Centripetal force, circular motion and universal gravitation: one idea, three questions

Centripetal force is not a new kind of force. It is the name for the net force pointing toward the centre of a circle, and something real must supply it: friction for a car on a curve, tension for a ball on a string, gravity for a satellite. Its size is mv²/r. Everything in uniform circular motion and orbital motion comes from writing ΣF toward the centre = mv²/r and asking what supplies the left side. Below, the three questions it answers, one orbit worked line by line, and the force students draw that does not exist.

Centripetal force is a job description, not a force

An object moving on a circle at a steady speed is still accelerating. Its speed is constant, but its velocity is not, because velocity has a direction and the direction is turning. The acceleration points toward the centre of the circle, and its size is a = v²/r: faster round the same circle means more acceleration, and the same speed round a tighter circle means more too. This is uniform circular motion, "uniform" meaning the speed does not change.

Newton's second law then says the net force on the object must point toward the centre and have size mv²/r. That net force is what "centripetal force" names. It is a job, filled by whatever real force happens to point inward: the tension in a string, the friction between tyre and road, the normal force from a banked track or a loop, gravity on a satellite. Sometimes two real forces share the job.

So on a free-body diagram you draw the real forces only, as in any free-body diagram, and never an arrow labelled "centripetal" or "centrifugal". Then you add the inward components of the real forces and set that sum equal to mv²/r. The equation is ΣF toward the centre = mv²/r, and the centripetal force is the left-hand side after you have added it up, not an extra term in it.

The three questions this one equation answers

Every circular-motion problem is one of three questions, and the same equation answers all of them.

1. What supplies the force? Draw the diagram, find the real force that points inward. Ball on a string: tension. Car on a flat curve: friction. Car on a banked curve: the horizontal part of the normal force, with friction helping. Satellite: gravity, and nothing else.

2. How fast before it fails? When the supplier has a ceiling, the maximum speed is the speed at which mv²/r reaches that ceiling. Take a car on a flat curve of radius r = 50.0 m with μₛ = 0.60 between tyres and road. Friction supplies the inward force, and its ceiling is μₛN = μₛmg on a flat road.

• μₛmg = mv²/r (friction at its maximum supplies the whole centripetal force)
• v_max = √(μₛgr) (the mass cancels from both sides)
• v_max = √(0.60 × 9.8 m/s² × 50.0 m) = √294 = 17.1 m/s (multiplying inside: m/s² × m = m²/s², so the root is in m/s)
• 17.1 m/s × 3.6 = 61.7 km/h, about 62 km/h (the conversion, for a sense of scale)

The mass cancelled, and that matters: a truck and a bicycle can take the same curve at the same top speed on the same road, because a heavier vehicle needs more inward force and gets more friction from its greater weight in exactly the same proportion.

3. How long does one trip take? The period T is the circumference divided by the speed, T = 2πr/v, because along the path the speed is constant and distance equals speed times time, the a = 0 case of the kinematics equations. The frequency is 1/T.

Newton's law of universal gravitation

Every two masses attract each other with a force F = Gm₁m₂/r², where G = 6.67 × 10⁻¹¹ N·m²/kg² and r is the distance between their centres. Not the gap between surfaces, and not the altitude: the centres.

The acceleration this force gives a small mass near a large one is g = GM/r², which is the force divided by the small mass. At Earth's surface, with M = 5.97 × 10²⁴ kg and r = 6.37 × 10⁶ m, that comes out at 9.8 m/s², which is where the familiar g comes from. Away from the surface, r is larger and g is smaller. So g = 9.8 m/s² is the surface value only, and an orbit problem must use GM/r² with the orbit's own r.

A working habit worth taking from the course: keep G, M and R as symbols until the last line. The algebra is cleaner, a cancelled mass is easier to see, and a symbolic answer such as √(GM/(R + h)) can be checked before a single number is multiplied.

Worked example: a satellite 400 km above Earth

A satellite is in a circular orbit 400 km above Earth's surface. Find its speed, its period, and the acceleration due to gravity at that height.

Givens: M = 5.97 × 10²⁴ kg, R = 6.37 × 10⁶ m, h = 4.00 × 10⁵ m, G = 6.67 × 10⁻¹¹ N·m²/kg². Three significant figures throughout.

Line 1, the orbit radius.

• r = R + h = 6.37 × 10⁶ m + 0.400 × 10⁶ m = 6.77 × 10⁶ m (the radius is measured from Earth's centre; the first wrong line on this topic is using R alone)

Line 2, what supplies the force. Gravity is the only force on the satellite, and it points toward Earth's centre, so it is the whole centripetal force.

• GMm/r² = mv²/r (gravity on the left, the centripetal requirement on the right)
• v² = GM/r (multiplying both sides by r and dividing by m; the satellite's mass m cancels)
• v = √(GM/r)

Line 3, the product GM.

• GM = 6.67 × 10⁻¹¹ N·m²/kg² × 5.97 × 10²⁴ kg = 3.98 × 10¹⁴ m³/s² (the units: N·m²/kg = kg·m/s² × m²/kg = m³/s²)

Line 4, the speed.

