IB DP Physics: HL复习笔记6.2.3 Gravitational Field Strength

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
  • A gravitational field is defined as:

A region of space where a test mass experiences a force due to the gravitational attraction of another mass

  •  The direction of the gravitational field is always towards the centre of the mass causing the field
    • Gravitational forces cannot be repulsive
  • Gravity has an infinite range, meaning it affects all objects in the universe
    • There is a greater gravitational force around objects with a large mass (such as planets)
    • There is a smaller gravitational force around objects with a small mass (almost negligible for atoms)

7.1.2-Gravitational-Attractive-Force

The Earth's gravitational field produces an attractive force. The force of gravity is always attractive

  • The gravitational field strength at a point is defined as:

The force per unit mass experienced by a test mass at that point

  • This can be written in equation form as:

7.1.2-Gravitational-Field-Strength-Equation_2

  • Where:
    • g = gravitational field strength (N kg−1)
    • F = force due to gravity, or weight (N)
    • m = mass of test mass in the field (kg)
  • This equation shows that:
    • 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 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 unable to fully stand up

13.1.1.1-gravitational-field-strength

A person’s weight on Jupiter would be so large that a human would be unable to fully stand up

  • 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
  • This can be shown by equating the equation F = mg with Newton's law of gravitation:

6.2.3-Resultant-g-strength-eq-1-1

  • Substituting the force F with the gravitational force mg leads to:

6.2.3-Resultant-g-strength-eq-2-1-e1642720169632

  • Cancelling the mass of the test mass, m, leads to the equation:

7.1.5-g-in-a-Radial-Field-Equation-2

  • Where:
    • G = Newton's Gravitational Constant
    • M = mass of the body causing the field (kg)
    • r = distance from the mass where you are calculating the field strength (m)
  • This equation shows that:
    • The gravitational field strength g depends only on the mass of the body M causing the field
    • Hence, objects with any mass m in that field will experience the same gravitational field strength
    • The gravitational field strength g is inversely proportional to the square of the radial distance, r2

Worked Example

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

WE-Gravitational-field-answer-image_1

Worked Example

The mean density of the Moon is 3/5 times the mean density of the Earth. The gravitational field strength on the Moon is 1/6 of the value on Earth.

Determine the ratio of the Moon's radius rM and the Earth's radius rE.

7.1.5-g-Radius-Field-Worked-Example-17.1.5-g-Radius-Field-Worked-Example-2

Exam Tip

There is a big difference between g and G (sometimes referred to as ‘little g’ and ‘big G’ respectively), g is the gravitational field strength and G is Newton’s gravitational constant. Make sure not to use these interchangeably! Remember the equation density ρ = mass m ÷ volume V, which may come in handy with some calculations

Resultant Gravitational Field Strength

  • In a similar way to other vectors, such as force or velocity, the gravitational field strength due to two bodies can be determined
    • This is because gravitational field strength is a vector, meaning it has both a magnitude and direction
  • The resultant gravitational field strength is, therefore, the vector sum of the gravitational field strength due to each body

Worked Example

A planet is equidistant from two stars in a binary system. Each star has a mass of 5.0 × 1030 kg and the planet is at a distance of 3.0 × 1012 m from each star. Calculate the magnitude of the resultant gravitational field strength at the position of the planet.

6-2-3-we-resultant-g-force

Step 1: List the known quantities

    • Mass of one star, M = 5.0 × 1030 kg
    • Distance between one star and the planet, r = 3.0 × 1012 m
    • Angle between the gravitational field strength and the planet, θ = 42°

Step 2: Write out the equation for gravitational field strength

7.1.5-g-in-a-Radial-Field-Equation-2Step 3: Calculate the gravitational field strength due to one star 

6.2.3-Resultant-g-strength-eq-3-1g = 3.7 × 10−5 N kg−1

Step 4: Resolve the vectors vertically

6.2.3-Resultant-G-Strength-1-1

    • The vertical component of each vector is g cos 42°

Step 5: Use vector addition to determine the resultant gravitational field strength

6.2.3-Resultant-G-Strength-2-1gresultant = g cos 42° + g cos 42° = 2g cos 42°

gresultant = 2 × (3.7 × 10−5) × cos 42°

gresultant = 5.5 × 10−5 N kg−1

Exam Tip

Don't worry, for calculation questions involving resultant gravitational field strength - only two bodies along a straight line will be tested!

 

 

 

 

 

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