass of each uniform sphere of density $\rho$ is:
$$M = \frac{4}{3}\pi R^3 \rho$$
Applying Newton’s Law of Gravitation:
$$F = G \frac{M \cdot M}{(2R)^2} = G \frac{\left(\frac{4}{3}\pi R^3 \rho\right)^2}{4 R^2} = \frac{G \left(\frac{16}{9}\pi^2 R^6 \rho^2\right)}{4 R^2} = \frac{4}{9}\pi^2 G \rho^2 \mathbf{R^4}$$
Hence, the force is proportional to $R^4$.
Correct Answer: (C)
Solution 3:
Let mass $m$ be placed at distance $x$ from $M$. The distance from $4M$ is then $(L – x)$.
For the net force on $m$ to be zero, the opposing gravitational attractions must be equal in magnitude:
$$\frac{G M m}{x^2} = \frac{G (4M) m}{(L – x)^2}$$
Canceling common terms $G M m$:
$$\frac{1}{x^2} = \frac{4}{(L – x)^2}$$
Taking the positive square root on both sides (since $0 < x < L$):
$$\frac{1}{x} = \frac{2}{L - x} \implies L - x = 2x \implies 3x = L \implies \mathbf{x = \frac{L}{3}}$$
Correct Answer: (C)
Solution 4:
Let the semicircle lie in the upper half-plane with its center at the origin $(0, 0)$.
Mass per unit length is:
$$\lambda = \frac{M}{\pi R}$$
Consider an element of the wire subtending angle $d\theta$ at angle $\theta$ with the horizontal ($-\pi/2 \le \theta \le \pi/2$).
Length of element $ds = R d\theta \implies dm = \lambda R d\theta = \frac{M}{\pi} d\theta$.
The elemental gravitational force on mass $m$ at the center is:
$$dF = \frac{G m dm}{R^2} = \frac{G M m}{\pi R^2} d\theta$$
By symmetry, components parallel to the chord cancel out, and only components along the axis of symmetry add up:
$$F_{\text{net}} = \int_{-\pi/2}^{\pi/2} dF \cos\theta = \frac{G M m}{\pi R^2} \int_{-\pi/2}^{\pi/2} \cos\theta d\theta = \frac{G M m}{\pi R^2} [\sin\theta]_{-\pi/2}^{\pi/2} = \frac{G M m}{\pi R^2} [1 – (-1)] = \mathbf{\frac{2 G M m}{\pi R^2}}$$
Correct Answer: (B)
Solution 5:
For a ring of mass $M$ and radius $R$, consider a point mass $m$ on the axis at distance $x$.
Every element $dM$ of the ring is at distance $r = \sqrt{R^2 + x^2}$ from $m$.
The axial component of gravitational force is:
$$F(x) = \frac{G M m x}{(R^2 + x^2)^{3/2}}$$
To find the maximum force, differentiate with respect to $x$ and set to zero:
$$\frac{dF}{dx} = G M m \left[ \frac{(R^2 + x^2)^{3/2} (1) – x \cdot \frac{3}{2}(R^2 + x^2)^{1/2} (2x)}{(R^2 + x^2)^3} \right] = 0$$
$$(R^2 + x^2)^{1/2} [ (R^2 + x^2) – 3x^2 ] = 0 \implies R^2 – 2x^2 = 0 \implies \mathbf{x = \frac{R}{\sqrt{2}}}$$
Correct Answer: (B)
Solution 6:
– (A) True: Gravitational force is strictly independent of the medium; there is no medium shielding.
– (B) True: In an optical lever, when a mirror turns through an angle $\theta$, the reflected beam turns through $2\theta$.
– (C) True: Gravitational forces act along the line joining the interacting bodies and are conservative.
– (D) False: By the Principle of Superposition, the pairwise force between two masses is independent of the presence of any other masses.
