MHT-CET Physics · Gravitation

How g Changes: Height, Depth, Density and Spin

Above the surface g falls as the inverse square of the distance from the centre, below it g falls linearly to zero at the centre; a planet's surface g is proportional to its radius times its density; and the earth's spin reduces the effective g everywhere except the poles, most at the equator.

All Gravitation notes MHT-CET Physics strategy

Why this matters

33 PYQs, one HARD — the largest page in the chapter, but mostly direct. Twenty-two put a body or a pendulum at a height or a depth: its weight, its period, the height where g falls to a fraction. Eight compare planets by radius and density, and three ask what the earth's spin does at the equator or a latitude. Three cards.

Concept 1 of 3: Height and Depth

At height h, g_h = g R²/(R + h)²: at h = R it is g/4, at 2R it is g/9. The small-height form g(1 − 2h/R) is for h much smaller than R. At depth d, only the sphere of radius R − d below still pulls, so g_d = g(1 − d/R), halved at R/2 and zero at the centre. A pendulum's period goes as 1/√g, so it slows at a height and at a depth; its frequency goes as √g. A capillary rise goes as 1/g, so it grows in a mine. 'Reduced BY 64%' means g becomes 36% of its value.

Definition

Height: $g_h = g\dfrac{R^2}{(R + h)^2}$ ( $h = R$ ⇒ g/4; $\tfrac{1}{9}$ ⇒ h = 2R; $\tfrac{1}{n}$ ⇒ $h = (\sqrt{n} - 1)R$ ). Small h: $g(1 - \tfrac{2h}{R})$ .
Depth: $g_d = g\left(1 - \dfrac{d}{R}\right)$ ( $\tfrac{g}{n}$ ⇒ $d = \dfrac{R(n - 1)}{n}$ ).
Weight at h = R/2: $\tfrac{4}{9}W$ (72 N ⇒ 32 N).
Pendulum: $T \propto \dfrac{1}{\sqrt{g}}$ (h = R ⇒ 2T; h = 2R ⇒ 3T); frequency at depth R/4 ⇒ $\dfrac{\sqrt{3}}{2}n$ .
Capillary rise ∝ 1/g: in a mine $\dfrac{Y}{X} = \dfrac{R}{R - d}$ .

Height and depth

g_h = g\frac{R^2}{(R + h)^2}, \qquad g_d = g\left(1 - \frac{d}{R}\right)

Worked example

At what height is g reduced by 64%?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 10th May Shift 2 · Q5Moderate

Example 1 · Gravitation · Variation of g with Depth, Altitude, Density, and Latitude

The height 'h' from the surface of the earth at which the value of 'g' will be reduced by 64% than the value at surface of the earth is (R = radius of the earth)

Using the small-height formula at large heights

g(1 − 2h/R) only works for h ≪ R; at h = R it would give −g. Use g R²/(R + h)² whenever h is comparable to R.

Reading 'reduced by' as 'reduced to'

Reduced BY 64% leaves 36%; reduced TO 64% leaves 64%. The two give 2R/3 and R/4.

Drill 21 more on height and depth

Concept 2 of 3: g on Other Planets

With M = (4/3)πR³ρ, the surface value g = GM/R² becomes (4/3)πGρR: proportional to radius times density. A planet of the same density and three times the radius has three times the earth's g; one with twice the density and the same g must have half the radius. Written with mass and radius instead, g ∝ M/R², so twice the mass and twice the radius give g/2. The same formula turned round gives the earth's density from g: ρ = 3g/(4πGR).

Definition

$g = \dfrac{4}{3}\pi G\rho R$ , so $g \propto \rho R$ ; $g \propto \dfrac{M}{R^2}$ .
Same ρ, radius 3R ⇒ 3g; same g, density 3ρ ⇒ radius R/3.
Mass and radius both doubled ⇒ g/2 ⇒ second's pendulum period $2\sqrt{2}$ s.
Earth's density: $\rho = \dfrac{3g}{4\pi GR}$ .

Surface gravity

g = \frac{GM}{R^2} = \frac{4}{3}\pi G\rho R

Worked example

A planet has half the earth's radius and twice its density. Its surface g?

Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 11th May Shift 1 · Q5Moderate

Example 2 · Gravitation · Variation of g with Depth, Altitude, Density, and Latitude

The acceleration due to gravity at the surface of the planet is same as that at the surface of the earth, but the density of planet is thrice that of the earth. If 'R' is the radius of the earth, the radius of the planet will be

Using g ∝ 1/R² at fixed density

1/R² holds for a fixed MASS. At a fixed density the mass grows as R³, so g grows as R.

Drill 7 more on g on other planets

Concept 3 of 3: The Earth's Spin and Latitude

A body on the spinning earth needs part of gravity to keep it moving in a circle, so the effective g at latitude λ is g − Rω²cos²λ: reduced most at the equator (by Rω²) and not at all at the poles. The difference between the equator and latitude 30° is Rω²(1 − cos²30°) = Rω²/4. A body at the equator would weigh nothing if Rω² = g, which needs ω = √(g/R) ≈ 1/800 rad/s.

Definition

$g_\lambda = g - R\omega^2\cos^2\lambda$ ; equator $g - R\omega^2$ , poles g.
$|g_{\text{eq}} - g_{30^\circ}| = \dfrac{1}{4}\omega^2R$ .
Weightless at equator: $\omega = \sqrt{\dfrac{g}{R}}$ ≈ 1/800 rad/s; weight $\tfrac{3}{5}$ ⇒ $\omega = \sqrt{\dfrac{2g}{5R}}$ .

Effective g with spin

g_\lambda = g - R\omega^2\cos^2\lambda

Worked example

How fast would the earth have to spin for a person at the equator to weigh half as much?

Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 1 · Q41Moderate

Example 3 · Gravitation · Variation of g with Depth, Altitude, Density, and Latitude

The speed with which the earth would have to rotate about its axis so that a person on the equator would weigh

\frac{3}{5}

th as much as at present weight is (

g =

gravitational acceleration,

R =

equatorial radius of the earth)

Using cos λ instead of cos²λ

The reduction is Rω²cos²λ: one cos for the smaller circle's radius, one for the component along gravity. At 30° that is 3/4 of Rω², not √3/2.

Drill 2 more on the earth's spin and latitude

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Height and Depth
Height and depth
$g_h = g\frac{R^2}{(R + h)^2}, \qquad g_d = g\left(1 - \frac{d}{R}\right)$
g on Other Planets
Surface gravity
$g = \frac{GM}{R^2} = \frac{4}{3}\pi G\rho R$
The Earth's Spin and Latitude
Effective g with spin
$g_\lambda = g - R\omega^2\cos^2\lambda$

Watch out for (4)

Using the small-height formula at large heights
→ Height and Depth
Reading 'reduced by' as 'reduced to'
→ Height and Depth
Using g ∝ 1/R² at fixed density
→ g on Other Planets
Using cos λ instead of cos²λ
→ The Earth's Spin and Latitude

Test yourself on Gravitation

20 past MHT-CET questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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