DSE Physics | Circular Motion and Gravitation Teaching Guide (for tutors) | From centripetal force to satellite periods

2026-11-03

Tutor lesson-prep guides · Secondary Physics

Students know F = mv squared over r but freeze when asked which force supplies it, and often treat centrifugal force as real. This lesson anchors centripetal force in three concrete situations.

Objectives — Students can: (1) compute 2.62 m/s for radius 0.5 m and period 1.2 s using omega = 2 pi / T and v = omega r; (2) compute centripetal acceleration 13.7 m/s squared and state that it points to the centre; (3) set gravitational force equal to centripetal force to get 7.55 km/s and a 97-minute period at r = 7.0 x 10^6 m; (4) explain why a geostationary satellite must sit above the equator at about 3.6 x 10^7 m with a period equal to Earth rotation.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Angular speed	What is omega when T = 2 s?	pi, about 3.14 rad/s
Square roots	Order of magnitude of the square root of 3.986e14 over 7.0e6?	About 7.5e3
Newton gravitation	In F = GMm/r squared, what is r measured between?	The two centres of mass
Vector direction	Does uniform circular motion have acceleration?	Yes, pointing to the centre

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-8 min	Hook	Slow-motion a coin on a spinning turntable and ask why it flies off	Describe the straight-line tendency
8-20 min	Angular and linear speed	Get omega and v from T, stress v proportional to r	Compute once with units
20-32 min	Centripetal acceleration	Compute 13.7 m/s squared and compare with g	State that it is perpendicular to v
32-44 min	Satellite orbits	Derive v = root(GM/r) from GMm/r squared = mv squared/r	Derive it themselves, no copying
44-54 min	Geostationary	Work back from 24 hours to the radius and compare with Earth radius	Explain why the equator is required
54-60 min	Exit check	Three quick questions	Answer orally or in writing

03 Teaching script (read it out)

Scene 1: Centripetal force is not a new force but the existing resultant pointing to the centre

Tension for a whirling string, gravity for the Moon, friction for a turning car: each supplies the centripetal force. Before writing F = mv squared over r, ask which force points to the centre. Miss that step and the rest of the question collapses. Centrifugal force exists only in a rotating frame; DSE uses an inertial frame, so it is not there.

Ask students to label the source of centripetal force in all three cases on their diagram.

Scene 2: Setting gravity equal to centripetal force gives satellite speed and period

GMm/r squared = mv squared/r cancels m and one r, leaving v = root(GM/r). At r = 7.0e+06 m, v = 7.55 km/s and T = 2 pi r / v = 5829 s, about 97 minutes. A larger radius means a smaller speed but a longer period; teach the two together so nobody thinks higher means faster.

Have students compare two radii and state the inverse relationships themselves.

Scene 3: Three geostationary conditions all follow from one equation

The period is fixed at 24 hours (86 164 s sidereal), so T = 2 pi root(r cubed / GM) gives r about 4.22 x 10^7 m and v about 3.07 km/s. The orbit must lie in the equatorial plane, otherwise the satellite drifts north and south relative to the ground. These are derived from being stationary relative to the ground, not memorised.

Ask why Hong Kong cannot have a geostationary satellite directly overhead; lead to latitude and the equatorial plane.

04 Question types and worked steps

Type	Example	Steps	Answer
Speed from period	r = 0.5 m, T = 1.2 s	omega = 2 pi / T, then v = omega r, units m/s	2.62 m/s
Centripetal acceleration	Same data	a = v squared / r or omega squared r, direction to the centre	13.7 m/s squared
Satellite speed	r = 7.0e+06 m	GMm/r squared = mv squared/r, v = root(GM/r), convert to km/s	7.55 km/s
Satellite period	Same data	T = 2 pi r / v, convert to minutes, check above 90 minutes	97 minutes

Sanity check (teach this): A low orbit of 97 minutes matches the familiar 90 minutes, and the geostationary period of 24.0 hours is close to 24, both sensible. Centripetal acceleration 13.7 m/s squared exceeds g, which is why a coin slides off the turntable edge. Every orbital radius is measured from the centre of the Earth, never from the surface.

05 Independent practice (3 questions)

Question	Answer	What to watch
r = 2 m, v = 6 m/s: centripetal acceleration	18 m/s squared	Using v squared over r, not v over r
How many seconds is a geostationary period?	86 164 s	Using the sidereal day, not exactly 24 hours
If r doubles, how does satellite speed change?	1 over root 2, about 0.707 times	Recognising v proportional to 1 over root r

06 Common mistakes and what to say

Mistake	Student writes	Say this
Treating centrifugal force as real	“The satellite is balanced by centrifugal force”	“There is no centrifugal force in an inertial frame. Gravity alone acts, and it is the centripetal force.”
Measuring r from the surface	“The altitude is 600 km, so r = 600 km”	“r starts at the centre: add Earth radius 6 400 km.”
Thinking higher orbits are faster	“Further out means it must move faster”	“v = root(GM/r) falls with r while the period rises; separate the two.”
Drawing acceleration along velocity	“Acceleration is in the same direction as velocity”	“Speed is constant, only direction changes, so acceleration points to the centre and is perpendicular to v.”

07 Homework (paper, section and question numbers)

Physics chapter index — Review forces and motion, then questions 1 and 3
Medical physics (circular motion application) — Question 2 uses centripetal acceleration
Electricity and circuits — Question 6 compares a charged particle circling in a magnetic field

08 Exit check (last 5 minutes)

Question	Answer	Decision
omega for T = 1.2 s	5.24 rad/s	Wrong: forgot 2 pi
a for r = 0.5 m, v = 2.62 m/s	13.7 m/s squared	Wrong: used v/r
Low orbit period in minutes	97 minutes	Wrong: answered 24 hours
Plane of a geostationary orbit	The equatorial plane	Wrong: said any plane

09 Tutor FAQ

Must students memorise GM?

The paper usually gives G and Earth mass, or GM directly; g = GM/R squared lets them derive it.

Why can a geostationary satellite not sit over the South Pole?

Centripetal force must point to the Earth centre within the equatorial plane, otherwise it cannot stay fixed relative to the ground.

When does a circular motion question need energy?

Vertical circles with height change need energy conservation; horizontal circles usually need only centripetal force.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.