Concept Card: Acceleration Due to Gravity and Its Variation with Altitude
1. Fundamental Definition of Acceleration Due to Gravity ($g$):
The acceleration experienced by an unconstrained test mass $m$ due to the gravitational attraction of a celestial body (such as Earth) in the absence of air resistance is termed the acceleration due to gravity ($g$).
Derivation at the Earth’s Surface:
By Newton’s Universal Law of Gravitation, the attractive force exerted by Earth of mass $M$ and mean radius $R$ on a point mass $m$ resting on its surface is:
$$F = G \frac{M m}{R^2}$$
By Newton’s Second Law of Motion, this gravitational attraction imparts an acceleration $g_0$:
$$F = m g_0 \implies \mathbf{g_0 = \frac{G M}{R^2}}$$
In terms of the mean mass density $\rho$ of Earth (approximating Earth as a uniform sphere of volume $V = \frac{4}{3}\pi R^3$):
$$M = \frac{4}{3}\pi R^3 \rho \implies g_0 = \frac{G \left(\frac{4}{3}\pi R^3 \rho\right)}{R^2} = \mathbf{\frac{4}{3}\pi G \rho R}$$
Standard Reference Value: At sea level, standard gravity is defined as $g_0 \approx 9.80665\text{ m/s}^2 \approx 9.8\text{ m/s}^2$.
2. Exact Variation with Altitude ($h$):
Consider a point mass $m$ elevated to an altitude $h$ above the Earth’s surface (distance from the center of Earth $r = R + h$).
By Newton’s Shell Theorem, the entire mass $M$ of Earth acts as if concentrated at its geometric center $O$. The gravitational acceleration at height $h$ is:
$$g(h) = \frac{G M}{(R + h)^2}$$
Factoring out $R^2$ from the denominator:
$$g(h) = \frac{G M}{R^2 \left(1 + \frac{h}{R}\right)^2}$$
Since $g_0 = \frac{G M}{R^2}$, we obtain the exact governing formula:
Exact Formula for Any Altitude $h \ge 0$:
$$\mathbf{g(h) = \frac{g_0}{\left(1 + \frac{h}{R}\right)^2}}$$
3. Binomial Approximation for Small Altitudes ($h \ll R$):
When the altitude is very small compared to the Earth’s radius ($h \ll R$, practically $h \le 320\text{ km}$ or $h/R \le 0.05$):
$$g(h) = g_0 \left(1 + \frac{h}{R}\right)^{-2}$$
Expanding via the binomial series $(1 + x)^n = 1 + n x + \frac{n(n-1)}{2!} x^2 + \dots$:
$$\left(1 + \frac{h}{R}\right)^{-2} = 1 – 2\left(\frac{h}{R}\right) + 3\left(\frac{h}{R}\right)^2 – \dots$$
Retaining only first-order terms in $(h/R)$:
First-Order Binomial Approximation ($h \ll R$):
$$\mathbf{g(h) \approx g_0 \left(1 – \frac{2h}{R}\right)}$$
Key Derived Quantities for $h \ll R$:
- Absolute Decrease in $g$:
$$\Delta g = g_0 – g(h) \approx \frac{2h}{R} g_0$$ - Fractional Decrease in $g$:
$$\frac{\Delta g}{g_0} = \frac{g_0 – g(h)}{g_0} \approx \frac{2h}{R}$$ - Percentage Decrease in $g$:
$$\% \text{ decrease} = \left(\frac{\Delta g}{g_0}\right) \times 100\% \approx \left(\frac{2h}{R}\right) \times 100\%$$ - Rate of Decrease with Altitude:
$$\frac{dg}{dh} = \frac{d}{dh} \left[ g_0 \left(1 + \frac{h}{R}\right)^{-2} \right] = -\frac{2 g_0}{R \left(1 + \frac{h}{R}\right)^3} \approx -\mathbf{\frac{2 g_0}{R}}$$
Numerically: $\frac{2 g_0}{R} \approx \frac{2(9.8)}{6.4 \times 10^6} \approx 3.06 \times 10^{-6}\text{ s}^{-2} \approx 0.0306\text{ m/s}^2\text{ per kilometer of altitude}$.
