Concept Card: Variation of $g$ with Depth and Latitude
1. Introduction to In-Depth and Rotational Perturbations of Gravity:
The acceleration due to gravity ($g$) is not a fixed universal constant; it varies systematically with geographic location on Earth. Beyond elevation (altitude), two primary physical mechanisms govern its variation:
- Depth ($d$): As one descends beneath the Earth’s surface into mines or boreholes, internal mass redistribution reduces the effective attracting mass.
- Latitude ($\lambda$): Due to Earth’s diurnal axial rotation, an outward centrifugal pseudo-force reduces the net effective gravitational pull toward the Earth’s center, reaching a maximum reduction at the equator and zero reduction at the geographic poles.
2. Mathematical Derivation: Variation of $g$ with Depth ($d$)
Consider Earth as a uniform solid sphere of total mass $M$, radius $R$, and uniform mass density $\rho = \frac{M}{\frac{4}{3}\pi R^3}$.
Let a test mass $m$ be located at depth $d$ below the Earth’s surface, corresponding to a distance $r = R – d$ from the center of the Earth ($0 \le d \le R$).
Application of Newton’s Shell Theorem:
Divide Earth into two concentric regions relative to radius $r$:
- Outer Spherical Shell (thickness $d$, from radius $r$ to $R$): By Newton’s Shell Theorem 2, the gravitational force exerted by a uniform spherical shell on any particle located inside its interior cavity is identically zero:
$$F_{\text{shell}} = 0$$ - Inner Solid Sphere (radius $r = R – d$): The test mass $m$ lies on the exterior boundary of this inner sphere. By Shell Theorem 1, its entire enclosed mass $M’$ acts as if concentrated at the center $O$.
Enclosed Mass & Effective Gravity:
The enclosed mass of the inner sphere of radius $r$ is:
$$M’ = \frac{4}{3}\pi r^3 \rho = \frac{4}{3}\pi r^3 \left(\frac{M}{\frac{4}{3}\pi R^3}\right) = M \left(\frac{r}{R}\right)^3 = M \left(\frac{R – d}{R}\right)^3$$
The gravitational force on mass $m$ at depth $d$ is:
$$F(d) = \frac{G M’ m}{r^2} = \frac{G \left[M \left(\frac{r}{R}\right)^3\right] m}{r^2} = \frac{G M m}{R^3} r$$
Hence, the acceleration due to gravity at depth $d$ is:
$$g(d) = \frac{F(d)}{m} = \frac{G M}{R^3} r = \left(\frac{G M}{R^2}\right) \frac{r}{R}$$
Substituting $g_0 = \frac{G M}{R^2}$ and $r = R – d$:
Exact Governing Formula for Depth ($0 \le d \le R$):
$$\mathbf{g(d) = g_0 \left(1 – \frac{d}{R}\right) = g_0 \left(\frac{r}{R}\right)}$$
Essential Characteristics of Depth Variation:
- Linearity: Unlike altitude variation (which follows an inverse-square law), gravity inside a uniform Earth decreases linearly with depth $d$.
- At the Center of Earth ($d = R \implies r = 0$):
$$g(R) = g_0 (1 – 1) = \mathbf{0}$$
A particle at the Earth’s center experiences complete gravitational equilibrium (zero net gravitational pull). - Fractional and Percentage Decrease:
$$\frac{\Delta g}{g_0} = \frac{g_0 – g(d)}{g_0} = \mathbf{\frac{d}{R}} \implies \% \text{ decrease} = \left(\frac{d}{R}\right) \times 100\%$$
3. Depth vs. Altitude Comparison: The “Factor of 2” Rule
For a small height $h \ll R$:
$$\left(\frac{\Delta g}{g_0}\right)_{\text{height}} \approx \frac{2h}{R}$$
For any depth $d \le R$:
$$\left(\frac{\Delta g}{g_0}\right)_{\text{depth}} = \frac{d}{R}$$
Equating the fractional decreases $\left(\frac{\Delta g}{g_0}\right)_{\text{height}} = \left(\frac{\Delta g}{g_0}\right)_{\text{depth}}$:
$$\frac{2h}{R} = \frac{d}{R} \implies \mathbf{d = 2h}$$
Crucial Conclusion: To achieve the same decrease in acceleration due to gravity, the required depth $d$ is twice the required height $h$ (valid strictly for $h \ll R$).
