4.1.1Tools and Inquiry — IB Physics skills
Newton’s Law of Gravitation — IB Physics SL Revision Notes & Topical Questions
5 min read 5 parts Lesson 30 of 33 Unit quiz
Curriculum Equation Key قوانين نيوتن للحركة - Newton Laws of Motion FormulasFast Exam Solution
F = m × a | W = m × g | F₁₂ = -F₂₁
Word Equation & Conditions: القوة المحصلة = الكتلة × التسارع | Force = Mass × Acceleration
Location & Specifications: الميكانيكا الكلاسيكية والتحصيلي وفيزياء 3 ثانوي و IGCSE Physics.
Section 1 of 5The Law
All masses attract each other, and the force depends on the masses and the distance between them.
Statement
- Statement The gravitational force between two point masses is proportional to the product of the masses and inversely proportional to the square of their separation.
Equation
- Equation F = GMm/r², where G = 6.67 × 10⁻¹¹ N m² kg⁻² and r is the distance between centres.
Always attractive
- Always attractive Gravity between masses is always attractive, and the forces on the two bodies are equal and opposite.
Spheres
- Spheres Uniform spherical bodies act as though all their mass is concentrated at their centres.
Section 2 of 5Field Strength
Dividing the force by the test mass gives a property of the field itself.
1
Gravitational field strength g is the force per unit mass experienced by a small mass placed in the field.
2
Dividing F = GMm/r² by m gives g = GM/r² for the field at distance r from a point mass or sphere of mass M.
3
The value of g falls with the square of distance from the centre, so doubling r reduces g to a quarter.
4
At the surface of the Earth g ≈ 9.8 N kg⁻¹, which is the same as an acceleration of 9.8 m s⁻².
5
Field lines for a sphere point radially inwards and become further apart as the distance increases.
Section 3 of 5Using the Law
Using the Law
Inverse square
- Inverse square The force falls to one quarter when the separation is doubled, and to one ninth when it is tripled.
Above the surface
- Above the surface Distances must be measured from the centre, so add the Earth's radius to any altitude.
Orbits
- Orbits For a circular orbit the gravitational force supplies the centripetal force.
Small forces
- Small forces Between everyday objects the force is tiny because G is so small.
| Separation | Field strength |
|---|---|
| R | g |
| 2R | g/4 |
| 3R | g/9 |
Section 4 of 5Worked Example
Question Calculate the gravitational force between the Earth (5.97 × 10²⁴ kg) and the Moon (7.35 × 10²² kg) when their centres are 3.84 × 10⁸ m apart.
- Write F = GMm/r².
- Numerator = 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × 7.35 × 10²² = 2.93 × 10³⁷.
- Denominator = (3.84 × 10⁸)² = 1.47 × 10¹⁷.
- F = 2.93 × 10³⁷ / 1.47 × 10¹⁷ = 2.0 × 10²⁰ N.
Answer F ≈ 2.0 × 10²⁰ N, acting on each body towards the other.
Section 5 of 5Exam Checklist & Tips
4.1.1 – Key Points for the Exam
- Gravitational constant G = 6.67 × 10⁻¹¹ N m² kg⁻², the constant in Newton's law.
- Gravitational field strength The force per unit mass at a point in a gravitational field.
- Point mass An object whose mass is treated as concentrated at one point.
- Inverse square law A law in which the quantity falls as 1/r².
Common Mistakes
- Measuring r from the surface instead of from the centre of the planet.
- Forgetting to square the distance.
- Believing that the larger body pulls with a larger force on the smaller one.
- Confusing the constant G with the field strength g.
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