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

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Written by: Ann Howell

Reviewed by: Caroline Carroll

Gravitational field strength equation

The magnitude of the gravitational field, $|\vec{g}|$ , created by a system of mass, $M$ , at a point in space is equal to the ratio of the gravitational force, $|F_{g}|$ , 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 $\vec{g}$
- The Moon is the test object of mass $m$ within the Earth's gravitational field
- Hence, $|\vec{g}|$ is equal to the ratio of $|F_{g}|$ 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:

$|\vec{g}| = \frac{|F_{g}|}{m}$

Where:
- $|\vec{g}| =$ magnitude of the gravitational field or gravitational field strength, measured in $N / kg$
- $|F_{g}| =$ gravitational force, measured in $N$
- $m =$ mass of test object, measured in $kg$
This is also equal to:

$|\vec{g}| = G \frac{M}{r^{2}}$

Where:
- $G =$ universal gravitational constant $= 6.67 \times 10^{- 11} N \cdot m^{2} / {kg}^{2}$
- $M =$ the mass of the system creating the gravitational field, measured in $kg$
- $r =$ the radius of the system creating the gravitational field, measured in $m$

Derived equation

The magnitude of the gravitational field, $|\vec{g}|$ , created by a system of mass, $M$ , at a point in space is equal to the ratio of the gravitational force, $|F_{g}|$ , exerted by the system on a test object of mass, $m$ , to the mass of the test object, $m$

$|\vec{g}| = \frac{|F_{g}|}{m} = G \frac{M}{r^{2}}$

Derivation:

Step 1: Identify the fundamental principles

Newton's second law

${\vec{F}}_{n e t} = m \vec{a}$

Newton's law of gravitation

$|F_{g}| = G \frac{m_{1} m_{2}}{r^{2}}$

Step 2: Apply the specific conditions

Newton's second law
- When the gravitational force is the only force exerted on an object the acceleration, $\vec{a}$ , is equal to the magnitude of the gravitational field, $\vec{g}$
- Hence, $\vec{a} = |\vec{g}|$

${\vec{F}}_{n e t} = |F_{g}| = m |\vec{g}|$

Newton's law of gravitation
- The mass of the system creating the gravitational field is $M$ which represents $m_{1}$
- The mass of the test mass in the gravitational field is $m$ which represents $m_{2}$

$|F_{g}| = G \frac{M m}{r^{2}}$

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

$m |\vec{g}| = |F_{g}| = G \frac{M m}{r^{2}}$

Divide by the test mass $m$
$\frac{m |\vec{g}|}{m} = \frac{|F_{g}|}{m} = G \frac{M m}{m r^{2}}$
Simplify

$|\vec{g}| = \frac{|F_{g}|}{m} = G \frac{M}{r^{2}}$

Last updated: 11 October 2024

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