Correct Answer: (A, B, C)
Solution 7:
– Force on mass $m_0$ at the origin:
$$F = G m m_0 \sum_{k=0}^\infty \frac{1}{(2^k)^2} = G m m_0 \left[ 1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \dots \right]$$
Using the sum of an infinite geometric progression with common ratio $r = 1/4$:
$$S = \frac{1}{1 – 1/4} = \frac{4}{3} \implies \mathbf{F = \frac{4}{3} G m m_0} \quad \text{(Option A is correct, B is incorrect).}$$
– Alternating signs:
$$F’ = G m m_0 \left[ 1 – \frac{1}{4} + \frac{1}{16} – \frac{1}{64} + \dots \right] = G m m_0 \left[ \frac{1}{1 – (-1/4)} \right] = \mathbf{\frac{4}{5} G m m_0} \quad \text{(Option C is correct).}$$
– Gravitational potential at the origin:
$$V = -G m \sum_{k=0}^\infty \frac{1}{2^k} = -G m \left[ 1 + \frac{1}{2} + \frac{1}{4} + \dots \right] = -G m \left[ \frac{1}{1 – 1/2} \right] = \mathbf{-2 G m} \quad \text{(Option D is correct).}$$
Correct Answer: (A, C, D)
Solution 8:
The cavity center $O’$ is at distance $d = R/2$ from the sphere center $O$.
– As proven in Example 3, the field inside any spherical cavity in a uniform solid sphere is uniform:
$$\vec{g} = -\frac{4}{3}\pi G \rho \vec{d}$$
(Option A is correct).
– Magnitude of this uniform field:
$$g = \frac{4}{3}\pi G \rho \left(\frac{R}{2}\right) = \mathbf{\frac{2}{3}\pi G \rho R} \quad \text{(Option B is correct).}$$
– Since the field is non-zero and uniform everywhere inside the cavity, a particle at the center experiences a non-zero force: $F = m g \ne 0$. (Option C is incorrect).
– At the center of the original sphere ($O$), the complete sphere produces zero field. The removed cavity (mass with center at $O’$) produces a field pulling away from $O’$ (i.e., directed opposite to $\vec{d}$, meaning away from the cavity center). (Option D is incorrect).
Correct Answer: (A, B)
Solution 9:
Three masses $m = 3.0\text{ kg}$ are at the vertices of an equilateral triangle of side $a = 1.0\text{ m}$.
The distance from each mass to the centroid is:
$$R = \frac{a}{\sqrt{3}}$$
On any one mass, the other two masses exert attractive forces of magnitude $F_0 = \frac{G m^2}{a^2}$ at an angle of $60^\circ$ to each other.
The resultant force points toward the centroid:
$$F_{\text{net}} = 2 F_0 \cos 30^\circ = 2 \left(\frac{G m^2}{a^2}\right) \frac{\sqrt{3}}{2} = \sqrt{3} \frac{G m^2}{a^2}$$
This provides the required centripetal force:
$$F_{\text{net}} = \frac{m v^2}{R} \implies \sqrt{3} \frac{G m^2}{a^2} = \frac{m v^2}{a/\sqrt{3}} = \frac{\sqrt{3} m v^2}{a}$$
Canceling $\sqrt{3} m / a$:
$$v^2 = \frac{G m}{a}$$
With $m = 3.0\text{ kg}$ and $a = 1.0\text{ m}$:
$$v^2 = \frac{G (3.0)}{1.0} = 3 G \implies v = \sqrt{3 G}$$
Hence, $n = 3$.
Correct Answer: 3
Solution 10:
The gravitational field due to mass $M_1$ at distance $x$ is $g_1 = \frac{G M_1}{x^2}$.
The gravitational field due to mass $M_2$ at distance $(d – x)$ is $g_2 = \frac{G M_2}{(d – x)^2}$.
For net field to be zero:
$$\frac{G M_1}{x^2} = \frac{G M_2}{(d – x)^2} \implies \frac{\sqrt{M_1}}{x} = \frac{\sqrt{M_2}}{d – x}$$
Substitute $M_1 = 1.0\text{ kg}, M_2 = 9.0\text{ kg}, d = 8.0\text{ m}$:
$$\frac{\sqrt{1}}{x} = \frac{\sqrt{9}}{8 – x} \implies \frac{1}{x} = \frac{3}{8 – x}$$
$$8 – x = 3x \implies 4x = 8 \implies \mathbf{x = 2.0\text{ m}}$$
Correct Answer: 2