• GM/r = 3.98 × 10¹⁴ m³/s² ÷ 6.77 × 10⁶ m = 5.88 × 10⁷ m²/s² (dividing)
• v = √(5.88 × 10⁷) = 7.67 × 10³ m/s, which is 7.67 km/s (the root)

Line 5, the period.

• T = 2πr/v = 2π × 6.77 × 10⁶ m ÷ 7.67 × 10³ m/s = 4.25 × 10⁷ m ÷ 7.67 × 10³ m/s = 5.55 × 10³ s (circumference over speed)
• 5.55 × 10³ s ÷ 60 s/min = 92.4 min (the conversion)

A little over an hour and a half per orbit, which is the kind of figure that lets you catch an error of a power of ten at once.

Line 6, g at that height. This is the line students find surprising.

• g = GM/r² = 3.98 × 10¹⁴ m³/s² ÷ (6.77 × 10⁶ m)² = 3.98 × 10¹⁴ ÷ 4.58 × 10¹³ = 8.68 m/s² (carrying an extra digit in GM gives 8.69; both are 8.7 to two figures)
• 8.68 ÷ 9.8 = 0.886, so gravity at 400 km is about 89 % of its surface value

Gravity is nearly as strong up there as it is on the ground. Astronauts on a station at that height are not weightless because gravity has gone. They are in free fall, and so is the station around them, both accelerating toward Earth at 8.7 m/s² while moving sideways fast enough that the ground curves away beneath them at the same rate.

The three wrong lines a marker sees

1. r = R instead of R + h. The symptom is a speed of 7.9 km/s, the value for an orbit skimming the surface: √(3.98 × 10¹⁴ ÷ 6.37 × 10⁶) = 7.91 × 10³ m/s. If your satellite speed is 7.9 km/s at any altitude, the altitude never went in.
2. A "centrifugal force" arrow drawn outward. It is then made to "balance" gravity, ΣF comes out zero, and the satellite has no acceleration. But the satellite is accelerating; that is what moving on a circle means. Nothing balances gravity in orbit. The Force Concept Inventory (Hestenes, Wells and Swackhamer, 1992) has items built around exactly this belief, that an object on a circle feels an outward force, and it is among the most persistent findings in the test's history.
3. g = 9.8 at orbital height, or a = v²/r and gravity added as two different things. The acceleration v²/r and the gravitational acceleration GM/r² are the same quantity in orbit, the two sides of one equation. Adding them counts the same thing twice. Using 9.8 m/s² ignores the altitude, as in line 1.

A fourth, briefer: forgetting to square r in F = GMm/r². The symptom is a force or an acceleration wrong by a factor of several million.

A sibling to try cold

Cover the example. The same satellite is moved to an orbit 1,000 km above the surface. Find its speed, its period, and g at that height. Before any line, write r with the altitude added, and write v = √(GM/r) in symbols. Then put the numbers in.

Check

• r = 6.37 × 10⁶ m + 1.00 × 10⁶ m = 7.37 × 10⁶ m
• v = √(3.98 × 10¹⁴ ÷ 7.37 × 10⁶) = √(5.40 × 10⁷) = 7.35 × 10³ m/s = 7.35 km/s
• T = 2π × 7.37 × 10⁶ ÷ 7.35 × 10³ = 6.30 × 10³ s = 105 min
• g = 3.98 × 10¹⁴ ÷ (7.37 × 10⁶)² = 7.33 m/s²

Notice which way each number moved. A higher orbit is slower, takes longer, and feels weaker gravity. If one of yours moved the other way, the sign of the error is in that line.

Where this sits in the course

This page covers three ideas of the 25 in SPH4U. Idea 4, Uniform circular motion and centripetal force, closes strand B right after idea 3, Friction and motion on inclined planes: the diagram is the one you drew for the incline, with the net force now required to point to the centre. Ideas 13 and 14, Newton's law of universal gravitation and Orbital motion and satellites, open strand D nine ideas later. The course teaches the circle first and brings gravity to it when the fields strand begins, which is why a satellite problem feels like two units meeting: it is.

After idea 14 comes idea 15, Gravitational potential energy and potential, where the same GM/r shape returns with one fewer power of r. Gravitational potential energy, the idea after this one has that section. The SPH4U course map shows all 25 ideas in teaching order, and every idea of the three science courses has a page like this one.

Centripetal force on one screen

• The job description. Centripetal force is the net force toward the centre, mv²/r, supplied by real forces: friction, tension, normal, gravity. Never an arrow of its own.
• The three questions. What supplies it; how fast before the supplier's ceiling (v_max = √(μₛgr) on a flat curve); how long a trip takes (T = 2πr/v).
• Gravitation. F = Gm₁m₂/r², r between centres; g = GM/r², so 9.8 m/s² is the surface value only.
• Orbits. v = √(GM/r), the satellite's mass cancels; T = 2πr/v; g at 400 km is about 89 % of 9.8.
• The three wrong lines. R for R + h; an outward arrow; 9.8 up there or v²/r and gravity counted twice.
• The method. Derive v = √(GM/r) from GMm/r² = mv²/r every time rather than memorising it; how to study physics is built on that habit, and a formula sheet you build by deriving each line teaches while you make it.