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First exams 2025

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Gravitational Potential Gradient (HL) (HL IB Physics)

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Physics

Gravitational Potential Gradient

  • A gravitational field can be defined in terms of the variation of gravitational potential at different points in the field:

The gravitational field at a particular point is equal to the negative gradient of a potential-distance graph at that point

  • The potential gradient is defined by the equipotential lines
    • These demonstrate the gravitational potential in a gravitational field and are always drawn perpendicular to the field lines
  • The potential gradient in a gravitational field is defined as:

The rate of change of gravitational potential with respect to displacement in the direction of the field

  • Gravitational field strength, g and the gravitational potential, V can be graphically represented against the distance from the centre of a planet, r 
g space equals space minus fraction numerator increment V subscript g over denominator increment r end fraction
  • Where:
    • g = gravitational field strength (N kg-1)
    • ΔVg = change in gravitational potential (J kg-1)
    • Δr = distance from the centre of a point mass (m)
  • The graph of Vg against r for a planet is:

Gravitational Potential and Distance Graph, downloadable AS & A Level Physics revision notes

The gravitational potential and distance graphs follow a -1/r relation
  • The key features of this graph are:
    • The values for Vg are all negative
    • As r increases, Vg against r follows a negative 1 over r relation
    • The gradient of the graph at any particular point is the value of g at that point
    • The graph has a shallow increase as r increases
  • To calculate g, draw a tangent to the graph at that point and calculate the gradient of the tangent
  • This is a graphical representation of the gravitational potential equation:

V subscript g space equals space minus fraction numerator G M over denominator r end fraction

where G and M are constant

Worked example

Determine the change in gravitational potential when travelling from 3 Earth radii (from Earth’s centre) to the surface of the Earth.

Take the mass of the Earth to be 5.97 × 1024 kg and the radius of the Earth to be 6.38 × 106 m.

Answer:

Step 1: List the known quantities

  • Mass of the Earth, ME = 5.97 × 1024 kg
  • Radius of the Earth, rE = 6.38 × 106 m
  • Initial distance, r1 = 3rE = 3 × (6.38 × 106) m = 1.914 × 107 m
  • Final distance, r2 = rE = 6.38 × 106 m
  • Gravitational constant, G = 6.67 × 10−11 m3 kg−1 s−2

Step 2: Write down the equation for potential difference

increment V subscript g space equals space minus G M subscript E open parentheses 1 over r subscript 2 minus 1 over r subscript 1 close parentheses

Step 3: Substitute the values into the equation

straight capital delta V subscript g space equals space minus left parenthesis 6.67 cross times 10 to the power of negative 11 end exponent right parenthesis space cross times space left parenthesis 5.97 cross times 10 to the power of 24 right parenthesis space cross times space open parentheses fraction numerator 1 over denominator 6.38 cross times 10 to the power of 6 end fraction minus fraction numerator 1 over denominator 1.914 cross times 10 to the power of 7 end fraction close parentheses

ΔVg = −4.16 × 107 J kg−1

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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.