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

Gravitational field strength equation

• The magnitude of the gravitational field, $\left|\overrightarrow{g}\right|$, created by a system of mass, $M$, at a point in space is equal to the ratio of the gravitational force, $\left|F_g\right|$, 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 $\overrightarrow{g}$

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

  • Hence, $\left|\overrightarrow{g}\right|$ is equal to the ratio of $\left|F_g\right|$ and $m$

Gravitational field strength of Earth on Moon

• This defines the gravitational field strength formula:

$\left|\overrightarrow{g}\right|=\frac{\left|F_g\right|}{m}$

• Where:

  • $\left|\overrightarrow{g}\right|=$magnitude of the gravitational field or gravitational field strength, measured in $\mathrm{N}/\mathrm{kg}$

  • $\left|F_g\right|=$gravitational force, measured in $\mathrm{N}$

  • $m=$mass of test object, measured in $\mathrm{kg}$

• This is also equal to:

$\left|\overrightarrow{g}\right|=G\frac{M}{r^2}$

• Where:

  • $G=$universal gravitational constant $=6.67\times {10}^{-11}\mathrm{N}·{\mathrm{m}}^2/{\mathrm{kg}}^2$

  • $M=$the mass of the system creating the gravitational field, measured in $\mathrm{kg}$

  • $r=$the radius of the system creating the gravitational field, measured in $\mathrm{m}$

Derived equation

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

$\left|\overrightarrow{g}\right|=\frac{\left|F_g\right|}{m}=G\frac{M}{r^2}$

Derivation:

Step 1: Identify the fundamental principles

• Newton's second law

${\overrightarrow{F}}_{net}=m\overrightarrow{a}$

• Newton's law of gravitation

$\left|F_g\right|=G\frac{m_1{m}_2}{r^2}$

Step 2: Apply the specific conditions

• Newton's second law

  • When the gravitational force is the only force exerted on an object the acceleration, $\overrightarrow{a}$, is equal to the magnitude of the gravitational field, $\overrightarrow{g}$

  • Hence, $\overrightarrow{a}=\left|\overrightarrow{g}\right|$

${\overrightarrow{F}}_{net}=\left|F_g\right|=m\left|\overrightarrow{g}\right|$

• 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 $m_2$

$\left|F_g\right|=G\frac{Mm}{r^2}$

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

$m\left|\overrightarrow{g}\right|=\left|F_g\right|=G\frac{Mm}{r^2}$

• Divide by the test mass $m$

  $\frac{\cancel{m}\left|\overrightarrow{g}\right|}{\cancel{m}}=\frac{\left|F_g\right|}{m}=G\frac{M\cancel{m}}{\cancel{m}{r}^2}$

• Simplify

$\left|\overrightarrow{g}\right|=\frac{\left|F_g\right|}{m}=G\frac{M}{r^2}$