AcademyGravitation
Academy
Universal Gravitation
Level 1 - Physics topic page in Gravitation.
Principle
Universal gravitation models the attraction between masses as a central inverse-square interaction.
Notation
\(G\)
gravitational constant
\(\mathrm{N\,m^{2}\,kg^{-2}}\)
\(m_1,m_2\)
interacting masses
\(\mathrm{kg}\)
\(M\)
source mass
\(\mathrm{kg}\)
\(r\)
separation from the source center
\(\mathrm{m}\)
\(\hat e_r\)
outward radial unit vector from source to test mass
1
\(\vec F_g\)
gravitational force on a test mass
\(\mathrm{N}\)
\(\vec g\)
gravitational field strength
\(\mathrm{N\,kg^{-1}}\)
\(g\)
gravitational field magnitude
\(\mathrm{N\,kg^{-1}}\)
Method
Derivation 1: Build the pairwise force law
Gravity depends on both masses and weakens with separation because the same central influence is spread over a larger spherical area as \(r\) increases.
Inverse-square scale
\[F_g\propto\frac{m_1m_2}{r^2}\]
Insert the constant
\[F_g=G\frac{m_1m_2}{r^2}\]
Give the force a direction
\[\vec F_g=-G\frac{m_1m_2}{r^2}\hat e_r\]
The minus sign makes the force attractive, back toward the source.
Derivation 2: Convert the source law into a field law
Once the source mass is fixed, the test mass only scales the force. Dividing by test mass defines the local gravitational field.
Start from the vector force law
\[\vec F_g=-G\frac{Mm}{r^2}\hat e_r\]
Define field strength
\[\vec g=\frac{\vec F_g}{m}\]
Radial field
\[\vec g=-G\frac{M}{r^2}\hat e_r\]
Field magnitude
\[g=G\frac{M}{r^2}\]
The graph shows the distance scaling only. A point twice as far from the same source feels one quarter of the field magnitude.
Because the curve is inverse-square, moving a little closer to the source matters much more than the same distance change far away.
Rules
These are the compact results from the method above.
Force magnitude
\[F_g=G\frac{m_1m_2}{r^2}\]
Force vector
\[\vec F_g=-G\frac{m_1m_2}{r^2}\hat e_r\]
Field vector
\[\vec g=-G\frac{M}{r^2}\hat e_r\]
Field magnitude
\[g=G\frac{M}{r^2}\]
Force from field
\[\vec F_g=m\vec g\]
Examples
Question
Two objects interact gravitationally. One mass is doubled while the separation is tripled. By what factor does the force magnitude change?
Answer
\[F_g\propto\frac{m_1m_2}{r^2}\]
\[\frac{F'_g}{F_g}=\frac{2}{3^2}=\frac{2}{9}\]
The new force is two ninths of the original force.Checks
- Gravity between ordinary masses is always attractive.
- The inverse-square law applies to point masses and to points outside spherically symmetric masses.
- \(\\hat e_r\) is defined outward, so the gravitational force vector carries a minus sign.
- Doubling distance reduces field and force magnitude by a factor of four, not two.