Gravitational acceleration is the acceleration experienced by a body due to the gravitational force exerted by a massive astronomical body. It acts radially towards the center of mass of the source body under standard central-field conditions.
Variation of g with altitude, depth, and latitude, along with basic pendulum and free-fall problems.
Non-inertial reference frames, centrifugal force effects due to Earth's rotation, spheroidal earth geometry, and interior density models.
Conditions: Spherical Earth of mass M, radius R, height h above surface
Result: g(h) = g(1 - 2h/R)
Conditions: Uniform density sphere of mass M, radius R, depth d below surface
Result: g(d) = g(1 - d/R)
| Formula / law | Equation | Variables and units | Conditions |
|---|---|---|---|
| Surface Gravitational Acceleration | $$g = \frac{GM}{R^2}$$ | G = Universal gravitational constant, M = Mass of Earth, R = Radius of Earth m s^-2 | Spherical, non-rotating homogeneous body |
| Altitude Correction (General) | $$g_h = \frac{GM}{(R+h)^2}$$ | h = Height above surface m s^-2 | Any height h |
| Altitude Correction (Approximate) | $$g_h \approx g\left(1 - \frac{2h}{R}\right)$$ | h = Height above surface m s^-2 | h << R |
| Depth Correction | $$g_d = g\left(1 - \frac{d}{R}\right)$$ | d = Depth below surface m s^-2 | Uniform density Earth model, depth d |
| Latitude Correction (Rotation) | $$g' = g - R\omega^2 \cos^2(\lambda)$$ | \omega = Angular speed of Earth, \lambda = Latitude m s^-2 | Rotating reference frame |
| Effective Gravity Vectorially | $$\vec{g} = \vec{g} - \vec{\omega} \times (\vec{\omega} \times \vec{r})$$ | \vec{r} = Position vector from rotation axis m s^-2 | General rotating frame |
| Density Form of Surface g | $$g = \frac{4}{3} \pi G \rho R$$ | \rho = Mean density of Earth m s^-2 | Uniform density assumption |
| Free Fall Velocity | $$v = \sqrt{2gh}$$ | h = Fall distance in constant field m s^-1 | Neglecting air resistance |
Draw: A continuous graph with r on the x-axis and g on the y-axis. Inside the Earth (r < R), a straight line starting from origin to g at r=R. Outside the Earth (r > R), a curved hyperbola falling as 1/r^2.
Interpret: Shows linear increase inside the earth transitioning smoothly to an inverse-square decrease outside the surface, peaking precisely at r = R.
Exam use: Frequently tested in objective questions asking to identify the correct qualitative variation of g across the boundary r = R.
Draw: A bar pendulum oscillating about a horizontal axis with adjustable knife edges, measuring time period T versus distance l from center of mass.
Interpret: Allows accurate experimental determination of local acceleration due to gravity using the relation $g = 4\pi^{2} / slope of T^{2} vs L graph$.
Exam use: Standard laboratory technique conceptualized in theoretical viva and conceptual viva questions.
Draw: Graphical representation of reversible pendulum time periods showing intersection of curves for equal periods.
Interpret: Eliminates radius of gyration errors to determine absolute local value of g.
Exam use: Advanced PGT level conceptual foundation.
Answer: $h = (2/3)R$
Answer: 1% decrease
Answer: Not possible at any latitude with current rotation speed (cos^2(lambda) > 1)
Common error: Gravitational acceleration is strictly constant everywhere on the Earth's surface.
Correct: It varies with latitude due to Earth's rotation and equatorial bulge.
Earth is an oblate spheroid (flattened at poles) and rotates, making centrifugal force and distance from center latitude-dependent.
Common error: g is maximum at the center of the Earth.
Correct: g is zero at the exact center of the Earth.
At the center, mass surrounding the point exerts symmetric gravitational pulls in all directions that cancel out completely.
Common error: Altitude formula g(h) = g(1 - 2h/R) can be used for any arbitrary height.
Correct: It is an approximation valid only when h << R.
Higher-order terms in the binomial expansion become significant when h is comparable to R.
| Feature | Altitude ($h$) | Depth ($d$) |
|---|---|---|
| Formula for small values | $g_h \approx g\left(1 - \frac{2h}{R}\right)$ | $g_d = g\left(1 - \frac{d}{R}\right)$ |
| Effect of increase | Decreases as distance from center increases | Decreases because enclosed mass decreases |
| Rate of change | Twice as fast as depth variation for small displacements | Linear decrease with depth |
| Location | Latitude ($\lambda$) | Effective Gravity ($g'$) |
|---|---|---|
| Equator | $0^\circ$ | Minimum ($g - R\omega^2$) |
| Poles | $90^\circ$ | Maximum ($g$) |
Choose an option to reveal the explanation.
Explanation: Gravitational acceleration is an acceleration, having dimensions of length per time squared, expressed dimensionally as $[L T^{-2}]$ or $[M^0 L T^{-2}]$.
Explanation: Using the depth formula $g(d) = g(1 - d/R)$, at $d = R$, the acceleration becomes $g(1 - 1) = 0$. Surrounding mass symmetrically cancels out gravitational pulls.
Explanation: Rotation creates an outward centrifugal term $R\omega^2 \cos^2(\lambda)$ which subtracts from $g$ at the equator. Stopping rotation removes this reduction, thus increasing effective gravity.