4. Work, Energy, and True Potential Energy at Altitude:
Because $g(h)$ is not constant over large vertical displacements, writing $\Delta U = m g_0 h$ introduces massive errors when $h$ is comparable to $R$.
- Gravitational Potential Energy at Distance $r = R + h$:
$$U(r) = -\frac{G M m}{r} = -\frac{G M m}{R + h}$$ - Exact Potential Energy Gain:
$$\Delta U = U(R + h) – U(R) = -\frac{G M m}{R + h} – \left(-\frac{G M m}{R}\right) = G M m \left(\frac{1}{R} – \frac{1}{R + h}\right)$$
$$\mathbf{\Delta U = \frac{G M m h}{R(R + h)} = \frac{m g_0 h}{1 + \frac{h}{R}}}$$
Verification: For $h \ll R$, $\frac{h}{R} \to 0$, giving $\Delta U \approx m g_0 h$ (recovering standard introductory mechanics).
5. Common Analytical Traps & Exam Pitfalls in JEE:
- Trap 1 (Inappropriate Use of Binomial Approximation):
Never use $g(h) \approx g_0 (1 – 2h/R)$ when $h$ is given as a fraction of $R$ (e.g., $h = R, R/2, R/4$).
Catastrophic Example: At $h = R$:
– Exact formula: $g(R) = \frac{g_0}{(1 + 1)^2} = \mathbf{\frac{g_0}{4}} = 0.25 g_0$.
– Binomial formula: $g(R) \approx g_0 (1 – 2) = \mathbf{-g_0}$ (gives negative gravity!). - Trap 2 (Altitude vs. Radial Distance from Center):
Always verify whether the question gives altitude $h$ above Earth’s surface or radial distance $r$ from Earth’s center ($r = R + h$). If $r = 2R$, then $h = R$, not $2R$. - Trap 3 (“Decreases by $x\%$ ” vs. “Becomes $x\%$ ”):
– If $g$ decreases by $1\%$: $g(h) = 0.99 g_0 \implies \frac{\Delta g}{g_0} = 0.01 \approx \frac{2h}{R} \implies h \approx \frac{0.01 R}{2} = 32\text{ km}$.
– If $g$ becomes $1\%$ of surface value: $g(h) = 0.01 g_0 = \frac{g_0}{100} \implies \left(1 + \frac{h}{R}\right)^2 = 100 \implies 1 + \frac{h}{R} = 10 \implies h = 9R$!
Solved Examples
Example 1 (Direct Conceptual & Threshold Comparison – Exact vs. Approximate Formulas):
(Take radius of Earth $R = 6400\text{ km}$ and $g_0 = 9.8\text{ m/s}^2$).
(a) At what height $h_1$ above the Earth’s surface does the acceleration due to gravity decrease by $1.0\%$?
(b) At what height $h_2$ above the Earth’s surface does the acceleration due to gravity become $25\%$ of its surface value?
(c) Compare the percentage error in $g$ obtained from the binomial approximation $g_{\text{approx}} = g_0(1 – 2h/R)$ versus the exact formula at altitudes $h = 64\text{ km}$ and $h = 640\text{ km}$.
