Circular Motion and Gravitation

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your JAMB UTME Physics paper.

Circular motion is the movement of an object at constant speed along a circular path of fixed radius. Although the speed is steady, the velocity direction keeps changing, so an inward centripetal acceleration is required to bend the path into a circle. That acceleration is supplied by a real centripetal force directed toward the centre, never outward.

Quantity	Formula	Meaning of symbols
Linear speed	v = 2πr / T	r = radius (m), T = period (s)
Centripetal acceleration	a_c = v² / r	v in m/s, r in m
Centripetal force	F_c = mv² / r = mω²r	m in kg, ω = 2π/T in rad/s
Newton’s gravitation	F = GMm / r²	G = 6.67 × 10⁻¹¹ N·m²/kg²
Gravitational field	g = GM / r²	g in m/s², r from centre

High-yield pointers:

For an orbiting satellite, gravity itself plays the role of centripetal force, giving v_o = √(GM/r).
Escape velocity from a body of mass M and radius r is v_e = √(2GM/r), exactly √2 times the orbital speed at the surface.
Use g = 10 m/s² when a JAMB question leaves it unspecified; convert any period given in minutes or hours to seconds before substitution.

🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months before JAMB UTME.

Uniform circular motion

An object undergoing uniform circular motion travels at constant speed v along a circle of radius r. The tangential velocity changes direction continuously, so even though |v| is constant the vector is not, producing an acceleration of magnitude a_c = v² / r pointing toward the centre. The corresponding centripetal force is F_c = mv² / r, and using v = ωr the same force can be written as mω²r, where ω = 2π/T is the angular velocity in rad/s and T is the period of one revolution.

The inward force is real: it can be tension in a string, gravity for a satellite, friction for a car rounding a bend, or the normal reaction of a banked road. The “outward pull” passengers feel in a turning car is a centrifugal pseudo-force that appears only in the rotating reference frame of the vehicle, not in an inertial frame.

Newton’s law of universal gravitation

Every point mass attracts every other with a force F = GMm/r² directed along the line joining their centres. G = 6.67 × 10⁻¹¹ N·m²/kg² is the gravitational constant. The associated gravitational field strength at distance r from a mass M is g = GM/r², which is exactly the acceleration of free fall a freely released body would experience there. At Earth’s surface g ≈ 9.8 m/s², but JAMB problems routinely use g = 10 m/s² when no value is stated.

Orbital mechanics and Kepler

For a satellite in a circular orbit, gravitational attraction provides the centripetal force:

GMm / r² = mv_o² / r ⇒ v_o = √(GM/r) and T = 2π√(r³/GM)

This last relation is Kepler’s third law (T² ∝ r³). Kepler’s other two laws state that planetary orbits are ellipses with the Sun at one focus, and that the line joining a planet to the Sun sweeps equal areas in equal times. Escape velocity v_e = √(2GM/r) is the minimum launch speed at which a projectile leaves the gravitational field without further propulsion.

Orbital quantity	Expression	Where it appears in JAMB
Orbital speed	v_o = √(GM/r)	Satellite period / speed questions
Escape speed	v_e = √(2GM/r)	“minimum speed to leave Earth”
Geostationary orbit	T = 24 h, orbit in equatorial plane	Communication-satellite MCQs

Common traps examiners exploit

Plugging the diameter (2r) instead of the radius into v²/r or GMm/r².
Forgetting to convert minutes or hours to seconds before substituting T.
Treating “weightless” astronauts as gravity-free; gravity is still nearly full strength at low orbit — both astronaut and craft fall together.

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer JAMB timeline who want full command of this 4%-weight topic.

Why the acceleration points inward

A particle moving with constant speed v sweeps an angle Δθ in time Δt, changing its velocity vector by Δv ≈ vΔθ perpendicular to the instantaneous velocity. Since a = Δv/Δt = v·(Δθ/Δt) = vω = v²/r, the acceleration is perpendicular to the path and therefore radial. This geometric argument is why circular motion demands a centre-seeking force regardless of whether the motion is at the end of a string, around a planet, or along a charged-particle track in a magnetic field (a separate but parallel formula F = qvB).

Connecting gravitation to field and potential

Gravitational field strength g = GM/r² is a vector field; gravitational potential V = −GM/r is a scalar from which g = −dV/dr. A body in circular orbit has constant V (conserved energy: ½mv² − GMm/r = constant). Setting v² = GM/r for circular motion gives total energy E = −GMm/(2r), showing bound orbits have negative energy while v ≥ v_e gives E ≥ 0 and the body escapes.

Edge cases and adjacent links

Banked curves and conical pendulums reduce to the same F_c = mv²/r with components resolved along horizontal/vertical or along/perp to the string.
Geostationary orbit requires T = 24 h AND the orbit to lie in the equatorial plane; the radius comes from T² = 4π²r³/(GM_E).
Variations of g with latitude (Earth’s rotation) and altitude: at height h, g_h = GM/(R+h)² ≈ g(1 − 2h/R) for small h.
Weightlessness in orbit is free fall, not absence of gravity; g at 400 km altitude ≈ 8.7 m/s², still ≈ 89% of surface value.
Newton’s vs Coulomb’s law: both are inverse-square, so v_o = √(GM/r) mirrors the Bohr orbital speed formula with k·e² replacing GMm.

Worked micro-example

A satellite orbits Earth at r = 7.0 × 10⁶ m. With M_E = 5.97 × 10²⁴ kg and G = 6.67 × 10⁻¹¹:

v_o = √(GM/r) = √((6.67×10⁻¹¹)(5.97×10²⁴)/(7.0×10⁶)) ≈ 7.54 × 10³ m/s; T = 2πr/v ≈ 5.83 × 10³ s ≈ 97 min.

Practice prompts

A 0.50 kg mass on the end of a 1.2 m string makes 60 revolutions in 30 s on a frictionless table. Find the tension in the string.
Show that the escape velocity from the surface of a planet of mass M and radius R is exactly √2 times the speed of a low circular orbit at the same radius.

Common mistakes to delete from your script

Writing F = mvr² or F = mv/r instead of mv²/r.
Mixing g = 9.8 and g = 10 within one calculation.
Treating centrifugal force as a real third-force instead of a frame artefact.
Using r = surface-to-surface distance instead of centre-to-centre in F = GMm/r².

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Circular Motion and Gravitation