4. Variation of $g$ with Latitude ($\lambda$) Due to Earth’s Diurnal Rotation
Earth rotates about its geographic polar axis with an angular speed:
$$\omega = \frac{2\pi}{T_{\text{day}}} = \frac{2\pi}{86400\text{ s}} \approx 7.292 \times 10^{-5}\text{ rad/s}$$
Consider a body of mass $m$ situated on the Earth’s surface at geographic latitude $\lambda$ (the angle subtended between the equatorial plane and the radius vector from the Earth’s center to the location, where $0^\circ \le \lambda \le 90^\circ$).
Kinematics in the Rotating Frame of Earth:
The body rotates in a horizontal circular path of radius:
$$r_{\text{rot}} = R \cos\lambda$$
In the non-inertial reference frame rotating with the Earth, the particle is subjected to:
- True Gravitational Attraction ($\vec{F}_g$): Directed straight toward the center of the Earth:
$$F_g = m g_0$$ - Centrifugal Pseudo-Force ($\vec{F}_{\text{cf}}$): Directed horizontally outwards, perpendicular to the axis of rotation:
$$F_{\text{cf}} = m \omega^2 r_{\text{rot}} = m \omega^2 R \cos\lambda$$
The angle between $\vec{F}_g$ (inward along radius) and $\vec{F}_{\text{cf}}$ (radially outward from axis) is $(180^\circ – \lambda)$.
The apparent weight $\vec{W}’ = m \vec{g}’$ is the vector sum:
$$m \vec{g}’ = m \vec{g}_0 + \vec{F}_{\text{cf}}$$
Taking the component along the true vertical (inward radial line):
$$g'(\lambda) = g_0 – \omega^2 R \cos\lambda \cos\lambda = \mathbf{g_0 – \omega^2 R \cos^2\lambda}$$
(Higher-order cross terms are negligible because $\frac{\omega^2 R}{g_0} \approx \frac{0.0337}{9.8} \approx 0.0034 \ll 1$).
Governing Formula for Latitude Variation:
$$\mathbf{g'(\lambda) = g_0 – \omega^2 R \cos^2\lambda}$$
Special Geographic Latitudes:
- At the Equator ($\lambda = 0^\circ \implies \cos\lambda = 1$):
$$\mathbf{g_{\text{equator}} = g_0 – \omega^2 R}$$
Gravity is at its global minimum. Numerically: $\omega^2 R \approx (7.292 \times 10^{-5})^2 (6.371 \times 10^6) \approx \mathbf{0.0339\text{ m/s}^2}$. - At the Geographic Poles ($\lambda = 90^\circ \implies \cos\lambda = 0$):
$$\mathbf{g_{\text{pole}} = g_0}$$
Gravity is at its global maximum because the rotation radius is zero ($r_{\text{rot}} = 0$), so no centrifugal force acts. - Difference Between Pole and Equator:
$$\Delta g_{\text{rot}} = g_{\text{pole}} – g_{\text{equator}} = \mathbf{\omega^2 R \approx 0.034\text{ m/s}^2}$$
Condition for Weightlessness at the Equator:
If Earth rotated sufficiently fast so that apparent gravity at the equator vanished ($g_{\text{equator}} = 0$):
$$g_0 – \omega_{\text{crit}}^2 R = 0 \implies \mathbf{\omega_{\text{crit}} = \sqrt{\frac{g_0}{R}}}$$
Numerically:
$$\omega_{\text{crit}} = \sqrt{\frac{9.8}{6.4 \times 10^6}} = \frac{1}{808}\text{ rad/s} \approx 1.24 \times 10^{-3}\text{ rad/s} \approx 17 \omega_{\text{present}}$$
The duration of the day would drop to:
$$T_{\text{crit}} = \frac{2\pi}{\omega_{\text{crit}}} = 2\pi \sqrt{\frac{R}{g_0}} \approx 5075\text{ seconds} \approx \mathbf{84.6\text{ minutes} \approx 1.41\text{ hours}}$$
5. Global Gravitational Profile from $r = 0$ to $r \to \infty$
The complete mathematical profile of $g$ as a function of distance $r$ from the Earth’s center is given by:
$$g(r) = \begin{cases} g_0 \left(\frac{r}{R}\right) = \frac{4}{3}\pi G \rho r, & 0 \le r \le R \quad (\text{Inside Earth, linear}) \\ g_0 \left(\frac{R}{r}\right)^2 = \frac{G M}{r^2}, & r \ge R \quad (\text{Outside Earth, inverse-square}) \end{cases}$$
- At the center ($r = 0$): $g = 0$.