Solution:
(a) Height for $1.0\%$ Decrease:
Since the decrease is small ($1.0\% \ll 100\%$), we can safely apply the first-order binomial formula:
$$\frac{\Delta g}{g_0} = \frac{2 h_1}{R} = 0.010$$
$$h_1 = \frac{0.010 \times R}{2} = \frac{0.010 \times 6400\text{ km}}{2} = \mathbf{32.0\text{ km}}$$
Verification with exact formula:
$$\frac{g}{g_0} = \frac{1}{(1 + 32/6400)^2} = \frac{1}{(1 + 0.005)^2} = \frac{1}{1.010025} \approx 0.990074 \implies \text{Decrease } = 0.9926\% \approx 1.0\%.$$
(b) Height where $g(h_2) = 0.25 g_0$:
Here the change is huge ($75\%$ reduction); the exact formula is strictly mandatory:
$$g(h_2) = \frac{g_0}{\left(1 + \frac{h_2}{R}\right)^2} = \frac{g_0}{4}$$
$$\left(1 + \frac{h_2}{R}\right)^2 = 4 \implies 1 + \frac{h_2}{R} = 2 \implies \frac{h_2}{R} = 1$$
$$\mathbf{h_2 = R = 6400\text{ km}}$$
(c) Percentage Error Comparison:
1. For $h = 64\text{ km}$ ($h/R = 64/6400 = 0.01$):
– Exact: $g_{\text{exact}} = \frac{g_0}{(1 + 0.01)^2} = \frac{g_0}{1.0201} \approx 0.980296 g_0$.
– Approx: $g_{\text{approx}} = g_0(1 – 2(0.01)) = 0.980000 g_0$.
– Percentage error: $\frac{|0.980296 – 0.980000|}{0.980296} \times 100\% = \mathbf{0.030\%}$ (negligible!).
2. For $h = 640\text{ km}$ ($h/R = 640/6400 = 0.10$):
– Exact: $g_{\text{exact}} = \frac{g_0}{(1 + 0.10)^2} = \frac{g_0}{1.21} \approx 0.8264 g_0$.
– Approx: $g_{\text{approx}} = g_0(1 – 2(0.10)) = 0.8000 g_0$.
– Percentage error: $\frac{|0.8264 – 0.8000|}{0.8264} \times 100\% = \mathbf{3.2\%}$ (significant discrepancy!).
Takeaway: For $h/R \le 0.01$ ($h \le 64\text{ km}$), the approximation error is under $0.03\%$. Beyond $h/R \ge 0.05$ ($h \ge 320\text{ km}$), only the exact inverse-square formula should be used.
Example 2 (Mathematical Integration – Vertical Projection Under Variable Gravity):
A projectile of mass $m$ is fired vertically upwards from the surface of Earth with an initial launch speed $v_0 = \sqrt{g_0 R}$. Neglecting atmospheric drag:
(a) Determine the maximum altitude $H$ attained by the projectile using the Work-Energy Theorem.
(b) Derive the expression for the velocity $v(h)$ as a function of altitude $h$.
(c) Compare this maximum altitude with the elementary introductory formula $H_{\text{naive}} = \frac{v_0^2}{2g_0}$.
Solution:
(a) Maximum Altitude $H$ via Mechanical Energy Conservation:
Total mechanical energy at Earth’s surface ($r = R$):
$$E_{\text{surface}} = \frac{1}{2} m v_0^2 – \frac{G M m}{R}$$
Since $g_0 = \frac{G M}{R^2} \implies G M = g_0 R^2$:
$$E_{\text{surface}} = \frac{1}{2} m (\sqrt{g_0 R})^2 – m g_0 R = \frac{1}{2} m g_0 R – m g_0 R = -\frac{1}{2} m g_0 R$$
At the apex of trajectory ($r = R + H$), instantaneous speed $v = 0$:
$$E_{\text{apex}} = 0 – \frac{G M m}{R + H} = -\frac{m g_0 R^2}{R + H}$$
By conservation of mechanical energy ($E_{\text{surface}} = E_{\text{apex}}$):
$$-\frac{1}{2} m g_0 R = -\frac{m g_0 R^2}{R + H}$$
$$\frac{1}{2} = \frac{R}{R + H} \implies R + H = 2R \implies \mathbf{H = R}$$
The projectile ascends to an altitude exactly equal to the Earth’s radius ($H = 6400\text{ km}$).
