Gravitation · Lesson 2 of 7
Universal Gravitation and the Measurement of G
“Develop Newton’s force law, superposition, spherical-shell results and the Cavendish experiment.”
• Write attractive gravitational force in scalar and vector form. • Add gravitational forces as vectors. • Apply the PDF’s spherical-shell results with their symmetry assumptions. • Explain how a torsion balance determines G.
The same interaction explains a fall and an orbit
The Moon does not need a force to maintain its speed along a straight line; it needs an inward force to bend its motion into an orbit. For an approximately circular lunar orbit, its inward acceleration is v²/r = 4π²r/T². Using the Moon’s distance 3.84×10⁸ m and period 27.3 days gives about 2.7×10⁻³ m/s². This is far below surface gravity. The distance is about 60 Earth radii, and 60² ≈ 3600 approximately matches the ratio of the accelerations. That comparison motivated an inverse-square distance dependence.
Attraction between two point masses
The force is attractive and acts on both bodies. Newton’s third law gives equal and opposite forces, not equal accelerations: the lighter body has larger acceleration for the same force magnitude. Doubling either mass doubles the force; doubling separation reduces it to one quarter. The separation is between point masses, or between centres only when the appropriate symmetry result justifies it.
G is a universal constant in this Newtonian model. Use the textbook value 6.67×10⁻¹¹ N m²/kg² for calculations unless a question specifies another rounding. Its dimensions follow by rearranging G = Fr²/(m1m2): M⁻¹ L³ T⁻². It is very different from the local acceleration g, whose units are m/s² and whose value depends on the attracting body and position.
Superposition and a symmetric triangle
Problem
Source Example 7.2 places masses m at the vertices A, B and C of an equilateral triangle and a test mass 2m at its centroid. Each vertex is 1 m from the centroid. Find the force before and after the mass at A is doubled.
- 1.Initially each force magnitude is F0 = Gm(2m)/(1 m)². The three directions are separated by 120°, so their vector sum is zero.
- 2.Choose +y toward A. The two lower forces have cancelling x components and total y component −F0, cancelling the upper +F0.
- 3.Doubling the mass at A adds one extra force F0 toward A; the original three still cancel. The new resultant is 2Gm²/(1 m)² toward A.
- 4.The test particle’s acceleration follows by dividing the resultant by 2m. Do not add the three force magnitudes as though their directions matched.
When a finite sphere behaves like a point mass
A real Earth contains distributed mass, so its attraction is formally the sum of attractions of all its elements. The source states two spherical-shell results without deriving their full calculus. A uniform thin spherical shell attracts an external particle as though all shell mass were at its centre. On a particle anywhere strictly inside that shell, the shell’s own net force is zero. Equal-distance reasoning works only at the centre, but the complete shell cancellation holds at every interior point.
A spherically symmetric solid body can be built from concentric uniform shells. Outside the body, every shell behaves as a central point mass, so the whole mass may be placed at the common centre for calculating attraction. Inside it, shells outside the test point contribute zero field and only enclosed mass contributes. Spherical symmetry is essential. Two arbitrary irregular bodies cannot generally be replaced by point masses at their centres of mass, and their resultant mutual force need not lie on the line joining those centres.
A hollow shell is not a gravitational shield
Zero force from a uniform shell’s own mass does not remove forces from other objects. A body inside the shell still responds to Earth, the Sun or nearby external matter. Newtonian gravity has no shielding analogue of a hollow conducting enclosure in electrostatics. Nor does zero field imply zero gravitational potential: field is related to spatial change of potential, not to its chosen numerical value.
Problem
In Exercises 7.10–7.11, a lower hemispherical shell has its flat opening in a horizontal plane. Determine the field direction at the opening centre C and at P in that opening plane.
- 1.At C, sideways contributions cancel by symmetry. All shell elements lie below the plane, so the remaining field points downward into the bowl: arrow c in the source.
- 2.To reason at P, add an identical upper hemisphere to complete a uniform spherical shell. The full shell’s interior field is zero.
- 3.Reflection across the opening plane makes the two hemispheres’ in-plane field components equal and their normal components opposite. Since their sum is zero, each in-plane component must vanish.
- 4.The lower half therefore gives a downward normal field at P as well: arrow e. This argument concerns P in the opening plane shown in the figure, not every arbitrary point in three-dimensional space.
Cavendish’s torsion balance
The apparatus has two small lead spheres at the ends of a light horizontal bar suspended by a fine wire. Two larger spheres are placed near opposite ends so their attractions form a turning couple. The wire twists until its restoring torque balances the gravitational torque. Repositioning the large spheres reverses the gravitational torque, allowing the small angular deflection to be measured. This demonstrates the attraction between laboratory-sized masses.
Each attractive force has a moment arm L/2, and the two moments add, giving FL rather than 2FL. In the textbook’s ideal arrangement the near-pair attractions dominate the measured couple. The calibration relates wire twist to torque; measuring only a deflection without knowing that calibration would not determine G. Once G is known, measured surface gravity and Earth’s radius can reveal Earth’s mass.
A force gradient explains tidal differences
The Sun’s total attraction on Earth is larger than the Moon’s, but tides depend on how gravitational acceleration varies across Earth rather than on a nearly common acceleration of the whole Earth. For a distant body, g(D)=GM/D² changes with distance at a rate of magnitude 2GM/D³. The differential effect across a fixed small length therefore scales as M/D³. This is why comparing total pulls, which scale as M/D², does not by itself compare tides. The nearby Moon can have the larger tidal effect despite the Sun’s larger total pull.
Check your understanding
Quiz
Doubling separation between two point masses makes their attraction
The force varies as 1/r².
Inside a uniform spherical shell, the shell’s own field is
The complete shell contributions cancel at every interior point.
Does that shell shield an external body’s gravity?
Superposition still includes the external body’s field.
The torsion-balance gravitational couple for bar length L is
Two forces each act at L/2 from the midpoint.
Tidal strength from a distant source scales approximately as
A tide is a field-gradient effect, adding another inverse distance.
Key Takeaways
• Gravitational force is attractive and adds vectorially. • Spherical-shell replacements require the stated symmetry. • The torsion balance measures G by balancing torques. • A shell’s zero own interior field does not shield other gravitational sources.