- At the surface ($r = R$): $g = g_0$ (global peak of the spherically symmetric field).
- As $r \to \infty$: $g \to 0$ asymptotically.
- Discontinuity of Derivative at $r = R$:
$$\left(\frac{dg}{dr}\right)_{r \to R^-} = +\frac{g_0}{R}, \quad \left(\frac{dg}{dr}\right)_{r \to R^+} = -\frac{2g_0}{R}$$
The slope experiences an abrupt jump from positive to negative, giving a sharp cusp at the surface.
Solved Examples
Example 1 (Direct Conceptual Application – Depth vs. Altitude Equivalence):
(Take radius of Earth $R = 6400\text{ km}$, $g_0 = 9.8\text{ m/s}^2$).
(a) Determine the depth $d_1$ below the Earth’s surface at which the acceleration due to gravity decreases by $2.5\%$.
(b) Find the altitude $h_1$ above the Earth’s surface at which the acceleration due to gravity decreases by the same percentage ($2.5\%$).
(c) Find the depth $d_2$ at which the acceleration due to gravity becomes one-fourth of its surface value ($g(d_2) = g_0 / 4$).
(d) Find the altitude $h_2$ at which $g(h_2) = g_0 / 4$.
Solution:
(a) Depth for $2.5\%$ Decrease:
Using the depth relation:
$$\frac{\Delta g}{g_0} = \frac{d_1}{R} = 0.025 \implies d_1 = 0.025 \times 6400\text{ km} = \mathbf{160\text{ km}}$$
(b) Altitude for $2.5\%$ Decrease:
Since the decrease is small ($2.5\% < 5\%$), we use the binomial approximation:
$$\frac{\Delta g}{g_0} \approx \frac{2 h_1}{R} = 0.025 \implies h_1 = \frac{0.025 \times 6400\text{ km}}{2} = \mathbf{80\text{ km}}$$
Notice that $d_1 = 2 h_1 = 160\text{ km}$, verifying the factor-of-2 rule.
(c) Depth for $g = g_0 / 4$:
$$g(d_2) = g_0 \left(1 – \frac{d_2}{R}\right) = \frac{g_0}{4} \implies 1 – \frac{d_2}{R} = \frac{1}{4} \implies \frac{d_2}{R} = \frac{3}{4}$$
$$\mathbf{d_2 = \frac{3}{4} R = 0.75 \times 6400\text{ km} = 4800\text{ km}}$$
(d) Altitude for $g = g_0 / 4$:
For large variations, the exact formula is required:
$$g(h_2) = \frac{g_0}{\left(1 + \frac{h_2}{R}\right)^2} = \frac{g_0}{4} \implies 1 + \frac{h_2}{R} = 2 \implies \mathbf{h_2 = R = 6400\text{ km}}$$
Notice that $d_2 = 4800\text{ km} \ne 2 h_2$! The factor-of-2 rule fails completely for large percentage changes.
Example 2 (Advanced SHM Mechanics – Particle Dropped Through a Tunnel in Earth):
A straight, smooth, frictionless tunnel is drilled entirely through the Earth along a diameter passing through the center $O$. A small particle of mass $m$ is released from rest at the surface entrance of the tunnel.
(a) Show that the particle executes Simple Harmonic Motion (SHM).
(b) Derive the time period $T$ of oscillation and calculate its numerical value.
(c) Calculate the maximum speed $v_{\text{max}}$ of the particle as it crosses the Earth’s center.
(d) Prove that if the tunnel were bored along an arbitrary chord (not passing through the center), the motion would still be SHM with the exact same time period $T$.
Solution:
(a) Proof of SHM:
Let $x$ be the displacement of the particle from the Earth’s center $O$ along the tunnel ($x = r$).
At distance $x$ from the center, the gravitational force directed toward $O$ is:
$$F(x) = -m g(x) = -m \left(g_0 \frac{x}{R}\right) = -\left(\frac{m g_0}{R}\right) x$$
This equation has the exact form of Hooke’s Law:
$$F = -k x \quad \text{where } k = \frac{m g_0}{R}$$
Since the restoring force is directly proportional to displacement and directed toward the center, the particle executes Simple Harmonic Motion.