Explanation: Using the binomial theorem on $g(h) = \frac{GM}{(R+h)^2} = g(1 + h/R)^{-2}$, for $h \ll R$, the approximation yields $g\left(1 - \frac{2h}{R}\right)$.
Explanation: The depth formula $g(d) = g(1 - d/R)$ shows that $g$ decreases linearly as depth $d$ increases.
Explanation: Effective gravity is given by $g' = g - R\omega^2 \cos^2(\lambda)$. At the poles, $\lambda = 90^\circ$, so $\cos(90^\circ) = 0$, yielding maximum effective gravity.
Explanation: Since $g \propto R^{-2}$, taking logarithms gives $\frac{\Delta g}{g} = -2 \frac{\Delta R}{R}$. A $1\%$ decrease in radius results in a $2\%$ increase in $g$.
Explanation: Substituting $M = \frac{4}{3}\pi R^3 \rho$ into $g = \frac{GM}{R^2}$ gives $g = \frac{4}{3} \pi G \rho R$.
Explanation: It is a common misconception that $g$ is uniform everywhere on Earth's surface; in reality, it varies due to Earth's rotation and oblate geometry.
Explanation: The centrifugal acceleration vector in a rotating frame is given by $-\vec{\omega} \times (\vec{\omega} \times \vec{r})$. Rotating reference frames require this correction.
Explanation: From $g(h) \approx g(1 - 2h/R)$, the fractional change $\frac{g(h) - g}{g} = -\frac{2h}{R}$. Magnitude-wise, it is twice the fractional change in height.
Explanation: Using $g(d) = g(1 - d/R) = g/2$, we get $1 - d/R = 1/2$, which yields $d = R/2$.
Explanation: Effective gravity at the equator is $g' = g - R\omega^2$. An increase in $\omega$ increases the centrifugal subtraction term, thus decreasing the effective gravity.
Explanation: The binomial expansion approximation is valid only when the height $h$ is much smaller than the radius of Earth ($h \ll R$).
Explanation: Gravitational acceleration has the same units as linear acceleration, which is meters per second squared ($\text{m s}^{-2}$).
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
Explanation: Using $g(h) = \frac{GM}{(R+h)^2} = \frac{g}{4}$, we get $(R+h)^2 = 4R^2$, so $R+h = 2R$, which gives $h = R$.
Explanation: Since $g = \frac{GM}{R^2}$, taking proportional variations gives $\frac{\Delta g}{g} = \frac{\Delta M}{M} - 2\frac{\Delta R}{R} = 1\% - 2(1\%) = -1\%$. Thus, it decreases by $1\%$.
Explanation: Using $g(d) = g(1 - d/R) = 0.75g$, we get $1 - d/R = 3/4$, which means $d/R = 1/4$, or $d = R/4$.
Explanation: Fractional change in $g$ is $\frac{\Delta g}{g} = \frac{2h}{R}$. Converting to percentage gives $\frac{2h}{R} \times 100\%$.
Explanation: At the equator, effective gravity is $g' = g - R\omega^2$. Setting $g' = 0$ gives $R\omega^2 = g$, so $\omega = \sqrt{\frac{g}{R}}$.
Explanation: Using $g = \frac{4}{3} \pi G \rho R$, $g$ is directly proportional to mean density $\rho$. Doubling $\rho$ doubles $g$.
Attempt these questions under a ten-minute limit using a separate timer.
Explanation: For small height, $\frac{\Delta g_h}{g} = \frac{2h}{R}$. For depth, $\frac{\Delta g_d}{g} = \frac{d}{R}$. Since $h = d$, the ratio is $2:1$.
Explanation: The reduction term is $R\omega^2 \cos^2(\lambda)$. Maximum reduction is at $\lambda = 0^\circ$ where $\cos^2(0) = 1$. Half reduction requires $\cos^2(\lambda) = 0.5$, meaning $\cos(\lambda) = 1/\sqrt{2}$, which corresponds to $\lambda = 60^\circ$.
Explanation: Using $g_{\text{planet}} = \frac{G(2M)}{(2R)^2} = \frac{2GM}{4R^2} = \frac{1}{2} \left(\frac{GM}{R^2}\right) = \frac{g}{2}$.
Explanation: Using $\frac{\Delta g}{g} = \frac{2h}{R} = \frac{2 \times 32}{6400} = \frac{64}{6400} = 0.01$, which is $1\%$.
Explanation: Using $g(d) = g(1 - d/R)$, at $d = R/2$, $g(R/2) = g(1 - 1/2) = g/2$. Thus, weight $mg$ becomes $W/2$.
Explanation: By definition $g = F/m = GM/R^2$, showing $g$ is independent of the test mass $m$.
Explanation: On a mountain top (altitude $h$), $g$ decreases. Since $T \propto 1/\sqrt{g}$, a decrease in $g$ causes the time period $T$ to increase.
Explanation: For a spherically symmetric, non-rotating Earth, $g = GM/R^2$ is constant everywhere on the surface. (Note: real Earth is oblate, making polar gravity higher).
Explanation: The exact formula for gravitational acceleration at any altitude $h$ is $g_h = \frac{GM}{(R+h)^2}$.
Explanation: Earth's oblate geometry (equatorial radius > polar radius) and its rotation (centrifugal effect) both combine to make $g$ lower at the equator than at the poles.