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

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

Exam code: 9702

Leander Oates

Last updated

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 Earth's 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 Earth's radius for an object within the Earth's 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

  • Therefore, 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, a distance of 8850 m above the Earth's surface.

Show that the value of g at the top of Everest is about 0.3% less than the value of g at the Earth's surface.

Mass of the Earth = 5.97 × 1024 kg

Radius of the Earth = 6370 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 6370 space plus space 8.85 space equals space 6378.85 space km space equals space 6378.85 cross times 10 cubed space straight m

Step 3: Substitute the known values to calculate

  • At the Earth's surface:

g subscript 0 space equals space fraction numerator open parentheses 6.67 cross times 10 to the power of negative 11 end exponent close parentheses cross times open parentheses 5.97 cross times 10 to the power of 24 close parentheses over denominator open parentheses 6370 cross times 10 cubed close parentheses squared end fraction space equals space 9.8134 space straight N space kg to the power of negative 1 end exponent

  • At the height of Mt Everest:

g apostrophe 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 5.97 cross times 10 to the power of 24 close parentheses over denominator open parentheses 6378.85 cross times 10 cubed close parentheses squared end fraction space equals space 9.7862 space straight N thin space kg to the power of negative 1 end exponent

Step 4: Calculate the percentage decrease

percent sign space g space equals space fraction numerator g subscript 0 space minus space g apostrophe over denominator g subscript 0 end fraction cross times 100 percent sign

percent sign space g space equals space fraction numerator 9.8134 space minus space 9.7862 over denominator 9.8134 end fraction cross times 100 space equals space 0.28 percent sign space almost equal to space 0.3 percent sign

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    Leander Oates

    Author: Leander Oates

    Expertise: Physics Content Creator

    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.