(b) Time Period of Oscillation:
The angular frequency of oscillation is:
$$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{m g_0 / R}{m}} = \mathbf{\sqrt{\frac{g_0}{R}}}$$
The time period $T$ of complete oscillation (surface $\to$ center $\to$ opposite surface $\to$ return) is:
$$\mathbf{T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{R}{g_0}}}$$
Substituting $R = 6.4 \times 10^6\text{ m}$ and $g_0 = 9.8\text{ m/s}^2$:
$$T = 2\pi \sqrt{\frac{6.4 \times 10^6}{9.8}} = 2\pi \sqrt{6.5306 \times 10^5} = 2\pi (808.12) \approx \mathbf{5077\text{ seconds} \approx 84.6\text{ minutes}}$$
The one-way transit time through the tunnel from one side of Earth to the other is:
$$t_{\text{transit}} = \frac{T}{2} \approx \mathbf{42.3\text{ minutes}}$$
(c) Maximum Velocity at the Center:
The amplitude of oscillation is the Earth’s radius $A = R$.
$$v_{\text{max}} = \omega A = \sqrt{\frac{g_0}{R}} \cdot R = \mathbf{\sqrt{g_0 R}}$$
Numerically:
$$v_{\text{max}} = \sqrt{(9.8)(6.4 \times 10^6)} = \sqrt{6.272 \times 10^7} \approx \mathbf{7919.6\text{ m/s} \approx 7.92\text{ km/s}}$$
(d) Arbitrary Non-Diametrical Chord Tunnel:
Let a straight tunnel be drilled along a chord at perpendicular distance $y_0$ from the center. Let $x$ be the coordinate along the tunnel measured from the midpoint of the chord.
The distance of the particle from the center is $r = \sqrt{x^2 + y_0^2}$.
The total gravitational force points toward the center $O$:
$$F_{\text{total}} = \frac{m g_0}{R} r$$
The component of force parallel to the tunnel (restoring force along $x$) is:
$$F_x = -F_{\text{total}} \cos\theta = -\left(\frac{m g_0}{R} r\right) \left(\frac{x}{r}\right) = -\left(\frac{m g_0}{R}\right) x$$
The normal force from the tunnel wall balances the transverse component $F_y$.
The equation of motion along the tunnel remains identically:
$$m \frac{d^2x}{dt^2} = -\left(\frac{m g_0}{R}\right) x \implies \mathbf{T = 2\pi \sqrt{\frac{R}{g_0}} \approx 84.6\text{ minutes}}$$
Remarkable Result: The transit time between any two cities on Earth connected by a frictionless straight tunnel is always exactly $42.3\text{ minutes}$, regardless of the tunnel length!
Example 3 (Rotational Dynamics & Plumb Line Deviation at Latitude $\lambda$):
A stationary observer on Earth holds a plumb line at latitude $\lambda = 45^\circ$.
(a) Determine the effective acceleration due to gravity $g'(\lambda)$ at this latitude. (Take $g_0 = 9.80\text{ m/s}^2, \omega = 7.29 \times 10^{-5}\text{ rad/s}, R = 6.4 \times 10^6\text{ m}$).
(b) Derive the expression for the angle of deflection $\theta$ of the plumb line from the true geometric vertical (direction toward the Earth’s center).
(c) If the Earth’s rotation speed accelerates so that objects at the equator experience weightlessness, find the new apparent weight of a $70\text{ kg}$ person at latitude $\lambda = 60^\circ$.
Solution:
(a) Effective Gravity at $\lambda = 45^\circ$:
$$\omega^2 R = (7.292 \times 10^{-5})^2 (6.4 \times 10^6) = (5.317 \times 10^{-9})(6.4 \times 10^6) = 0.0340\text{ m/s}^2$$
$$g'(45^\circ) = g_0 – \omega^2 R \cos^2(45^\circ) = 9.80 – 0.0340 \left(\frac{1}{\sqrt{2}}\right)^2 = 9.80 – \frac{0.0340}{2} = 9.80 – 0.0170 = \mathbf{9.783\text{ m/s}^2}$$
(b) Angular Deviation of Plumb Line:
The plumb line hangs along the direction of effective gravity $\vec{g}’$.
