DSE Physics | Gravity and Satellite Orbits: Lesson-Prep Guide for Tutors | Gravitation, Orbital Speed and Geostationary Satellites

2026-11-01

Tutor lesson-prep guides · Secondary Physics

Students can quote F = GMm/r squared, but ask why a satellite does not fall and they answer that there is no gravity up there. This lesson joins gravity to centripetal force with one equation, and every orbit question afterwards is algebra.

Objectives — Students can: (1) compute surface gravity from gravitation; (2) derive orbital speed and period by treating gravity as the centripetal force; (3) invert the period to find the geostationary radius and explain why it must sit above the equator.

01 Prerequisites (3-minute check)

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

Skill	Quick check	Pass mark
Circular motion	Centripetal acceleration formula	a equals v squared over r
Units	Kilometres to metres	multiply by 1000
Orders of magnitude	Rough size of 6.67e-11 times 5.97e24	about 4e14

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-10 min	Gravitation	F equals GMm over r squared, inverse square	Double the distance, quarter the force
10-22 min	Surface gravity	g equals GM over R squared	9.81 metres per second squared
22-36 min	Orbital speed	Gravity is the centripetal force	7.54 kilometres per second
36-46 min	Period	T equals two pi r over v	97.2 minutes
46-56 min	Geostationary	Period of one day, invert for radius	altitude 3.58 times ten to the seven metres
56-60 min	Exit check	Three questions marked on the spot	Done independently

03 Teaching script (read it out)

Scene 1: Gravity is the centripetal force: one equation for everything

A satellite in a circular orbit has exactly one force on it, the pull of the Earth. So GMm/r squared equals mv squared over r. Cancel m and one r and you get v = square root of GM/r. Notice that m has vanished: orbital speed does not depend on the mass of the satellite, so a heavy one moves at the same speed as a light one.
With G = 6.67e-11, M = 5.97e+24 and r = 7.0 million metres, GM is 3.982e+14, so v is the square root of 5.689e+07, which is 7,542 metres per second, or 7.54 kilometres per second.

Ask the student: why a larger orbit is slower (r sits in the denominator, so bigger r means smaller v).

Scene 2: Period and altitude: invert T for r

T is two pi r over v. With r = 7.0 million metres and v = 7,542 metres per second, T is 5,831 seconds, about 97.2 minutes.
A geostationary satellite has a period of one sidereal day, 86,164 seconds. From r cubed equals GM T squared over four pi squared, r is 4.215e+07 metres, about 4.21 times ten to the seven metres. Subtract the Earth radius of 6.37 million metres and the altitude is about 3.58 times ten to the seven metres, roughly 36,000 kilometres.

Ask the student to compute surface gravity: GM over R squared is 9.81 metres per second squared, close to the textbook 9.8, with the gap coming from the value of R used.

Scene 3: Three results that are always examined

First, a higher orbit means slower speed and a longer period, because v goes as one over the square root of r and T goes as r to the power 1.5.
Second, a geostationary satellite must sit above the equator, at a fixed altitude near 3.58 times ten to the seven metres, moving west to east. Miss any one condition and it drifts instead of hovering over one point.
Third, escape velocity is the square root of 2GM/R, which is 11,181 metres per second, about 11.2 kilometres per second, or root two times the orbital value.

Reminder: a satellite stays up because it keeps falling forwards, not because gravity is absent.

04 Question types and worked steps

Type	Example	Steps	Answer
Calculate	Orbital speed at 7.0 million metres	v equals root GM over r	7.54 kilometres per second
Calculate	Period of that orbit	T equals two pi r over v	97.2 minutes
Calculate	Geostationary altitude	r cubed equals GM T squared over four pi squared	3.58 times ten to the seven metres
Explain	Why geostationary orbits must be equatorial	Orbit plane must pass through the centre and match rotation	Otherwise it drifts

Sanity check (teach this): Surface g must land between 9.7 and 9.9; an answer near 98 means the radius was not squared. A low Earth orbit takes about 90 minutes, here 97 minutes, so an answer in seconds means a unit slip. The geostationary altitude must exceed five Earth radii, here 5.6 times, or the algebra went wrong.

05 Independent practice (3 questions)

Question	Answer	What to watch
Surface g with R = 6.37 million metres	9.81 metres per second squared	GM over R squared
Orbital speed at r = 7.0 million metres	7.54 kilometres per second	Root GM over r
Geostationary altitude	3.58 times ten to the seven metres	r cubed equals GM T squared over four pi squared

06 Common mistakes and what to say

Mistake	Student writes	Say this
Claiming there is no gravity in orbit	Says nothing holds the satellite so it does not fall	Gravity is always there; the satellite keeps falling forwards.
Forgetting to cancel m	Substitutes the satellite mass into the speed formula	v equals root GM over r, and m cancels.
Using altitude as the radius	Puts 6.37e5 metres into g	r is measured from the centre, not from the ground.
Placing geostationary satellites anywhere	Says one can hover over any city	It must be equatorial, or it drifts.

07 Homework (paper, section and question numbers)

Secondary Physics chapter index — All chapters
DSE Physics practice paper (mechanics and space) — Section B gravity questions

08 Exit check (last 5 minutes)

Question	Answer	Decision
Orbital speed at 7.0 million metres	7.54 kilometres per second	All three correct: move to fields
Surface gravity of the Earth	9.81 metres per second squared	One wrong: reteach inverse square
Condition for geostationary orbit	Above the equator at a fixed altitude	Two wrong: restart scene three

09 Tutor FAQ

Why is orbital speed independent of mass?

Gravity and the required centripetal force both scale with mass, so the mass cancels.

Is a higher orbit harder to maintain?

No. Gravity is weaker there and the required speed is lower, but launching to it costs more energy, and those are separate points.

How does a polar orbit differ from geostationary?

A polar orbit crosses both poles and covers the globe but never hovers; geostationary hovers over one equatorial point.

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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.