Gravitational Potential Gradient (DP IB Physics: HL): Revision Note

Katie M

Written by: Katie M

Reviewed by: Caroline Carroll

Updated on

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 = Vgr

  • 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:

7BStox9R_7-2-3-gravitational-potential-and-distance-graph

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 (because the graph is drawn below the horizontal r axis)

    • As r increases, Vg against r follows a 1r relation

    • The gradient of the graph at any particular point is the value of g at that point, g = Vg × 1r = Vgr

    • 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:

    Vg = GMr

    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

Vg = GME(1r21r1)

Step 3: Substitute the values into the equation

ΔVg = (6.67×1011) × (5.97×1024) × (16.38×10611.914×107)

ΔVg = −4.16 × 107 J kg−1

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Katie M

Author: Katie M

Expertise: Curriculum Expert

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.

Caroline Carroll

Reviewer: Caroline Carroll

Expertise: Head of Content Delivery

Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about delivering high-quality resources to help students achieve their full potential.