Resolving forces perpendicular and parallel to the true vertical:
– Perpendicular (tangential) component of centrifugal force: $F_{\text{cf}, \perp} = m \omega^2 R \cos\lambda \sin\lambda$.
– Radial component: $F_{\text{rad}} \approx m g_0$.
The deflection angle $\theta$ is:
$$\tan\theta = \frac{m \omega^2 R \cos\lambda \sin\lambda}{m g_0 – m \omega^2 R \cos^2\lambda} \approx \frac{\omega^2 R \sin(2\lambda)}{2 g_0}$$
For $\lambda = 45^\circ \implies \sin(2\lambda) = \sin 90^\circ = 1$:
$$\theta \approx \frac{\omega^2 R}{2 g_0} = \frac{0.0340}{2(9.8)} = \frac{0.0340}{19.6} \approx 1.73 \times 10^{-3}\text{ rad} \approx \mathbf{0.099^\circ \approx 5.96\text{ arcminutes}}$$
(c) Apparent Weight When Equator is Weightless:
Condition for weightlessness at the equator:
$$g_0 – \omega_{\text{new}}^2 R = 0 \implies \omega_{\text{new}}^2 R = g_0$$
At latitude $\lambda = 60^\circ$:
$$g'(60^\circ) = g_0 – \omega_{\text{new}}^2 R \cos^2(60^\circ) = g_0 – g_0 \left(\frac{1}{2}\right)^2 = g_0 – \frac{1}{4} g_0 = \mathbf{\frac{3}{4} g_0}$$
For a person of mass $m = 70\text{ kg}$:
$$W’ = m g'(60^\circ) = \frac{3}{4} m g_0 = \frac{3}{4}(70)(9.8) = \mathbf{514.5\text{ N}}$$
(The person retains exactly $75\%$ of their normal weight!).
Example 4 (Non-Uniform Density Model – Core-Mantle Boundary Gravity):
A simplified geophysical model represents a planet of total radius $R$ as consisting of two concentric uniform layers:
1. A dense central core of radius $R_c = R/2$ and uniform mass density $\rho_c = 2\rho_0$.
2. An outer mantle extending from $r = R/2$ to $r = R$ of uniform density $\rho_m = \rho_0$.
(a) Calculate the total mass $M$ of the planet in terms of $\rho_0$ and $R$.
(b) Find the acceleration due to gravity $g_{\text{surface}}$ at the surface $r = R$.
(c) Find the acceleration due to gravity $g_{\text{core}}$ at the core-mantle boundary $r = R/2$.
(d) Determine whether gravity increases or decreases as one descends from the surface to the core-mantle boundary.
Solution:
(a) Total Mass of the Planet:
– Mass of the core:
$$M_{\text{core}} = \frac{4}{3}\pi \left(\frac{R}{2}\right)^3 (2\rho_0) = \frac{4}{3}\pi \frac{R^3}{8} (2\rho_0) = \frac{4}{3}\pi R^3 \rho_0 \left(\frac{1}{4}\right)$$
– Mass of the mantle:
$$M_{\text{mantle}} = \frac{4}{3}\pi \left[ R^3 – \left(\frac{R}{2}\right)^3 \right] \rho_0 = \frac{4}{3}\pi R^3 \left(1 – \frac{1}{8}\right) \rho_0 = \frac{4}{3}\pi R^3 \rho_0 \left(\frac{7}{8}\right)$$
Total mass:
$$M = M_{\text{core}} + M_{\text{mantle}} = \frac{4}{3}\pi R^3 \rho_0 \left(\frac{2}{8} + \frac{7}{8}\right) = \mathbf{\frac{9}{8} \left(\frac{4}{3}\pi R^3 \rho_0\right)}$$
(b) Surface Gravity ($r = R$):
$$g_{\text{surface}} = \frac{G M}{R^2} = \frac{G \left[\frac{9}{8} \left(\frac{4}{3}\pi R^3 \rho_0\right)\right]}{R^2} = \mathbf{\frac{9}{8} \left(\frac{4}{3}\pi G \rho_0 R\right) = \frac{3}{2}\pi G \rho_0 R}$$
(c) Gravity at Core-Mantle Boundary ($r = R/2$):
By Newton’s Shell Theorem, the mantle exerts zero force on any point inside it. Only the core attracts:
$$g_{\text{core}} = \frac{G M_{\text{core}}}{(R/2)^2} = \frac{G \left[\frac{4}{3}\pi \frac{R^3}{4} \rho_0\right]}{\frac{R^2}{4}} = \mathbf{\frac{4}{3}\pi G \rho_0 R}$$
(d) Comparison:
$$\frac{g_{\text{core}}}{g_{\text{surface}}} = \frac{\frac{4}{3}}{\frac{9}{8} \cdot \frac{4}{3}} = \frac{8}{9} \approx 0.889$$
$g_{\text{core}} < g_{\text{surface}}$. Gravity decreases as one descends from the surface into the mantle, but because of the dense core, it only drops by $11.1\%$ at the halfway mark (compared to a $50\%$ drop in a completely uniform planet!).