(b) Velocity $v(h)$ at Intermediate Altitude $h$:
$$\frac{1}{2} m v(h)^2 – \frac{m g_0 R^2}{R + h} = -\frac{1}{2} m g_0 R$$
$$\frac{1}{2} v(h)^2 = \frac{g_0 R^2}{R + h} – \frac{1}{2} g_0 R = g_0 R \left( \frac{R}{R + h} – \frac{1}{2} \right) = g_0 R \left( \frac{R – h}{2(R + h)} \right)$$
$$\mathbf{v(h) = \sqrt{\frac{g_0 R (R – h)}{R + h}}}$$
Sanity Check: At $h = 0 \implies v(0) = \sqrt{g_0 R}$. At $h = R \implies v(R) = 0$. Matches perfectly!
(c) Contrast with Naive Uniform-Gravity Calculation:
If one mistakenly assumed gravity remained constant at $g_0$:
$$H_{\text{naive}} = \frac{v_0^2}{2g_0} = \frac{(\sqrt{g_0 R})^2}{2g_0} = \frac{g_0 R}{2g_0} = \frac{R}{2} = 0.5 R$$
The true altitude ($H = R$) is double the naive calculation! Because $g(h)$ weakens as the projectile climbs, gravity removes kinetic energy at a progressively lower rate, allowing the projectile to travel twice as high.
Example 3 (Multi-Concept Linkage – Satellite Orbit at Substantial Altitude):
A scientific research satellite of mass $m = 1200\text{ kg}$ circles Earth in a stable circular orbit at an altitude equal to half the Earth’s radius ($h = R/2 = 3200\text{ km}$). (Take $R = 6400\text{ km}$, $g_0 = 9.8\text{ m/s}^2$).
(a) Find the effective acceleration due to gravity $g(h)$ at the satellite’s orbital altitude.
(b) Calculate the orbital velocity $v_{\text{orb}}$ of the satellite.
(c) Determine the orbital time period $T$ of the satellite.
(d) What is the true gravitational weight of the satellite at this orbit, and what is its apparent weight registered by an onboard spring balance?
Solution:
(a) Acceleration Due to Gravity at $h = R/2$:
The radial distance from the center of Earth is $r = R + h = R + 0.5R = 1.5R$.
$$g(h) = \frac{g_0}{\left(1 + \frac{h}{R}\right)^2} = \frac{g_0}{\left(1 + 0.5\right)^2} = \frac{g_0}{(1.5)^2} = \frac{g_0}{2.25} = \frac{4}{9} g_0$$
Substituting $g_0 = 9.8\text{ m/s}^2$:
$$g(h) = \frac{4}{9}(9.8) = \mathbf{4.356\text{ m/s}^2}$$
(b) Orbital Speed $v_{\text{orb}}$:
The gravitational force provides the necessary centripetal acceleration:
$$\frac{m v_{\text{orb}}^2}{r} = m g(h) \implies v_{\text{orb}} = \sqrt{r g(h)} = \sqrt{(1.5 R) \left(\frac{4}{9} g_0\right)} = \sqrt{\frac{2}{3} g_0 R}$$
Substitute $R = 6.4 \times 10^6\text{ m}$ and $g_0 = 9.8\text{ m/s}^2$:
$$v_{\text{orb}} = \sqrt{\frac{2}{3}(9.8)(6.4 \times 10^6)} = \sqrt{4.1813 \times 10^7} = \mathbf{6.466 \times 10^3\text{ m/s} = 6.47\text{ km/s}}$$
(c) Orbital Period $T$:
$$T = \frac{2\pi r}{v_{\text{orb}}} = \frac{2\pi (1.5 \times 6.4 \times 10^6\text{ m})}{6466\text{ m/s}} = \frac{6.032 \times 10^7}{6466} \approx \mathbf{9329\text{ seconds} \approx 2.59\text{ hours}}$$
(d) True Weight vs. Apparent Weight:
– True Gravitational Weight: The actual gravitational pull exerted by Earth:
$$W_{\text{true}} = m g(h) = (1200\text{ kg})(4.356\text{ m/s}^2) = \mathbf{5227\text{ N}}$$
(compared to surface weight $W_0 = 1200 \times 9.8 = 11760\text{ N}$).
