Skip to lesson content

Lesson 3 of 7

Gravitation · Lesson 3 of 7

Earth’s Gravity and Its Variation with Height and Depth

“Derive surface gravity and the altitude and uniform-density depth relations.”

Learning Objectives

• Derive g = GME/RE² and distinguish it from G. • Use the exact height formula and its limited approximation. • Derive the inside-Earth field with enclosed mass. • Apply force ratios and estimate Earth’s mass.

Surface gravity from the universal law

Approximate Earth as spherically symmetric. For a test mass at its surface, the shell result allows the attracting Earth mass to be placed at its centre. Equating gravitational force GME m/RE² to ma cancels the test mass and gives its acceleration. The cancellation explains why ideal freely falling bodies have the same gravitational acceleration at the same location, even though their gravitational forces differ.

Surface gravityLaTeX
This model describes gravitational acceleration for a spherical Earth; rotation, latitude effects and Earth’s actual density profile are not developed as separate topics in the supplied PDF.

G does not change when the test body or planet changes. Surface g does, because it depends on the planet’s mass and radius. A heavier test body has greater weight mg but not greater free-fall acceleration in this model. Earth’s radius in these expressions is centre-to-surface distance, not its diameter.

Above Earth: exact inverse-square dependence

Gravity at height hLaTeX
The radial distance is r=RE+h. This expression does not require h to be small.

As altitude grows, the same attracting mass is farther away. The inverse-square relation therefore decreases smoothly without becoming abruptly zero at any finite altitude. At height h = RE, the centre distance doubles and the acceleration is g0/4, not g0/2. Confusing height with centre distance is the most common mistake in these questions.

Small-height approximationLaTeX
Apply the binomial approximation (1+x)^−2 ≈ 1−2x only for small x. Use the exact formula for large altitude.
Weight at half an Earth radius above the surface

Problem
Source Exercise 7.15 gives a surface weight of 63 N and altitude h=RE/2. Find the gravitational force there.

  1. 1.The centre distance is RE+RE/2 = 3RE/2.
  2. 2.F/F0 = [RE/(3RE/2)]² = 4/9.
  3. 3.F = 63×4/9 = 28 N. The small-height approximation is inappropriate because h/RE=0.5 is not small.

Inside a uniform Earth: only enclosed mass contributes

Imagine a point at radius r inside Earth. All shells outside r exert zero net gravitational field there. The remaining sphere of radius r acts, at its surface, as a central mass Mr. To calculate Mr, the textbook assumes uniform density throughout Earth. Then mass is proportional to volume, so Mr/ME = r³/RE³. Substituting into GMr/r² leaves a field proportional to r rather than an inverse-square increase toward the centre.

Enclosed-mass derivationLaTeX
The linear relation requires uniform density. General spherical symmetry still gives g(r)=GMr/r², but Mr(r) need not scale as r³.
Gravity at depth dLaTeX
Depth is measured downward from the surface, with 0≤d≤RE.

At the centre the forces balance and g=0. In this uniform-density model, g increases linearly from the centre to the surface and decreases outside as 1/r², so its maximum is at the surface. This statement is a conclusion of the model, not a general claim about the true interior of a layered planet. Also distinguish a uniform hollow shell, whose entire cavity has zero own field, from a filled uniform sphere, whose off-centre interior field is nonzero.

g₀R2R3R0rg ∝ rg ∝ 1/r²Uniform-density Earth; r is measured from the centre
Gravity inside and outside a uniform spherical Earth— The interior straight line and sampled exterior inverse-square curve meet at r=R. Labels refer to centre distance; height outside is r−R and depth inside is R−r.
Halfway down to the centre

Problem
In source Exercise 7.16, a body weighs 250 N on the surface. Find its gravitational force halfway down to the centre of a uniform-density Earth.

  1. 1.The depth is d=RE/2, so r=RE/2.
  2. 2.The force ratio equals the acceleration ratio 1−d/RE = 1/2.
  3. 3.The gravitational force is 125 N toward the centre. It decreases rather than increasing when the enclosing mass is treated correctly.

Weighing Earth with a surface measurement

Earth mass from surface gravity

Problem
Use source Example 7.6 data g0=9.81 m/s², RE=6.37×10⁶ m and G=6.67×10⁻¹¹ N m²/kg² to estimate Earth’s mass.

  1. 1.Rearrange g0=GME/RE² to get ME=g0RE²/G.
  2. 2.Substitution gives 9.81×(6.37×10⁶)²/(6.67×10⁻¹¹) ≈ 5.97×10²⁴ kg.
  3. 3.The test body’s mass never appears. A laboratory measurement of G turns the local acceleration measurement into an astronomical mass estimate.
  4. 4.The satellite lesson gives an independent estimate using the Moon’s orbit, as the source example requests.

Check your understanding

Quiz

Quick check

Which mass cancels when deriving surface g?

Explanation

Gravity and inertia are both proportional to the test mass.

Quick check

At height h=RE, g is

Explanation

The centre distance doubles.

Quick check

The relation g(h)≈g0(1−2h/RE) is valid for

Explanation

It is the first-order binomial approximation.

Quick check

Inside a uniform-density Earth, g is proportional to

Explanation

Enclosed mass scales as r³ and division by r² leaves r.

Quick check

At depth RE/2, a surface weight 250 N becomes

Explanation

The uniform-density depth ratio is 1/2.

Key Takeaways

Key Takeaways

• Surface g depends on the attracting planet, not the test mass. • Use r=RE+h above the surface and r=RE−d below it. • Linear interior gravity depends on the uniform-density assumption. • Use the exact height expression when the height is not small.

Previous · Lesson 2

Universal Gravitation and the Measurement of G

Next · Lesson 4

Gravitational Potential and Potential Energy