Practice Worksheet: Variation of $g$ with Depth and Latitude
Section A: Single Correct Questions (Q1 to Q5)
Q1. If $R$ is the radius of the Earth, at what depth $d$ below the Earth’s surface will the acceleration due to gravity become $\frac{1}{n}$ of its value at the surface? (Assume uniform Earth density)
(A) $\frac{R}{n}$
(B) $R\left(1 – \frac{1}{n}\right)$
(C) $\frac{R}{n^2}$
(D) $R\left(1 – \frac{1}{n^2}\right)$
Q2. At what altitude $h$ above the Earth’s surface does the acceleration due to gravity experience the same decrease as that observed at a depth $d = 20\text{ km}$ below the Earth’s surface? (Assume $h \ll R$)
(A) $5\text{ km}$
(B) $10\text{ km}$
(C) $20\text{ km}$
(D) $40\text{ km}$
Q3. If the Earth suddenly speeds up its rotation such that bodies on the equator fly off or experience complete weightlessness, the required angular speed $\omega$ of the Earth is:
(A) $\sqrt{\frac{g_0}{R}}$
(B) $\sqrt{\frac{2g_0}{R}}$
(C) $\sqrt{\frac{g_0}{2R}}$
(D) $\frac{g_0}{R}$
Q4. If the Earth rotates at the critical angular speed such that the apparent weight of a body at the equator is zero, what will be the apparent weight of the same body at a latitude of $\lambda = 60^\circ$?
(A) Zero
(B) $\frac{1}{4} m g_0$
(C) $\frac{1}{2} m g_0$
(D) $\frac{3}{4} m g_0$
Q5. A particle of mass $m$ is released from rest at the surface entrance of a straight frictionless tunnel drilled through the center of the Earth. The time taken by the particle to travel from one end of the tunnel to the other is approximately:
(A) $21.1\text{ minutes}$
(B) $42.3\text{ minutes}$
(C) $84.6\text{ minutes}$
(D) $169.2\text{ minutes}$
Section B: One or More than One Correct Questions (Q6 to Q8)
Q6. Which of the following statements regarding the graph of acceleration due to gravity $g$ versus distance $r$ from the center of a uniform Earth ($0 \le r < \infty$) is/are correct?
(A) For $r \le R$, the graph is a straight line passing through the origin with slope $\frac{g_0}{R}$.
(B) For $r \ge R$, the graph is a hyperbola asymptotically approaching the $r$-axis.
(C) The maximum value of $g$ occurs precisely at the surface $r = R$.
(D) The slope of the graph is continuous at $r = R$.
Q7. If the diurnal rotation of the Earth were to stop completely, which of the following consequences would occur?
(A) The value of $g$ at the geographic poles would remain completely unchanged.
(B) The value of $g$ at the equator would increase by $\omega^2 R$.
(C) The value of $g$ at latitude $\lambda = 45^\circ$ would increase by $\frac{1}{2}\omega^2 R$.
(D) The length of the day would remain $24\text{ hours}$.
Q8. Consider two identical test masses $m$: Mass 1 is taken to height $h$ ($h \ll R$) and Mass 2 is taken to depth $d = h$. Which of the following statements is/are correct?
(A) The decrease in $g$ experienced by Mass 1 is twice the decrease experienced by Mass 2.
(B) Mass 1 weighs less than Mass 2.
(C) The work done in moving Mass 1 from the surface to height $h$ is approximately $m g_0 h$.
(D) The work done by gravity in moving Mass 2 from the surface to depth $d = h$ is $m g_0 h \left(1 – \frac{h}{2R}\right)$.