– Apparent Weight (Registered on Scale): Inside the freely falling frame of the satellite, the normal contact reaction force is identically zero:
$$N = m (g(h) – a_{\text{centripetal}}) = m (g(h) – g(h)) = \mathbf{0\text{ N}}$$
The satellite and all its contents are in a state of continuous weightlessness (free fall).
Example 4 (Planetary Scaling & Equi-Gravity Altitudes):
Planet Exoplanet-$X$ has twice the mass of Earth ($M_X = 2 M_E$) and twice the radius of Earth ($R_X = 2 R_E$).
(a) Determine the acceleration due to gravity $g_{X, 0}$ at the surface of Planet-$X$ in terms of Earth’s surface gravity $g_0$.
(b) Find the mean mass density $\rho_X$ of Planet-$X$ relative to Earth’s mean density $\rho_E$.
(c) At what altitude $h_X$ above the surface of Planet-$X$ will the acceleration due to gravity equal the gravity experienced at an altitude $h_E = R_E$ above Earth’s surface?
Solution:
(a) Surface Gravity of Planet-$X$:
$$g_{X, 0} = \frac{G M_X}{R_X^2} = \frac{G (2 M_E)}{(2 R_E)^2} = \frac{2 G M_E}{4 R_E^2} = \frac{1}{2} \left(\frac{G M_E}{R_E^2}\right) = \mathbf{\frac{1}{2} g_0}$$
Surface gravity on Planet-$X$ is exactly half of Earth’s surface gravity.
(b) Mean Density Ratio:
$$\rho = \frac{M}{\frac{4}{3}\pi R^3} \implies \frac{\rho_X}{\rho_E} = \left(\frac{M_X}{M_E}\right) \left(\frac{R_E}{R_X}\right)^3 = (2) \left(\frac{1}{2}\right)^3 = 2 \times \frac{1}{8} = \mathbf{\frac{1}{4}}$$
The mean density of Planet-$X$ is one-fourth that of Earth.
(c) Altitude for Identical Gravitational Acceleration:
At altitude $h_E = R_E$ above Earth:
$$g_E(h_E) = \frac{G M_E}{(R_E + h_E)^2} = \frac{G M_E}{(2 R_E)^2} = \frac{G M_E}{4 R_E^2}$$
For Planet-$X$ at altitude $h_X$:
$$g_X(h_X) = \frac{G M_X}{(R_X + h_X)^2} = \frac{G (2 M_E)}{(2 R_E + h_X)^2}$$
Equating $g_X(h_X) = g_E(h_E)$:
$$\frac{2 G M_E}{(2 R_E + h_X)^2} = \frac{G M_E}{4 R_E^2}$$
$$\frac{2}{(2 R_E + h_X)^2} = \frac{1}{4 R_E^2} \implies (2 R_E + h_X)^2 = 8 R_E^2$$
Taking the square root:
$$2 R_E + h_X = \sqrt{8} R_E = 2\sqrt{2} R_E$$
$$\mathbf{h_X = (2\sqrt{2} – 2) R_E = 2(\sqrt{2} – 1) R_E \approx 0.828 R_E = 0.414 R_X}$$
Practice Worksheet: Acceleration Due to Gravity & Altitude Variation
Section A: Single Correct Questions (Q1 to Q5)
Q1. At what altitude $h$ above the surface of Earth does the acceleration due to gravity reduce to $\frac{1}{9}$ of its value at the Earth’s surface? (Radius of Earth = $R$)
(A) $\frac{R}{3}$
(B) $R$
(C) $2R$
(D) $3R$
Q2. At what altitude $h$ above the Earth’s surface does the acceleration due to gravity decrease by exactly $2.0\%$ compared to its surface value? (Assume $R = 6400\text{ km}$)
(A) $32\text{ km}$
(B) $64\text{ km}$
(C) $128\text{ km}$
(D) $256\text{ km}$
Q3. A simple pendulum clock keeps accurate time at the Earth’s surface. If this clock is transported to an altitude $h$ ($h \ll R$) above the Earth’s surface, the fractional loss of time per day $\frac{\Delta T}{T_{\text{day}}}$ experienced by the clock is approximately:
(A) $\frac{h}{2R}$
(B) $\frac{h}{R}$
(C) $\frac{2h}{R}$
(D) $\frac{4h}{R}$
Q4. The ratio of the weight of a body measured at an altitude $h = R$ above the Earth’s surface to its weight measured at the surface of Earth is:
(A) $1 : 2$
(B) $1 : 4$
(C) $1 : 8$
(D) $1 : 16$
Q5. What minimum vertical projection speed $v$ must be imparted to a body at Earth’s surface so that it ascends to a maximum altitude equal to the Earth’s radius ($h = R$)?
