Gravitational Field Strength (OCR A Level Physics)

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Gravitational Field Strength

  • There is a universal force of attraction between all matter with mass

    • This force is known as the ‘force due to gravity’ or the weight

  • The Earth’s gravitational field is responsible for the weight of all objects on Earth

  • The gravitational field strength g at a point is defined as force F per unit mass m of an object at that point:

Gravitational Field Strength Equation_2
  • Where:

    • g = gravitational field strength (N kg–1)

    • F = force due to gravity, or weight (N)

    • m = mass (kg)

 

  • This equation shows that:

    • The larger the mass of an object, the greater its pull on another mass

    • On planets with a large value of g, the gravitational force per unit mass is greater than on planets with a smaller value of g

  • An object's mass remains the same at all points in space

    • However, on planets such as Jupiter, the weight of an object will be a lot greater than on a less massive planet, such as Earth

    • This means the gravitational force would be so high that humans, for example, would not be able to fully stand up (or, even worse...)

gravitational field strength, downloadable AS & A Level Physics revision notes

The weight force on Jupiter would be so large that even standing upright would be difficult

  • Factors that affect the gravitational field strength at the surface of a planet are:

    • The radius (or diameter) of the planet

    • The mass (or density) of the planet

Worked Example

Calculate the mass of an object with weight 10 N on Earth.

Answer:

Gravitational Field Strength in a Radial Field

  • In a radial field (due to a point mass M), the gravitational field lines get further apart from each other

    • This indicates that the strength of the gravitational field decreases with distance from the centre of mass of M

  • The gravitational field strength g in a radial field, due to some mass M, is given by the equation:

g equals negative fraction numerator G M over denominator r squared end fraction

  • Where: 

    • g = gravitational field strength (N kg–1)

    • G = Newton's Gravitational constant (N m2 kg–2)

    • M = mass of the object causing the gravitational field (kg)

    • r = radial distance from the centre of mass of M (m)

  • Note:

    • The negative sign in this equation indicates that the gravitational field is attractive

    • In other words, the direction of the gravitational field lines is towards the mass M

  • On the Earth’s surface, g has a constant value of 9.81 N kg-1

  • However far outside the Earth’s surface, g is not constant

    • g decreases as r increases by a factor of 1/r2

    • This is an inverse square law relationship with distance

  • When the magnitude of g is plotted against the distance from the centre of a planet, r has two parts:

    • When r < R (the radius of the planet), g is directly proportional to r

    • When r > R, g is inversely proportional to r2 

g v R graph on Earth (1), downloadable AS & A Level Physics revision notes

The magnitude of gravitational field strength g against distance r from the Earth's surface follows a 1/r2 relationship

Gravitational Field Strength Close to the Earth's Surface

  • Near the Earth's surface, the gravitational field is uniform

    • Hence, the gravitational field lines are parallel and evenly spaced

  • This means the gravitational field strength is constant at every point near the Earth's surface

    • Numerically, the gravitational field strength near Earth's surface is equal to the acceleration due to gravity, g = 9.81 m s–2 

Worked Example

Determine the distance from the Earth's surface at which the gravitational field strength decreases by a factor of 0.5. 

(The radius of the Earth is 6400 km and its mass is 6.0 × 1024 kg)

Answer:

Step 1: Write the known quantities

  • Radius of the Earth RE = 6400 km = 6400 × 103 m

  • Mass of the earth ME = 6.0 × 1024 kg

  • Gravitational constant G = 6.67 × 10–11 N m2 kg–2

Step 2: Recall the value of the gravitational field strength at the Earth's surface

  • The gravitational field strength at the Earth's surface g = 9.81 N kg–1

Step 3: Write the equation for gravitational field strength in a radial field

  • The Earth creates a radial gravitational field (far from its surface) therefore the equation for gravitational field strength g is: 

g equals negative fraction numerator G M over denominator r squared end fraction

Step 4: Determine the distance at which the field strength reduces by a factor of 0.5

  • If the field strength decreases by a factor of 0.5, then g × 0.5 = 9.81 × 0.5 = 4.905 N kg–1

  • Therefore, ignoring the negative sign (as we only want a magnitude): 

4.905 = fraction numerator left parenthesis 6.67 cross times 10 to the power of negative 11 end exponent right parenthesis cross times left parenthesis 6 cross times 10 to the power of 24 right parenthesis over denominator r squared end fraction

r2fraction numerator left parenthesis 6.67 cross times 10 to the power of negative 11 end exponent right parenthesis cross times left parenthesis 6 cross times 10 to the power of 24 right parenthesis over denominator 4.905 end fraction

rsquare root of fraction numerator left parenthesis 6.67 cross times 10 to the power of negative 11 end exponent right parenthesis cross times left parenthesis 6 cross times 10 to the power of 24 right parenthesis over denominator 4.905 end fraction end root= 9.0 × 106 m

Step 5: Determine the distance from the Earth's surface

5HxVla3q_5-7-3-grav-field-strength-we_ocr-al-physics
  • The value r = 9.0 × 106 m is the radial distance from the Earth's centre of mass

  • Therefore, the gravitational field strength reduces by a factor 0.5 at a distance r – RE

– RE = (9.0 × 106) – (6400 × 103) = 2.6 × 106 m

Examiner Tips and Tricks

The equation for the gravitational field strength in a radial field is in terms of the distance r from the centre of mass of mass M. If the exam question is about a planet, remember that you might have to take the planet's radius into account, which is the distance between its centre of mass and its surface! As ever, drawing a labelled diagram of the distances in question really helps. 

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Katie M

Author: Katie M

Expertise: Physics

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.