Section C: Numerical Value Type Questions (Q9 to Q10)
Q9. At an altitude $h = 1600\text{ km}$ above the Earth’s surface, the acceleration due to gravity has a certain value. Find the depth $d$ (in kilometers) below the Earth’s surface at which the acceleration due to gravity has the exact same value. (Take $R = 6400\text{ km}$).
Q10. If a body placed at the equator is to experience an apparent weight reduction of exactly $50\%$ due to Earth’s rotation, the angular speed of Earth’s rotation must become $\omega = \sqrt{\frac{g_0}{n R}}$. Find the integer value of $n$.
Comprehensive Solutions and Answer Key
Answer Key
| Question | Correct Answer | Core Physics Concept |
|---|---|---|
| Q1 | (B) | Depth formula inversion: $g(d) = g_0/n \implies d = R(1 – 1/n)$ |
| Q2 | (B) | Factor-of-2 Rule: $2h/R = d/R \implies h = d/2 = 10\text{ km}$ |
| Q3 | (A) | Critical rotation speed for equatorial weightlessness: $\omega = \sqrt{g_0/R}$ |
| Q4 | (D) | Latitude weight under critical spin: $g’ = g_0 – g_0\cos^2(60^\circ) = \frac{3}{4}g_0$ |
| Q5 | (B) | Transit time through Earth tunnel: $T/2 = \pi\sqrt{R/g_0} \approx 42.3\text{ min}$ |
| Q6 | (A, B, C) | Complete $g(r)$ profile: linear inside, inverse-square outside, slope jump |
| Q7 | (A, B, C) | Rotational cessation effects: pole unaffected, equatorial increase |
| Q8 | (A, B, C, D) | Comparative dynamics: altitude vs depth changes and work integrals |
| Q9 | 2304 | Equating altitude & depth gravities: $d = 0.36 R = 2304\text{ km}$ |
| Q10 | 2 | $50\%$ weight reduction condition: $\omega^2 R = 0.5 g_0 \implies n = 2$ |
Detailed Explanations
Solution 1:
The acceleration due to gravity at depth $d$ is:
$$g(d) = g_0 \left(1 – \frac{d}{R}\right) = \frac{g_0}{n}$$
$$1 – \frac{d}{R} = \frac{1}{n} \implies \frac{d}{R} = 1 – \frac{1}{n} \implies \mathbf{d = R\left(1 – \frac{1}{n}\right)}$$
Correct Answer: (B)
Solution 2:
For small altitudes $h \ll R$, the fractional decrease is:
$$\left(\frac{\Delta g}{g_0}\right)_{\text{height}} \approx \frac{2h}{R}$$
For depth $d$, the fractional decrease is:
$$\left(\frac{\Delta g}{g_0}\right)_{\text{depth}} = \frac{d}{R}$$
Equating both fractional decreases:
$$\frac{2h}{R} = \frac{d}{R} \implies h = \frac{d}{2} = \frac{20\text{ km}}{2} = \mathbf{10\text{ km}}$$
Correct Answer: (B)
Solution 3:
Apparent acceleration due to gravity at the equator is:
$$g_{\text{equator}} = g_0 – \omega^2 R$$
For weightlessness, $g_{\text{equator}} = 0$:
$$g_0 – \omega^2 R = 0 \implies \omega^2 R = g_0 \implies \mathbf{\omega = \sqrt{\frac{g_0}{R}}}$$
Correct Answer: (A)
Solution 4:
When the equator experiences weightlessness:
$$\omega^2 R = g_0$$
The apparent gravity at latitude $\lambda = 60^\circ$ is:
$$g'(60^\circ) = g_0 – \omega^2 R \cos^2(60^\circ) = g_0 – g_0 \left(\frac{1}{2}\right)^2 = g_0 – \frac{1}{4}g_0 = \mathbf{\frac{3}{4} g_0}$$
The apparent weight is:
$$W’ = m g'(60^\circ) = \mathbf{\frac{3}{4} m g_0}$$
Correct Answer: (D)
Solution 5:
The particle executes SHM with period:
$$T = 2\pi \sqrt{\frac{R}{g_0}} \approx 84.6\text{ minutes}$$
The transit time from one surface to the other is half of the period:
$$t_{\text{transit}} = \frac{T}{2} = \pi \sqrt{\frac{R}{g_0}} = \frac{84.6\text{ min}}{2} = \mathbf{42.3\text{ minutes}}$$
Correct Answer: (B)
Solution 6:
– (A) For $r \le R$, $g(r) = \frac{g_0}{R} r$, which is a straight line through the origin with slope $g_0/R$. (True).