(A) $\sqrt{g_0 R}$
(B) $\sqrt{2 g_0 R}$
(C) $\sqrt{\frac{g_0 R}{2}}$
(D) $\sqrt{\frac{3}{2} g_0 R}$
Section B: One or More than One Correct Questions (Q6 to Q8)
Q6. Let $r$ be the radial distance from the center of Earth ($r \ge R$). The acceleration due to gravity $g(r)$ outside Earth satisfies which of the following mathematical properties?
(A) $\frac{dg}{dr} = -\frac{2g(r)}{r}$
(B) $\frac{d^2 g}{dr^2} = +\frac{6 G M}{r^4} > 0$ (the function $g(r)$ is strictly convex upward for $r \ge R$)
(C) $\lim_{r \to \infty} g(r) = 0$
(D) The slope of $g$ versus $r$ is discontinuous across the boundary $r = R$ between the interior and exterior of Earth.
Q7. Let $g_{\text{exact}} = \frac{g_0}{(1 + h/R)^2}$ and $g_{\text{approx}} = g_0\left(1 – \frac{2h}{R}\right)$. Which of the following statements is/are correct?
(A) For all altitudes $h > 0$, $g_{\text{exact}} > g_{\text{approx}}$.
(B) The difference $(g_{\text{exact}} – g_{\text{approx}})$ is proportional to $(h/R)^2$ to leading order.
(C) At $h = R/2$, $g_{\text{approx}}$ predicts that gravity becomes zero.
(D) The fractional error $\frac{g_{\text{exact}} – g_{\text{approx}}}{g_{\text{exact}}}$ exceeds $5\%$ when $h/R \ge 0.15$.
Q8. A rocket of mass $m$ ascends vertically from the Earth’s surface with a constant speed $v_0$. Which of the following statements is/are correct?
(A) The rate of work done by the rocket engines against gravity decreases continuously as altitude increases.
(B) The instantaneous power delivered against gravity at altitude $h$ is $P(h) = \frac{m g_0 v_0}{(1 + h/R)^2}$.
(C) The total mechanical work done against gravity in elevating the rocket from $h = 0$ to $h = R$ is $\frac{1}{2} m g_0 R$.
(D) If the rocket had moved with constant upward acceleration $a$, its weight would appear to increase without bound.
Section C: Numerical Value Type Questions (Q9 to Q10)
Q9. At an altitude $h$ above the Earth’s surface, the acceleration due to gravity is exactly $64\%$ of its value at the Earth’s surface. Given that the radius of Earth is $R = 6400\text{ km}$, calculate the value of altitude $h$ in kilometers ($\text{km}$).
Q10. At a height of $h = 32\text{ km}$ above the Earth’s surface, the fractional decrease in the acceleration due to gravity, $\frac{g_0 – g(h)}{g_0}$, is equal to $x \times 10^{-3}$, where $R = 6400\text{ km}$. Find the integer value of $x$.
