Skip to main content
GIOPHYSICS
Menu

IB Physics HL · guided topic map

Gravitation for IB Physics HL

Gravitation for IB Physics HL, organized into 1 syllabus topic and 3 mapped concept guides.

Syllabus topics
1
Mapped concept guides
3
Educational level
IB Diploma Physics Higher Level

Syllabus to lesson

Choose the exact concept

Work in order or jump to the concept named in your specification, course outline, or assignment.

D.1

Gravitational fields

Fields

3 guides
  1. 01Gravitational fields and the inverse-square lawSL + HL extension
  2. 02Circular orbits and centripetal accelerationSL + HL extension
  3. 03Gravitational potential and potential energySL + HL extension

Diagrams

Gravitation as IB Physics HL draws it

The figures from the IB Physics HL practice papers that sit on these syllabus points — the apparatus, circuits and graphs an exam question actually puts in front of you.

01Figure 2Gravitational fieldsIB
A satellite in a circular orbit about a planetcircular orbitplanet of mass Msatellite of mass mvr

Figure comment

Figure 2A planet is drawn as a large shaded circle labelled 'planet of mass M'. A dashed circle, concentric with the planet and well outside it, is labelled 'circular orbit'. A small rectangle sitting on the dashed circle, above and to the right of the planet, is labelled 'satellite of mass m'. A short arrow from the satellite, drawn along the tangent to the orbit and labelled v, shows the direction in which it travels. A line runs from the centre of the planet out to the satellite and is labelled r.

Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.

D.1 Gravitational fieldsOpen IB question · Fields Q4
Guided questions 5 parts

Reading cue. r runs from the planet's centre to the satellite, not from its surface, and v is tangential, at right angles to r, so gravity does no work and the speed is constant.

  1. aState State the direction of the resultant force on the satellite at the instant drawn, and state its effect on the satellite's speed.

    recall2 marks

    Check answer 2 marks
    1. the force acts along the line marked r, directed towards the centre of the planet
    2. it is perpendicular to v, so it changes the direction of the velocity but not the speed

  2. bShow (that) Show that the orbital speed of the satellite is v = sqrt(GM/r), and hence show that the square of its orbital period is proportional to the cube of the radius marked.

    routine4 marks

    Check answer 4 marks
    1. gravitational attraction supplies the centripetal force: GMm / r² = mv² / r
    2. the satellite mass m cancels, giving v² = GM / r and so v = sqrt(GM/r)
    3. the period is T = 2π r / v, so T² = 4π² r² / (GM/r) = 4π² r³ / (GM)
    4. hence T² is proportional to r³, with a constant that does not contain m

  3. cDetermine The planet has mass M = 6.4 × 10²³ kg and the orbit radius marked is r = 9.4 × 10⁶ m. Determine the orbital speed of the satellite and the time it takes to complete one orbit.

    demanding4 marks

    Check answer 4 marks
    1. v = sqrt(GM/r) = sqrt(6.67 × 10⁻¹¹ x 6.4 × 10²³ / 9.4 × 10⁶)
    2. v = 2.1 × 10³ m s⁻¹
    3. T = 2π r / v = 2π x 9.4 × 10⁶ / 2.1 × 10³
    4. T = 2.8 × 10⁴ s, about 7.7 hours

  4. dDiscuss A thruster fires briefly along the direction of the arrow v, and the satellite settles into a new circular orbit of larger radius. Discuss why its orbital speed in the new orbit is lower than in the orbit drawn, even though the thruster did positive work on it.

    top of the paper4 marks

    Check answer 4 marks
    1. in any circular orbit v = sqrt(GM/r), so a larger radius gives a smaller speed
    2. the kinetic energy GMm/(2r) therefore falls, while the potential energy -GMm/r rises, becoming less negative
    3. the potential energy gains twice as much as the kinetic energy loses
    4. so the total energy -GMm/(2r) still increases, and the thruster's work equals that increase: the satellite trades speed for height

Transfer challenge

A moon completes one circular orbit of a different planet every 1.77 days, at a mean radius of 4.22 × 10⁸ m. Determine the mass of that planet, and state one assumption you have made.

Check answer 5 marks
  1. T = 1.77 × 24 × 3600 = 1.53 × 10⁵ s
  2. rearranging T² = 4π² r³ / (GM) gives M = 4π² r³ / (G T²)
  3. M = 4π² x (4.22 × 10⁸)³ / (6.67 × 10⁻¹¹ x (1.53 × 10⁵)²)
  4. M = 1.9 × 10²⁷ kg
  5. assume the orbit is circular and that the moon's mass is negligible compared with the planet's