Gravitational Field Strength Equation (College Board AP® Physics 1: Algebra-Based)

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Ann Howell

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Gravitational field strength equation

  • The magnitude of the gravitational field, open vertical bar g with rightwards arrow on top close vertical bar, created by a system of mass, M, at a point in space is equal to the ratio of the gravitational force, open vertical bar F subscript g close vertical bar, exerted by the system on a test object of mass, m, to the mass of the test object, m

  • Consider the Moon in orbit around the Earth:

    • The Earth is the system of mass M that creates the gravitational field of magnitude g with rightwards arrow on top

    • The Moon is the test object of mass m within the Earth's gravitational field

    • Hence, open vertical bar g with rightwards arrow on top close vertical bar is equal to the ratio of open vertical bar F subscript g close vertical bar and m

Gravitational field strength of Earth on Moon

The Earth has a mass M and the Moon a mass m, illustrating the gravitational force from Earth on the Moon, labeled as |Fg|.
The Earth of mass M is the system that creates the gravitational force Fg on the moon which is a test mass m
  • This defines the gravitational field strength formula:

open vertical bar g with rightwards arrow on top close vertical bar space equals space fraction numerator open vertical bar F subscript g close vertical bar over denominator m end fraction

  • Where:

    • open vertical bar g with rightwards arrow on top close vertical bar space equals spacemagnitude of the gravitational field or gravitational field strength, measured in straight N divided by kg

    • open vertical bar F subscript g close vertical bar space equals spacegravitational force, measured in straight N

    • m space equals spacemass of test object, measured in kg

  • This is also equal to:

open vertical bar g with rightwards arrow on top close vertical bar space equals space G M over r squared

  • Where:

    • G space equals spaceuniversal gravitational constant equals space 6.67 space cross times space 10 to the power of negative 11 end exponent space straight N times straight m squared space divided by space kg squared

    • M space equals spacethe mass of the system creating the gravitational field, measured in kg

    • r space equals spacethe radius of the system creating the gravitational field, measured in straight m

Derived equation

  • The magnitude of the gravitational field, open vertical bar g with rightwards arrow on top close vertical bar, created by a system of mass, M, at a point in space is equal to the ratio of the gravitational force, open vertical bar F subscript g close vertical bar, exerted by the system on a test object of mass, m, to the mass of the test object, m

open vertical bar g with rightwards arrow on top close vertical bar space equals space fraction numerator open vertical bar F subscript g close vertical bar over denominator m end fraction space equals space G M over r squared

Derivation:

Step 1: Identify the fundamental principles

F with rightwards arrow on top subscript n e t end subscript space equals space m a with rightwards arrow on top

open vertical bar F subscript g close vertical bar space equals space G fraction numerator m subscript 1 m subscript 2 over denominator r squared end fraction

Step 2: Apply the specific conditions

  • Newton's second law

    • When the gravitational force is the only force exerted on an object the acceleration, a with rightwards arrow on top, is equal to the magnitude of the gravitational field, g with rightwards arrow on top

    • Hence, a with rightwards arrow on top space equals space open vertical bar g with rightwards arrow on top close vertical bar

F with rightwards arrow on top subscript n e t end subscript space equals space open vertical bar F subscript g close vertical bar space equals space m open vertical bar g with rightwards arrow on top close vertical bar

  • Newton's law of gravitation

    • The mass of the system creating the gravitational field is M which represents m subscript 1

    • The mass of the test mass in the gravitational field is m which represents m subscript 2

open vertical bar F subscript g close vertical bar space equals space G fraction numerator M m over denominator r squared end fraction

Step 3: Combine and simplify the two conditions

  • The gravitational force from Newton's second law equals the gravitational force from Newton's law of gravitation

m open vertical bar g with rightwards arrow on top close vertical bar space equals space open vertical bar F subscript g close vertical bar space equals space G fraction numerator M m over denominator r squared end fraction

  • Divide by the test mass m

    fraction numerator up diagonal strike m open vertical bar g with rightwards arrow on top close vertical bar over denominator up diagonal strike m end fraction space equals space fraction numerator open vertical bar F subscript g close vertical bar space over denominator m end fraction equals space G fraction numerator M up diagonal strike m over denominator up diagonal strike m r squared end fraction

  • Simplify

open vertical bar g with rightwards arrow on top close vertical bar equals space fraction numerator open vertical bar F subscript g close vertical bar space over denominator m end fraction equals space G M over r squared

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Ann Howell

Author: Ann Howell

Expertise: Physics Content Creator

Ann obtained her Maths and Physics degree from the University of Bath before completing her PGCE in Science and Maths teaching. She spent ten years teaching Maths and Physics to wonderful students from all around the world whilst living in China, Ethiopia and Nepal. Now based in beautiful Devon she is thrilled to be creating awesome Physics resources to make Physics more accessible and understandable for all students, no matter their schooling or background.