Comprehensive Solutions and Answer Key
Answer Key
| Question | Correct Answer | Core Physics Concept |
|---|---|---|
| Q1 | (C) | Exact Inverse-Square Relation: $g(h) = g_0/9 \implies h = 2R$ |
| Q2 | (B) | Binomial Fractional Decrease: $\frac{2h}{R} = 0.02 \implies h = 64\text{ km}$ |
| Q3 | (B) | Pendulum Clock Time Dilatation: $\frac{\Delta T}{T} = \frac{h}{R}$ |
| Q4 | (B) | Weight Ratio at $h = R$: $W(R)/W_0 = (1 + 1)^{-2} = 1/4$ |
| Q5 | (A) | Energy Conservation with Altitude: $v = \sqrt{g_0 R}$ |
| Q6 | (A, B, C, D) | Calculus & Convexity Analysis of Gravitational Field |
| Q7 | (A, B, C) | Higher-Order Taylor Expansion & Approximation Limits |
| Q8 | (A, B, C) | Variable Power and Integrated Gravitational Work |
| Q9 | 1600 | Exact Formula Inversion: $h = 0.25 R = 1600\text{ km}$ |
| Q10 | 10 | Fractional Decrease Evaluation: $\frac{2(32)}{6400} = 10 \times 10^{-3}$ |
Detailed Explanations
Solution 1:
Using the exact formula:
$$g(h) = \frac{g_0}{\left(1 + \frac{h}{R}\right)^2} = \frac{g_0}{9}$$
$$\left(1 + \frac{h}{R}\right)^2 = 9 \implies 1 + \frac{h}{R} = 3 \implies \mathbf{h = 2R}$$
Correct Answer: (C)
Solution 2:
Since the decrease is small ($2\% = 0.02$), we use the binomial approximation:
$$\frac{\Delta g}{g_0} \approx \frac{2h}{R} = 0.02$$
$$h = \frac{0.02 R}{2} = 0.01 R = 0.01 (6400\text{ km}) = \mathbf{64\text{ km}}$$
Correct Answer: (B)
Solution 3:
The time period of a simple pendulum is:
$$T = 2\pi \sqrt{\frac{L}{g}} \implies T \propto g^{-1/2}$$
Taking fractional variations:
$$\frac{\Delta T}{T} = -\frac{1}{2} \frac{\Delta g}{g_0}$$
For an altitude $h \ll R$, $\frac{\Delta g}{g_0} \approx -\frac{2h}{R}$ (since $g$ decreases):
$$\frac{\Delta T}{T} = -\frac{1}{2} \left(-\frac{2h}{R}\right) = +\frac{h}{R}$$
Since the time period increases, each oscillation takes longer, meaning the clock ticks slower and loses time.
The fractional loss of time per day is $\frac{\Delta T}{T} = \mathbf{\frac{h}{R}}$.
(In seconds per day: $\Delta t_{\text{lost}} = \frac{h}{R} \times 86400\text{ s}$).
Correct Answer: (B)
Solution 4:
Weight of a body is given by $W(h) = m g(h)$.
At $h = R$:
$$W(R) = m g(R) = m \frac{g_0}{\left(1 + \frac{R}{R}\right)^2} = m \frac{g_0}{2^2} = \frac{1}{4} m g_0 = \frac{1}{4} W_0$$
$$\frac{W(R)}{W_0} = \mathbf{\frac{1}{4}}$$
Correct Answer: (B)
Solution 5:
By conservation of mechanical energy between Earth’s surface ($r = R$) and the maximum height ($r = 2R$ where $v = 0$):
$$\frac{1}{2} m v^2 – \frac{G M m}{R} = 0 – \frac{G M m}{2R}$$
$$\frac{1}{2} m v^2 = \frac{G M m}{R} – \frac{G M m}{2R} = \frac{G M m}{2R}$$
Multiply both sides by $\frac{2}{m}$:
$$v^2 = \frac{G M}{R} = g_0 R \implies \mathbf{v = \sqrt{g_0 R}}$$
Correct Answer: (A)
Solution 6:
For $r \ge R$, $g(r) = \frac{G M}{r^2}$.
