Physics · Pearson Edexcel 9PH0 · Topic 12

Gravitational fields

Gravitational fields, Newton's law of gravitation and orbits.

  • 7 sub-topics
  • Taught in Year 13
  • 44 practice questions
  • 7 animated lessons
Practice questions (44) Lessons Worksheets All Edexcel Physics topics

What the specification says

Every sub-topic Pearson Edexcel can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. 174Gravitational fieldsY13
  2. 175Gravitational field strength, g = F/mY13
  3. 176Newton's law of universal gravitationY13
  4. 177Field due to a point massY13
  5. 178Gravitational potentialY13
  6. 179Comparing electric and gravitational fieldsY13
  7. 180OrbitsY13

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Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 44 questions in this topic, free, no account needed.

Gravitational fields

Not needed for Year 12Easy

Gravitational fields · Year 13 · 3 marks

The Earth is surrounded by a gravitational field.
  1. (a)
    Define a gravitational field.
    (1)
  2. (b)
    Sketch the field lines around the Earth and near its surface.
    (2)
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Part (a)1 mark

  1. A region in which a mass experiences a force.

Part (b)2 marks

  1. Radial lines pointing towards the centre of the Earth;
  2. near the surface they are approximately parallel and equally spaced (a uniform field).

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Gravitational fields

Not needed for Year 12Medium

Gravitational field strength · Year 13 · 3 marks

The Moon has mass 7.3×10227.3 \times 10^{22} kg and radius 1.7×1061.7 \times 10^{6} m.
  1. (a)
    Calculate the gravitational field strength at the surface of the Moon.
    (2)
  2. (b)
    Calculate the field strength at a height equal to one radius above the surface.
    (1)

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Part (a)2 marks

  1. g=GMr2g = \frac{GM}{r^2}
  2. g=1.68 N kg−1g = 1.68\ \mathrm{N\,kg^{-1}}

Part (b)1 mark

  1. g=0.421 N kg−1g = 0.421\ \mathrm{N\,kg^{-1}}

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Gravitational fields

Not needed for Year 12Hard

Geostationary orbit · Year 13 · 4 marks

A geostationary satellite orbits the Earth once every 24 hours. Mass of the Earth =5.97×1024= 5.97 \times 10^{24} kg, radius of the Earth =6.37×106= 6.37 \times 10^{6} m, GG=6.67×10−11 N m2 kg−2{}= 6.67 \times 10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}.
  1. Show that the height of the satellite above the Earth's surface is about 3.6×1073.6 \times 10^{7} m.
    (4)
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  1. Gravity provides the centripetal force that keeps the satellite in its circle.GMmr2\dfrac{GMm}{r^2}=mr(2πT)2{}= mr\left(\dfrac{2\pi}{T}\right)^2
  2. Cancel mm, rearrange for r3r^3, and convert 24 hours to seconds.r3=GMT24π2r^3 = \dfrac{GMT^2}{4\pi^2}, with T=86 400T = 86\,400 s
  3. Take the cube root. This is measured from the centre of the Earth.r=4.22×107 mr = 4.22 \times 10^{7}\ \mathrm{m}
  4. Subtract the Earth's radius to get the height above the surface.Height =4.22×107= 4.22 \times 10^{7}−6.37×106{}- 6.37 \times 10^{6}=3.59×107 m{}= 3.59 \times 10^{7}\ \mathrm{m}

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