Gravitational Potential Energy (OCR A Level Physics) : Revision Note

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Gravitational Potential Energy

  • The gravitational potential Vg at a point is given by:

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

  • Recall that potential V is defined as the energy required to bring unit mass m from infinity to a defined point in the field

    • Recall that the gravitational field is usually caused by 'big mass', M

  • Therefore, the potential energy E  is given by:

EmVg

  • Substituting the equation for gravitational potential Vg gives the equation for the gravitational potential energy E between two masses M and m:

E equals fraction numerator negative G M m over denominator r end fraction

  • Where: 

    • G = Universal Gravitational Constant (N m2 kg–2)

    • M = mass causing the field (kg)

    • m = mass moving within the field of M (kg)

    • r = distance between the centre of m and M (m)

Calculating Changes in Gravitational Potential Energy

  • When a mass is moved against the force of gravity, work is required

    • This is because gravity is attractive, therefore, energy is needed to work against this attractive force

  • The work done (or energy transferred) ∆W to move a mass m between two different points in a gravitational potential ∆V is given by:

W = m V

  • Where:

    • W = work done or energy transferred (J)

    • m = mass (kg)

    • V = change in gravitational potential (J kg-1)

 

  • This work done, or energy transferred, is the change in gravitational potential energy (G.P.E) of the mass

    • When ∆V = 0, then the the change in G.P.E = 0

  • The change in G.P.E, or work done, for an object of mass m at a distance r1 from the centre of a larger mass M, to a distance of r2 further away can be written as:

  • Where:

    • M = mass that is producing the gravitational field (e.g., a planet) (kg)

    • m = mass that is moving in the gravitational field (e.g., a satellite) (kg)

    • r1initial distance of m from the centre of M (m)

    • r2 = final distance of m from the centre of (m)

  • Work is done when an object in a planet's gravitational field moves against the gravitational field lines, i.e., away from the planet

    • This is, again, because gravity is attractive

    • Therefore, energy is required to move against gravitational field lines

Change in GPE, downloadable AS & A Level Physics revision notes

Gravitational potential energy increases as a satellite leaves the surface of the Moon

Worked Example

A spacecraft of mass 300 kg leaves the surface of Mars to an altitude of 700 km. Calculate the work done by the spacecraft.

Radius of Mars = 3400 km

Mass of Mars = 6.40 × 1023 kg

Answer:

Step 1: Write down the work done (or change in G.P.E) equation

Worked Example G.P.E equation

Step 2Determine values for r1 and r2

r1 is the radius of Mars = 3400 km = 3400 × 103 m

r2 is the radius + altitude = 3400 + 700 = 4100 km = 4100 × 103 m

Step 3: Substitute in values

Worked Example G.P.E Calculation

ΔG.P.E =  643.076 × 10= 640 MJ (2 s.f.)

Examiner Tips and Tricks

Make sure to not confuse the ΔG.P.E equation with

ΔG.P.E = mgΔh

The above equation is only relevant for an object lifted in a uniform gravitational field (close to the Earth’s surface). The new equation for G.P.E will not include g, because this varies for different planets and is no longer a constant (decreases by 1/r2) outside the surface of a planet.

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Ashika

Author: Ashika

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Ashika graduated with a first-class Physics degree from Manchester University and, having worked as a software engineer, focused on Physics education, creating engaging content to help students across all levels. Now an experienced GCSE and A Level Physics and Maths tutor, Ashika helps to grow and improve our Physics resources.

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