– (B) For $r \ge R$, $g(r) = g_0 \frac{R^2}{r^2}$, which is an inverse-square hyperbola. (True).
– (C) The peak value $g = g_0$ occurs at the boundary $r = R$. (True).
– (D) The left-hand derivative is $+g_0/R$ and the right-hand derivative is $-2g_0/R$. The slope is discontinuous. (Statement D is false).
Statements (A, B, C) are correct.
Correct Answer: (A, B, C)
Solution 7:
Before stopping: $g'(\lambda) = g_0 – \omega^2 R \cos^2\lambda$.
– (A) At poles ($\lambda = 90^\circ$): $\cos 90^\circ = 0 \implies g’ = g_0$. The poles experience zero centrifugal force, so their gravity is unaffected. (True).
– (B) At the equator ($\lambda = 0^\circ$): $g’ = g_0 – \omega^2 R$. When rotation stops, $g$ becomes $g_0$, an increase of $\omega^2 R$. (True).
– (C) At $\lambda = 45^\circ$: $g’ = g_0 – \omega^2 R (1/\sqrt{2})^2 = g_0 – \frac{1}{2}\omega^2 R$. Upon stopping, $g$ increases by $\frac{1}{2}\omega^2 R$. (True).
– (D) Day length would become infinite (or 1 year relative to the Sun). (False).
Statements (A, B, C) are correct.
Correct Answer: (A, B, C)
Solution 8:
– (A) For small $h$: $\Delta g_{\text{height}} \approx \frac{2h}{R} g_0$, whereas $\Delta g_{\text{depth}} = \frac{h}{R} g_0$. The height decrease is twice the depth decrease. (True).
– (B) Since $g(h) < g(d)$ for $d = h$, Mass 1 weighs less than Mass 2. (True).
- (C) For $h \ll R$, gravity is approximately constant at $g_0$, so work done is $m g_0 h$. (True).
- (D) Work done by gravity in moving from $r = R$ to $r = R - h$:
$$W = \int_{R-h}^R m g(r) dr = \int_{R-h}^R \frac{m g_0}{R} r dr = \frac{m g_0}{R} \left[ \frac{r^2}{2} \right]_{R-h}^R = \frac{m g_0}{2R} [R^2 - (R - h)^2]$$
$$W = \frac{m g_0}{2R} [R^2 - (R^2 - 2Rh + h^2)] = \frac{m g_0}{2R} (2Rh - h^2) = m g_0 h \left(1 - \frac{h}{2R}\right)$$
(True).
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 9:
At altitude $h = 1600\text{ km}$:
$$\frac{h}{R} = \frac{1600\text{ km}}{6400\text{ km}} = 0.25 = \frac{1}{4}$$
The acceleration due to gravity is:
$$g(h) = \frac{g_0}{\left(1 + \frac{h}{R}\right)^2} = \frac{g_0}{(1 + 0.25)^2} = \frac{g_0}{(1.25)^2} = \frac{g_0}{1.5625} = 0.64 g_0$$
At depth $d$:
$$g(d) = g_0 \left(1 – \frac{d}{R}\right)$$
Equating the two:
$$g_0 \left(1 – \frac{d}{R}\right) = 0.64 g_0 \implies 1 – \frac{d}{R} = 0.64 \implies \frac{d}{R} = 0.36$$
$$d = 0.36 R = 0.36(6400\text{ km}) = \mathbf{2304\text{ km}}$$
Correct Answer: 2304
Solution 10:
Apparent gravity at the equator is:
$$g’ = g_0 – \omega^2 R$$
For an apparent weight reduction of $50\%$:
$$g’ = 0.50 g_0 \implies g_0 – \omega^2 R = 0.50 g_0 \implies \omega^2 R = 0.50 g_0 = \frac{g_0}{2}$$
$$\omega = \sqrt{\frac{g_0}{2R}}$$
Comparing with $\omega = \sqrt{\frac{g_0}{n R}}$:
$$\mathbf{n = 2}$$
Correct Answer: 2