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Gravitational Force Between Point Masses (CIE A Level Physics)

Revision Note

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Newton's law of gravitation

  • The gravitational force between two bodies outside a uniform field, e.g. between the Earth and the Sun, is defined by Newton’s Law of Gravitation
    • Recall that the mass of a uniform sphere can be considered to be a point mass at its centre
  • Newton’s Law of Gravitation states that:

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

  • In equation form, this can be written as:

F subscript G space equals fraction numerator space 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
    • m1 and m2 = two points masses (kg)
    • r = distance between the centre of the two masses (m)

Newton's law of gravitation

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

  • Although planets are not point masses, their separation is much larger than their radius
    • Therefore, 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 of mass 6500 kg is orbiting the Earth at 2000 km above the Earth's surface.

The gravitational force between them is 37 kN. The radius of the Earth is 6400 km.

Calculate the mass of the Earth. 

Answer:

Step 1: List the known quantities

  • Mass of satellite, m1 = 6500 kg
    • m1 and m2 can be either way around
  • Distance of satellite above Earth's surface = 2000 km
  • Gravitational force, FG = 37 kN
  • Radius of Earth = 6400 km

Step 2: State the equation for Newton's Law of Gravitation and rearrange for the mass of the Earth

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

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

Step 3: Calculate the distance, r

  • r is the distance between the centre of the Earth and the satellite
  • r = distance of satellite above Earth's surface + radius of Earth

13-2-2-we-newtons-law-of-gravitation-answer--cie-new

r space equals space 2000 space plus space 6400 space equals space 8400 cross times 10 cubed space straight m

Step 4: Substitute the known values into Newton's Law of Gravitation to calculate the mass of the Earth

m subscript 2 space equals space fraction numerator open parentheses 8400 cross times 10 cubed close parentheses squared space cross times space open parentheses 37 cross times 10 cubed close parentheses over denominator open parentheses 6.67 cross times 10 to the power of negative 11 end exponent close parentheses space cross times space 6500 end fraction

m subscript 2 space equals space 6.0 cross times 10 to the power of 24 space end exponent kg space open parentheses 2 space straight s. straight f. close parentheses

Examiner Tip

A common mistake in exams is to forget to add together the distance from the surface of the planet and its radius to obtain the value of r. The distance r is measured from the centre of the mass, which is from the centre of the planet.

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Leander

Author: Leander

Expertise: Physics

Leander graduated with First-class honours in Science and Education from Sheffield Hallam University. She won the prestigious Lord Robert Winston Solomon Lipson Prize in recognition of her dedication to science and teaching excellence. After teaching and tutoring both science and maths students, Leander now brings this passion for helping young people reach their potential to her work at SME.