Mechanics

Circular Motion and Gravitation

5th Year · 6th Year (Leaving Cert)

  • By the end of this lesson students will be able to define and apply the concept of centripetal force.
  • By the end of this lesson students will be able to define and calculate angular velocity and its relationship to linear velocity.
  • By the end of this lesson students will be able to state and apply Newton's Law of Universal Gravitation.
  • By the end of this lesson students will be able to explain the motion of satellites and apply Kepler's Third Law (HL).

Key concepts

Circular Motion

When an object moves in a circular path at a constant speed, its velocity is continuously changing because its direction is changing. This change in velocity means there is an acceleration, known as centripetal acceleration, which is always directed towards the centre of the circle.

Centripetal Force

The force required to keep an object moving in a circular path. It is always directed towards the centre of the circle. Without this force, the object would move in a straight line tangent to the circle.

F = mv²/r
Angular Velocity (ω)

The rate of change of angular displacement. It is the angle swept out per unit time by the radius vector. Its unit is radians per second (rad s⁻¹).

ω = v/r (also ω = 2π/T or ω = 2πf)
Newton's Law of Universal Gravitation

States that every particle in the Universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres. G is the universal gravitational constant (6.67 × 10⁻¹¹ N m² kg⁻²).

F = Gm₁m₂/r²
Gravitational Field Strength (g)

The force per unit mass experienced by a small test mass placed in the field. It can also be expressed in terms of the mass of the central body (M) and the distance from its centre (r).

g = F/m (or g = GM/r²)
Satellites (HL)

Objects that orbit a larger body due to gravitational attraction. For a satellite in a stable orbit, the gravitational force provides the necessary centripetal force. Geostationary satellites are a special type of satellite that remain in a fixed position relative to a point on the Earth's surface, orbiting directly above the equator with a period of 24 hours.

mv²/r = GMm/r² (for orbital motion)
Kepler's Third Law (HL)

For any planet (or satellite) orbiting a central body, the square of its orbital period (T) is directly proportional to the cube of its average orbital radius (R). This means that T²/R³ is a constant for all objects orbiting the same central mass.

T²/R³ = constant (or T²/R³ = 4π²/GM)

Key facts to remember

  • 1Centripetal force is always directed towards the centre of the circular path and is responsible for changing the direction of velocity.
  • 2Angular velocity (ω) is measured in radians per second (rad s⁻¹), not degrees per second.
  • 3Newton's Law of Universal Gravitation describes the attractive force between any two masses in the Universe.
  • 4The universal gravitational constant, G, has a value of 6.67 × 10⁻¹¹ N m² kg⁻².
  • 5For a satellite in orbit, the gravitational force provides the necessary centripetal force.
  • 6Kepler's Third Law (HL) states that T²/R³ is constant for all objects orbiting the same central body.
  • 7Geostationary satellites (HL) have an orbital period of 24 hours and orbit above the Earth's equator.

Worked examples

Example 1

A car of mass 1200 kg travels around a circular bend of radius 50 m at a speed of 15 m/s. Calculate the centripetal force required to keep the car on the bend.

IIdentify the given values: mass (m) = 1200 kg, radius (r) = 50 m, speed (v) = 15 m/s.
IIState the formula for centripetal force: F = mv²/r.
IIISubstitute the values into the formula: F = (1200 kg)(15 m/s)² / (50 m).
IVCalculate the result: F = (1200)(225) / 50 = 270000 / 50 = 5400 N.

Answer

The centripetal force required is 5400 N.

This force is typically provided by friction between the tyres and the road.

Example 2

Calculate the gravitational force between the Earth (mass = 5.97 × 10²⁴ kg) and the Moon (mass = 7.35 × 10²² kg) if the average distance between their centres is 3.84 × 10⁸ m. (Universal gravitational constant G = 6.67 × 10⁻¹¹ N m² kg⁻²).

IIdentify the given values: m₁ = 5.97 × 10²⁴ kg, m₂ = 7.35 × 10²² kg, r = 3.84 × 10⁸ m, G = 6.67 × 10⁻¹¹ N m² kg⁻².
IIState Newton's Law of Universal Gravitation formula: F = Gm₁m₂/r².
IIISubstitute the values into the formula: F = (6.67 × 10⁻¹¹)(5.97 × 10²⁴)(7.35 × 10²²) / (3.84 × 10⁸)².
IVCalculate the product of the masses and G: (6.67 × 10⁻¹¹)(5.97 × 10²⁴)(7.35 × 10²²) ≈ 2.923 × 10³⁷.
VCalculate the square of the distance: (3.84 × 10⁸)² ≈ 1.47456 × 10¹⁷.
VIDivide the product by the squared distance: F ≈ (2.923 × 10³⁷) / (1.47456 × 10¹⁷) ≈ 1.98 × 10²⁰ N.

Answer

The gravitational force between the Earth and the Moon is approximately 1.98 × 10²⁰ N.

Example 3

(HL) A satellite orbits a planet at an average radius of 2.0 × 10⁷ m with a period of 1.0 × 10⁴ s. Another satellite orbits the same planet at an average radius of 3.0 × 10⁷ m. Calculate the period of the second satellite.

IState Kepler's Third Law for two satellites orbiting the same central body: T₁²/R₁³ = T₂²/R₂³.
IIIdentify the given values: R₁ = 2.0 × 10⁷ m, T₁ = 1.0 × 10⁴ s, R₂ = 3.0 × 10⁷ m.
IIIRearrange the formula to solve for T₂: T₂² = T₁² * (R₂³/R₁³).
IVSubstitute the values: T₂² = (1.0 × 10⁴ s)² * ((3.0 × 10⁷ m)³ / (2.0 × 10⁷ m)³).
VCalculate the terms: T₂² = (1.0 × 10⁸) * (2.7 × 10²² / 8.0 × 10²¹).
VISimplify the ratio of radii cubed: T₂² = (1.0 × 10⁸) * (3.375).
VIICalculate T₂²: T₂² = 3.375 × 10⁸.
VIIITake the square root to find T₂: T₂ = √(3.375 × 10⁸) ≈ 1.84 × 10⁴ s.

Answer

The period of the second satellite is approximately 1.84 × 10⁴ s.

Ensure units are consistent (e.g., metres and seconds).

Common mistakes

  • Confusing centripetal force (a real force towards the centre) with centrifugal force (a fictitious outward force experienced in a rotating frame of reference).
  • Using degrees per second instead of radians per second for angular velocity calculations, leading to incorrect results.
  • Forgetting to square the distance 'r' in the denominator of Newton's Law of Gravitation (F = Gm₁m₂/r²).
  • Using the radius of the Earth instead of the orbital radius (distance from the centre of the Earth to the satellite) for satellite problems.
  • Not converting units (e.g., kilometres to metres, minutes or hours to seconds) before performing calculations.

Exam tips

  • Always draw a clear diagram to visualise the forces and motion involved in circular motion and gravitation problems.
  • List all known quantities with their correct SI units before starting calculations to avoid errors.
  • Show all steps in your calculations clearly, as marks are often awarded for method even if the final answer is incorrect.
  • Pay close attention to units and ensure your final answer has the correct units. For HL questions, practice the derivation of Kepler's Third Law.

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