– (A) $\frac{dg}{dr} = -\frac{2 G M}{r^3} = -\frac{2}{r} \left(\frac{G M}{r^2}\right) = -\frac{2g(r)}{r}$. (True).
– (B) $\frac{d^2 g}{dr^2} = +\frac{6 G M}{r^4} > 0$. The second derivative is strictly positive, meaning the curve is convex upward. (True).
– (C) $\lim_{r \to \infty} g(r) = \lim_{r \to \infty} \frac{G M}{r^2} = 0$. (True).
– (D) Inside Earth ($r \le R$), $g_{\text{in}}(r) = \frac{G M r}{R^3}$, so $\left(\frac{dg}{dr}\right)_{r \to R^-} = +\frac{G M}{R^3}$. Outside, $\left(\frac{dg}{dr}\right)_{r \to R^+} = -\frac{2 G M}{R^3}$. The slope abruptly jumps from positive to negative, exhibiting a slope discontinuity at $r = R$. (True).
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 7:
Using binomial series:
$$g_{\text{exact}} = g_0 \left(1 + \frac{h}{R}\right)^{-2} = g_0 \left(1 – \frac{2h}{R} + 3\left(\frac{h}{R}\right)^2 – 4\left(\frac{h}{R}\right)^3 + \dots\right)$$
– (A) Since the alternating series has positive second term $3(h/R)^2$, $g_{\text{exact}} > g_0(1 – 2h/R) = g_{\text{approx}}$ for all $h > 0$. (True).
– (B) $(g_{\text{exact}} – g_{\text{approx}}) = 3 g_0 (h/R)^2 + \mathcal{O}(h^3/R^3)$, which is second order in $(h/R)$. (True).
– (C) At $h = R/2$: $g_{\text{approx}} = g_0(1 – 2(0.5)) = 0$, predicting zero gravity. (True).
Statements (A, B, C) are correct.
Correct Answer: (A, B, C)
Solution 8:
– (A) Because $g(h)$ decreases with altitude, the gravitational force $F_g = m g(h)$ drops, so the required upward force to maintain constant speed decreases. Hence, rate of work decreases. (True).
– (B) Instantaneous power against gravity is $P = \vec{F}_g \cdot \vec{v} = m g(h) v_0 = \frac{m g_0 v_0}{(1 + h/R)^2}$. (True).
– (C) Total work done against gravity:
$$W = \int_0^R m g(h) dh = \int_0^R \frac{m g_0 dh}{(1 + h/R)^2} = m g_0 R \left[ -\frac{1}{1 + h/R} \right]_0^R = m g_0 R \left( 1 – \frac{1}{2} \right) = \mathbf{\frac{1}{2} m g_0 R}$$
(True).
Statements (A, B, C) are correct.
Correct Answer: (A, B, C)
Solution 9:
Given $g(h) = 0.64 g_0$:
$$\frac{g_0}{\left(1 + \frac{h}{R}\right)^2} = 0.64 g_0 \implies \frac{1}{\left(1 + \frac{h}{R}\right)^2} = 0.64$$
Taking the positive square root:
$$\frac{1}{1 + \frac{h}{R}} = \sqrt{0.64} = 0.8 = \frac{4}{5}$$
$$1 + \frac{h}{R} = \frac{5}{4} = 1.25 \implies \frac{h}{R} = 0.25$$
$$h = 0.25 R = 0.25(6400\text{ km}) = \mathbf{1600\text{ km}}$$
Correct Answer: 1600
Solution 10:
At $h = 32\text{ km}$, since $h \ll R$ ($32/6400 = 0.005$):
$$\frac{\Delta g}{g_0} \approx \frac{2h}{R} = \frac{2(32\text{ km})}{6400\text{ km}} = \frac{64}{6400} = \frac{1}{100} = 0.01$$
Writing this in the form $x \times 10^{-3}$:
$$0.01 = 10 \times 10^{-3} \implies \mathbf{x = 10}$$
Correct Answer: 10