Acceleration due to Gravity — Revision Notes

⚡ 30-Second Revision

Definition: — Acceleration due to gravity, $g$, is the acceleration of an object due to gravitational force.
Standard Value: — $g approx 9.8,\text{m/s}^2$ on Earth's surface.
Formula: — $g = \frac{GM}{R^2}$ (where $M$ is planet mass, $R$ is planet radius).
Independence: — $g$ is independent of the mass of the falling object.
Variation with Altitude ($h$): — $g_h = \frac{g}{(1 + h/R_E)^2}$. For $h \ll R_E$, $g_h \approx g(1 - \frac{2h}{R_E})$.
Variation with Depth ($d$): — $g_d = g(1 - \frac{d}{R_E})$. At center ($d=R_E$), $g_d = 0$.
Variation with Latitude ($\lambda$): — $g' = g - R_E \omega^2 \cos^2\lambda$. Max at poles ($\lambda=90^\circ$), Min at equator ($\lambda=0^\circ$).
Relationship: — $g_{\text{pole}} > g_{\text{equator}}$.

2-Minute Revision

Acceleration due to gravity, $g$, is the acceleration an object experiences solely due to a planet's gravitational pull. On Earth's surface, its average value is $9.8,\text{m/s}^2$. The fundamental formula is $g = GM/R^2$, where $M$ is the planet's mass and $R$ is its radius.

Crucially, $g$ does not depend on the mass of the falling object. However, $g$ is not constant. It decreases with increasing altitude ($h$) above the surface, following $g_h = g/(1 + h/R_E)^2$, which approximates to $g(1 - 2h/R_E)$ for small $h$.

It also decreases with increasing depth ($d$) below the surface, given by $g_d = g(1 - d/R_E)$, becoming zero at the Earth's center. Furthermore, Earth's rotation and its oblate shape cause $g$ to be maximum at the poles and minimum at the equator.

Remember to distinguish $g$ (local acceleration) from $G$ (universal constant) and apply the correct formulas for different scenarios.

5-Minute Revision

Acceleration due to gravity, $g$, is the specific acceleration an object undergoes due to the gravitational force of a celestial body. It is a vector quantity, directed towards the center of the attracting mass. Its average value on Earth's surface is $9.8,\text{m/s}^2$. The derivation $mg = GMm/R^2$ leads to the core formula $g = GM/R^2$, which clearly shows $g$ is independent of the falling object's mass. This is a critical concept for NEET.

Variations of $g$ are highly testable:

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With Altitude ($h$): — As height increases, distance from Earth's center ($r = R_E + h$) increases. The exact formula is $g_h = GM_E/(R_E + h)^2 = g/(1 + h/R_E)^2$. For small $h$ ($h ll R_E$), use the binomial approximation: $g_h approx g(1 - 2h/R_E)$. This means $g$ decreases with height.
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With Depth ($d$): — As depth increases, the effective mass pulling the object decreases. The formula is $g_d = g(1 - d/R_E)$. This shows a linear decrease in $g$ with depth. At the Earth's center ($d=R_E$), $g_d = 0$.
3
With Latitude ($lambda$): — Earth's rotation and its oblate shape cause $g$ to vary. The effective $g'$ at latitude $lambda$ is $g' = g - R_E omega^2 cos^2lambda$. This implies $g$ is maximum at the poles ($lambda=90^circ$, $coslambda=0$) and minimum at the equator ($lambda=0^circ$, $coslambda=1$).

Key Points for NEET:

Always differentiate $g$ (acceleration) from $G$ (universal constant).
Understand the graphical representation: $g$ increases linearly from $0$ at the center to $g$ at the surface, then decreases as $1/r^2$ outside.
Practice problems involving percentage changes in $g$ due to small changes in $M$, $R$, $h$, or $d$. For small changes, $Delta g/g = Delta M/M - 2Delta R/R$ is useful.
Remember that at maximum height in projectile motion, velocity is zero, but acceleration is still $g$ downwards.

Prelims Revision Notes

Acceleration due to Gravity ($g$):

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Definition: — Acceleration of an object solely due to gravitational force. Vector quantity, directed towards center of mass.
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Value on Earth: — Average $9.8,\text{m/s}^2$ (or $10,\text{m/s}^2$ for approximation).
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Fundamental Formula: — $g = \frac{GM}{R^2}$, where $G$ is universal gravitational constant, $M$ is planet mass, $R$ is planet radius.
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Independence of Object Mass: — $g$ does NOT depend on the mass of the falling object ($m$ cancels out in $mg = GMm/R^2$).
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**Variation with Altitude ($h$):**

* Exact: $g_h = \frac{GM}{(R_E + h)^2} = \frac{g}{(1 + h/R_E)^2}$ * Approximate (for $h \ll R_E$): $g_h \approx g(1 - \frac{2h}{R_E})$ * $g$ decreases with increasing height.

1
**Variation with Depth ($d$):**

* Formula: $g_d = g(1 - \frac{d}{R_E})$ (assuming uniform density) * $g$ decreases linearly with increasing depth. * At Earth's center ($d=R_E$), $g_d = 0$.

1
**Variation with Latitude ($\lambda$):**

* Effective $g'$: $g' = g - R_E \omega^2 \cos^2\lambda$ (where $\omega$ is angular velocity of Earth). * At Equator ($\lambda=0^\circ$, $\cos\lambda=1$): $g'_{\text{equator}} = g - R_E \omega^2$ (minimum $g$). * At Poles ($\lambda=90^\circ$, $\cos\lambda=0$): $g'_{\text{pole}} = g$ (maximum $g$). * $g_{\text{pole}} > g_{\text{equator}}$ due to both rotation and Earth's oblate shape.

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Graphical Representation:

* Inside Earth (center to surface): $g$ increases linearly with distance from center ($g \propto r$). * Outside Earth (surface to infinity): $g$ decreases as inverse square of distance from center ($g \propto 1/r^2$).

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Percentage Change (for small changes):

* If $g = GM/R^2$, then $\frac{\Delta g}{g} = \frac{\Delta M}{M} - 2\frac{\Delta R}{R}$.

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Common Misconceptions: — Don't confuse $g$ with $G$. Don't assume $g$ is constant everywhere on Earth's surface. Velocity is zero at max height, but acceleration is still $g$ downwards.

Vyyuha Quick Recall

GRAVITY: G-M-R-Squared, Altitude 2H, Depth D-R, Rotate Cos-Squared.