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Lesson 7 of 7

Gravitation · Lesson 7 of 7

Chapter Summary and Practice

“Connect the PDF’s gravitational laws, assumptions and numerical applications in one reference and practice lesson.”

Learning Objectives

• Choose a force, field, energy or period relation for the question. • Keep centre distance, height and depth distinct. • State symmetry, density and reference assumptions. • Practise the PDF’s connected gravitational applications.

Choose the right quantity

QuestionRelationCondition
Force between massesF=Gm1m2/r²Point masses or justified spherical replacement
Field or accelerationg=F/mVector quantity; test mass cancels
Potential energyU=−GMm/rExterior point-source model; zero at infinity
Potentialφ=−GM/rEnergy per unit test mass
Circular orbital motionv²=GM/r; T²=4π²r³/(GM)Dominant source, circular radius
Escape thresholdve²=2GM/r0Ideal coasting escape, no other sources or losses
Inside a uniform Earthg=g0r/REUniform density is essential

Formula reference and dimensions

Kepler and NewtonLaTeX
Position vector r points from the attracting mass to the test body.
Gravity above and below the surfaceLaTeX
The height approximation requires h≪RE. The depth formula requires uniform Earth density.
Energy and escapeLaTeX
Use the same zero of potential throughout each calculation.
QuantitySI unitDimensions
GN m²/kg²M⁻¹ L³ T⁻²
ForceNM L T⁻²
Gravitational potential energyJM L² T⁻²
Gravitational potentialJ/kgL² T⁻²
Gravitational intensityN/kg or m/s²L T⁻²

Practice from the PDF

Practice Problems

Source-based Practice
  1. Use the equal-area law to compare perihelion and aphelion speeds and the two traversal times in source Example 7.1.
  2. Find the centroid force in the source’s equilateral triangle before and after doubling the mass at A.
  3. Explain why a spherical shell is not a gravitational shield, and derive the hemisphere field directions c and e.
  4. Show why the Moon can have a larger tidal effect than the Sun even though its total pull on Earth is smaller.
  5. Compare g above and below Earth using the exact height expression and uniform-density depth expression; state where each model applies.
  6. Count the six square-pair energies and distinguish system energy from potential at the centre.
  7. Find the limiting escape speed between the M and 4M spheres; explain the finite-time qualification at the neutral point.
  8. Use Phobos data to weigh Mars, lunar data to weigh Earth, Io data to weigh Jupiter and Earth’s orbit to weigh the Sun.
  9. For the galaxy model in Exercise 7.5, specify the assumed enclosed mass before applying the circular-period expression.
  10. Check every sign in the 2RE-to-4RE satellite energy transfer and calculate escape energy for the 400 km orbit.
  11. Solve the two-star head-on approach in Exercise 7.20, distinguishing each star’s speed from relative closing speed.
  12. Calculate force and potential between the equal 100 kg spheres and test stability along the joining line.

A worked two-body energy application

Two stars approach for a head-on collision

Problem
Source Exercise 7.20 gives two equal stars of mass M=2×10³⁰ kg and radius Rs=10⁴ km, initially almost at rest with centre separation d0=10⁹ km. Find the speed when they first touch, assuming no distortion.

  1. 1.At contact their centre separation is 2Rs=2×10⁷ m; initially d0=10¹² m. Equal masses and zero total initial momentum give equal opposite speeds v in the centre-of-mass frame.
  2. 2.Initial energy is −GM²/d0. Final energy is 2(½Mv²)−GM²/(2Rs).
  3. 3.Therefore v²=GM[1/(2Rs)−1/d0], giving v≈2.58×10⁶ m/s for each star.
  4. 4.The relative closing speed is 2v≈5.17×10⁶ m/s. The source answer of about 2.6×10⁶ m/s refers to each star’s speed, not their relative speed.
  5. 5.This is the exercise’s ideal undeformed-sphere model. Count the interaction potential once for the pair and both stars’ kinetic energies.

Distinctions that prevent common errors

G is a universal constant; g is a local field or acceleration. Gravitational force and field are vectors; potential and potential energy are scalars. A body’s zero net field does not require zero potential. The surface value mgh is not the general inverse-distance energy with its zero at infinity. A satellite is not outside gravity merely because it is weightless. A particle’s linear momentum need not be constant while it follows a central-force orbit; its angular momentum and mechanical energy can be conserved. For the full isolated source-plus-particle system, total linear momentum is conserved as well.

The area law applies to any central force, but the particular period–size relation follows from inverse-square gravity for a shared dominant source. An ellipse’s semi-major axis is not the same as its changing Sun–planet distance. Energy and angular momentum determine different aspects of orbital motion. When a question changes source body, density assumptions, reference level or distance definition, identify that change before substituting numbers.

Check the physical model and the answer

Write the system, source masses and relevant symmetry. Convert kilometres and days to SI when using SI G. Replace altitude by r=R+h and depth by r=R−d. Use the uniform-density assumption only for the inside-Earth formula. Add gravitational force components with signs but add scalar potentials directly. Verify dimensions, energy signs and limiting cases: g should fall at large distance, U should tend to zero from below, and escape speed should decline as launch radius grows. Recompute a printed answer if substitution fails; the audit records the sign and rounding corrections found in the supplied source.

Check your understanding

Quiz

Quick check

For a spherical source, orbit height h corresponds to centre distance

Explanation

Add the source radius to altitude.

Quick check

A zero-field point between two equal masses has potential

Explanation

Field cancellation does not cancel scalar potential.

Quick check

For a comet in an ideal elliptical orbit, which pair remains constant?

Explanation

Gravity is central and conservative.

Quick check

If identical stars collide head-on with speeds v and −v, their relative closing speed is

Explanation

The relative velocity magnitude is |v−(−v)|=2v.

Quick check

Which potential-energy change is correct for a satellite raised from 2RE to 4RE?

Explanation

−GMm/r becomes less negative as r increases.

Key Takeaways

Key Takeaways

• Select the relation only after stating the geometry and assumptions. • Preserve vector directions, scalar signs and a common energy reference. • Check numerical answers against the defining equations rather than copying printed signs.