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Forces & Inverse-Square Law Behaviour (DP IB Physics: HL)

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Katie M

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Katie M

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Physics

Forces & Inverse-Square Law Behaviour

  • The gravitational force between two bodies outside a uniform field (for example, between the Earth and the Sun) is defined by Newton’s Law of Gravitation which states that:

The gravitational force between two point masses is proportional to the product of the masses and inversely proportional to the square their separation

  • In equation form, this can be written as:

F subscript G equals fraction numerator G m subscript 1 m subscript 2 over denominator r squared end fraction

  •  Where:
    • FG = gravitational force between two masses (N)
    • G = Newton’s gravitational constant
    • m1m2 = two points masses (kg)
    • r = distance between the centre of the two masses (m)

Newton's law of gravitation, downloadable AS & A Level Physics revision notes

The gravitational force between two masses outside a uniform field is defined by Newton’s Law of Gravitation

  • The mass of a uniform sphere can be considered to be a point mass at its centre
    • The point mass approximation is a valid assumption if the separation between two objects is much larger than their radii
    • This is why Newton’s law of gravitation applies to planets orbiting the Sun

  • The 1/r2 relation is called the ‘inverse square law
    • This means that when a mass is twice as far away from another, its force due to gravity reduces by (½)2 = ¼

Worked example

A satellite with a mass of 6500 kg is orbiting the Earth at 2000 km above the Earth's surface. The gravitational force between them is 37 kN.

Calculate the mass of the Earth.

Radius of the Earth = 6400 km.

Coulomb's Law

  • All charged particles produce an electric field around it
    • This field exerts a force on any other charged particle within range

  • The electrostatic force between two charges is defined by Coulomb’s Law
    • A charge of a uniform spherical conductor can also be considered as a point charge at its centre

  • Coulomb’s Law states that:

The electrostatic force between two point charges is proportional to the product of the charges and inversely proportional to the square of their separation

  • The Coulomb equation is defined as:
F subscript e equals k fraction numerator Q q over denominator r squared end fraction equals fraction numerator Q q over denominator 4 πε subscript 0 r squared end fraction

  • Where:
    • Fe = electrostatic force between two charges (N)
    • Q (or Q1) and q (or Q2) = two point charges (C)
    • k = Coulomb's constant (8.99 x 109 N m2 C-2
    • ε0 = permittivity of free space
    • r = distance between the centre of the charges (m)

Coulombs Law diagram, downloadable AS & A Level Physics revision notes

The electrostatic force between two charges is defined by Coulomb’s Law

  • The 1/r2 relation is called the inverse square law
  • This means that when a charge is twice as far as away from another, the electrostatic force between them reduces by (½)2 = ¼

  • If there is a positive and negative charge, then the electrostatic force is negative, this can be interpreted as an attractive force
  • If the charges are the same, the electrostatic force is positive, this can be interpreted as a repulsive force
    • Since uniformly charged spheres can be considered as point charges, Coulomb’s law can be applied to find the electrostatic force between them as long as the separation is taken from the centre of both spheres

Worked example

An alpha particle is situated 2.0 mm away from a gold nucleus in a vacuum. Assuming them to be point charges, calculate the magnitude of the electrostatic force acting on each of the charges. Atomic number of helium = 2, Atomic number of gold = 79, Charge of an electron = −1.60 × 10-19 C.

Step 1: Write down the known quantities

    • Distance, r = 2.0 mm =2.0 × 10-3 m
    • The charge of one proton = +1.60 × 10-19 C
    • An alpha particle (helium nucleus) has 2 protons
    • Charge of alpha particle, Q1 = 2 × 1.60 × 10-19 = +3.2 × 10-19 C
    • The gold nucleus has 79 protons
    • Charge of gold nucleus, Q2 = 79 × 1.60 × 10-19 = +1.264 × 10-17 C

Step 2: The electrostatic force between two point charges is given by Coulomb’s Law

F subscript e equals fraction numerator k cross times Q cross times q over denominator r squared end fraction equals fraction numerator Q cross times q over denominator 4 cross times straight pi cross times straight epsilon subscript 0 cross times r squared end fraction

Step 3: Substitute values into Coulomb's Law

F subscript e equals fraction numerator open parentheses 3.2 cross times 10 to the power of negative 19 end exponent close parentheses cross times open parentheses 1.264 cross times 10 to the power of negative 17 end exponent close parentheses over denominator open parentheses 4 cross times straight pi cross times 8.85 cross times 10 to the power of negative 12 end exponent close parentheses cross times open parentheses 2.0 cross times 10 to the power of negative 3 end exponent close parentheses squared end fraction equals 9.092 cross times 10 to the power of negative 21 end exponent equals 9.1 cross times 10 to the power of negative 21 end exponent space N

Exam Tip

A common mistake in exams is to forget to add together the radius of the planet (which is the distance from the centre of mass to the surface) and then, the height above the surface of the planet.

Sketching a diagram will remind you of these two distances and is really a few seconds well spent!

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Katie M

Author: Katie M

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.