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A LevelPhysicsCambridge (CIE)Revision Notes13. Gravitational FieldsGravitational Field of a Point MassThe Value of g on Earth

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First teaching 2023

First exams 2025

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The Value of g on Earth (Cambridge (CIE) A Level Physics)

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Leander Oates

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The value of g on earth

  • Gravitational field strength g is approximately constant for relatively small changes in height near the Earth’s surface

    • Close to the Earth's surface, within the Earth's atmosphere, g = 9.81 N kg−1

  • g can be approximated as constant because the Earth's radius, R is much larger than the distance between the Earth's surface and the position of an object in the Earth's atmosphere, h

R space much greater-than space h

  • Where small changes in height do not affect the total height of an object, h

    • Consider an object orbiting at a height, h of 150 km (1.5 × 105 m) above the Earths surface

    • The radius of the Earth, R is 6400 km = 6.4 × 106 m

    • So the total orbital radius, r is (6.4 × 106) + (1.5 × 105) = 6.55 × 106 m

    • Hence 6.4 × 106 ≅ 6.55 × 106

Diagram of orbital height

6-1-2-worked-example-solution-cie-igcse-23-rn

Orbital radius, r is roughly equal to the Earths radius for an object within the Earths atmosphere

g space equals fraction numerator space G M over denominator r squared end fraction

  • Gravitational field strength, g, and orbital radius, r, have an inverse square law relationship:

    g space proportional to fraction numerator space 1 over denominator r squared end fraction

    • So small changes in r result in small changes in g

g space equals fraction numerator space G M over denominator open parentheses R space plus space h close parentheses squared end fraction space asymptotically equal to space fraction numerator G M over denominator R squared end fraction

Worked Example

The highest point above the Earth's surface is at the peak of Mount Everest.

Given that this is 8800 m above the Earth's surface, show that g decreases by 0.7%.

Mass of the Earth = 6.0 × 1024 kg.

Radius of the Earth = 6400 km.

Answer: 

Step 1: Gravitational field strength equation

g space equals fraction numerator space G M over denominator r squared end fraction

Step 2: Determine the value of r

r space equals space radius space of space Earth space plus space height space of space Mount space Everest

r space equals space 6400 space plus space 8.800 space equals space 6408.8 space km space equals space 6408.8 cross times 10 cubed space straight m

Step 3: Substitute the known values to calculate

g space equals space fraction numerator open parentheses 6.67 cross times 10 to the power of negative 11 end exponent close parentheses space cross times space open parentheses 6.0 cross times 10 to the power of 24 close parentheses over denominator open parentheses 6408.8 cross times 10 cubed close parentheses squared end fraction

g space equals space 9.74 space straight N space kg to the power of negative 1 end exponent space open parentheses 3 space straight s. straight f. close parentheses

Step 4: Calculate he percentage decrease

  • g on Earth's surface = 9.81 N kg−1

percent sign space decrease space equals fraction numerator space decrease over denominator original end fraction cross times 100

percent sign space decrease space equals space fraction numerator 9.81 space minus space 9.74 over denominator 9.74 end fraction space cross times space 100 space equals space 0.7 percent sign

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    Leander Oates

    Author: Leander Oates

    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.