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Gravitational Potential (CIE A Level Physics)

Revision Note

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Leander

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Leander

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Physics

Gravitational Potential

  • The gravitational potential energy is the energy an object has when lifted off the ground given by the familiar equation:

Ep = mgh{"language":"en","fontFamily":"Times New Roman","fontSize":"18","autoformat":true}

  • The gravitational potential energy on the surface of the Earth is taken to be 0
    • This means work is done to lift the object

  • However, outside the Earth’s surface, G.P.E can be defined as:

The energy an object possess due to its position in a gravitational field

  • The gravitational potential at a point is the gravitational potential energy per unit mass at that point
  • Therefore, the gravitational potential is defined as:

The work done per unit mass in bringing a test mass from infinity to a defined point

Calculating Gravitational Potential

  • The equation for gravitational potential ɸ is defined by the mass M and distance r:

ϕ space equals space minus fraction numerator G M over denominator r end fraction

Calculating Gravitational Potential RN equation 1

  • Where:
    • ɸ = gravitational potential (J kg-1)
    • G = Newton’s gravitational constant
    • M = mass of the body producing the gravitational field (kg)
    • r = distance from the centre of the mass to the point mass (m)

  • The gravitational potential is negative near an isolated mass, such as a planet, because the potential when r is at infinity is defined as 0
  • Gravitational forces are always attractive so as r decreases, positive work is done by the mass when moving from infinity to that point
    • When a mass is closer to a planet, its gravitational potential becomes smaller (more negative)
    • As a mass moves away from a planet, its gravitational potential becomes larger (less negative) until it reaches 0 at infinity

  • This means when the distance (r) becomes very large, the gravitational force tends rapidly towards 0 at a point further away from a planet

Gravitational Potential of a Meteor

Gravitational potential diagram, downloadable AS & A Level Physics revision notes

Gravitational potential increases and decreases depending on whether the object is travelling towards or against the field lines from infinity

Worked example

A planet has a diameter of 7600 km and a mass of 3.5 × 1023 kg. A rock of mass 528 kg accelerates towards the planet from infinity.

At a distance of 400 km above the planet’s surface, calculate the gravitational potential of the rock.

Answer:

Step 1: Write the gravitational potential equation

ϕ space equals space minus fraction numerator G M over denominator r end fraction

Step 2: Determine the value of r

  • r is the distance from the centre of the planet
  • Radius of the planet = planet diameter ÷ 2 = 7600 ÷ 2  = 3800 km

r = 3800 + 400 = 4200 km = 4.2 × 106 m

Step 3:  Substitute in values

ϕ space equals space minus 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 3.5 cross times 10 to the power of 23 close parentheses over denominator 4.2 cross times 10 to the power of 6 end fraction space equals space minus 5.6 cross times 10 to the power of 6 space straight J space kg to the power of negative 1 end exponent

Exam Tip

Remember to keep the negative sign in your solution for gravitational potential. However, if you’re asked for the ‘change in’ gravitational potential, no negative sign should be included since you are finding a difference in values (between 0 at infinity and the gravitational potential from your calculation).

Gravitational Potential Energy Between Two Point Masses

  • The gravitational potential energy (G.P.E) at point in a gravitational field is defined as:

 The work done in bringing a mass from infinity to that point

  • The equation for G.P.E of two point masses m and M at a distance r is:

E subscript p space equals space minus fraction numerator G M m over denominator r end fraction

  • The change in G.P.E is given by:

ΔG.P.E = mgΔh

  • Where:
    • m = mass of the object (kg)
    • ɸ = gravitational potential at that point (J kg-1)
    • Δh = change in height (m)

 

  • Recall that at infinity, ɸ = 0 and therefore G.P.E = 0
  • It is more useful to find the change in G.P.E e.g. a satellite lifted into space from the Earth’s surface
  • The change in G.P.E from for an object of mass m at a distance r1 from the centre of mass M, to a distance of r2 further away is:

increment G P E space equals fraction numerator space G M m over denominator r subscript 2 end fraction space minus space open parentheses negative fraction numerator G M m over denominator r subscript 1 end fraction close parentheses space equals space G M m space open parentheses 1 over r subscript 1 space minus fraction numerator space 1 over denominator r subscript 2 end fraction close parentheses

  • The change in potential Δɸ is the same, without the mass of the object m:

increment ϕ space equals space minus fraction numerator G M over denominator r subscript 2 end fraction space minus open parentheses fraction numerator space minus G M over denominator r subscript 1 end fraction close parentheses space equals space G M open parentheses 1 over r subscript 1 space minus fraction numerator space 1 over denominator r subscript 2 end fraction close parentheses

Change in Gravitational Potential Energy of a Satellite

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

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

Maths tip

  • Multiplying two negative numbers equals a positive number, for example:

negative open parentheses negative fraction numerator G M over denominator r end fraction close parentheses space equals space plus fraction numerator G M over denominator r end fraction

Worked example

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

Calculate the change in gravitational potential energy of the spacecraft.

Radius of Mars = 3400 km

Mass of Mars = 6.40 x 1023 kg

Answer: 

Step 1: Difference in gravitational potential energy equation

increment G P E space equals space G M m open parentheses 1 over r subscript 1 minus 1 over r subscript 2 close parentheses

Step 2: Determine 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

increment G P E space equals space open parentheses 6.67 cross times 10 to the power of negative 11 end exponent close parentheses space cross times space open parentheses 6.40 cross times 10 to the power of 23 close parentheses space cross times space 300 space cross times space open parentheses fraction numerator 1 over denominator 3400 cross times 10 cubed end fraction space minus fraction numerator space 1 over denominator 4100 cross times 10 cubed end fraction close parentheses

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

Exam Tip

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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Leander

Author: Leander

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.