Newton’s law of universal gravitation states that every mass attracts every other mass. The force is proportional
to the product of the masses and inversely proportional to the square of the distance between their centers.
1. Scalar form
For two point masses or spherically symmetric bodies,
\[
\boxed{
F=G\frac{m_1m_2}{r^2}
}.
\]
Here:
- \(F\) is the gravitational force in newtons,
- \(G\) is the gravitational constant,
- \(m_1\) and \(m_2\) are the two masses,
- \(r\) is the center-to-center distance.
2. Gravitational constant
\[
\boxed{
G=6.67430\times10^{-11}\ \mathrm{N\,m^2/kg^2}
}.
\]
The value of \(G\) is very small, so gravitational attraction between everyday objects is usually tiny.
It becomes important when at least one mass is astronomical, such as a planet, moon, or star.
3. Inverse-square law
The distance \(r\) appears squared in the denominator:
\[
F\propto\frac{1}{r^2}.
\]
Therefore:
\[
\text{if } r \text{ doubles, then } F \text{ becomes } \frac14 \text{ as large}.
\]
\[
\text{if } r \text{ triples, then } F \text{ becomes } \frac19 \text{ as large}.
\]
4. Vector form
Gravitational force has direction. It points along the line connecting the two masses and is always attractive.
\[
\boxed{
\vec F_{12}
=
G\frac{m_1m_2}{|\vec r|^3}\vec r
}.
\]
In two dimensions, for a test mass at \((x_0,y_0)\) and a source mass at \((x_i,y_i)\),
\[
r_i=\sqrt{(x_i-x_0)^2+(y_i-y_0)^2}.
\]
The component form is
\[
F_{ix}
=
G\frac{m_0m_i}{r_i^2}
\frac{x_i-x_0}{r_i},
\qquad
F_{iy}
=
G\frac{m_0m_i}{r_i^2}
\frac{y_i-y_0}{r_i}.
\]
5. Multiple bodies
If several bodies pull on the same object, calculate each gravitational force separately and add the vectors:
\[
\boxed{
\vec F_{\mathrm{net}}
=
\sum_i \vec F_i
}.
\]
In component form,
\[
F_x=\sum_iF_{ix},
\qquad
F_y=\sum_iF_{iy}.
\]
The magnitude and direction are
\[
F_{\mathrm{net}}
=
\sqrt{F_x^2+F_y^2},
\qquad
\theta=\tan^{-1}\left(\frac{F_y}{F_x}\right).
\]
6. Solving for an unknown
Newton’s law can be rearranged. To solve for \(m_1\),
\[
\boxed{
m_1=\frac{Fr^2}{Gm_2}
}.
\]
To solve for \(m_2\),
\[
\boxed{
m_2=\frac{Fr^2}{Gm_1}
}.
\]
To solve for distance \(r\),
\[
\boxed{
r=\sqrt{\frac{Gm_1m_2}{F}}
}.
\]
7. Connection with weight
Near Earth’s surface, the gravitational force on a mass is its weight:
\[
W=mg.
\]
This comes from Newton’s law:
\[
F=G\frac{M_{\oplus}m}{R_{\oplus}^2}.
\]
Therefore,
\[
g=G\frac{M_{\oplus}}{R_{\oplus}^2}.
\]
8. Worked example: person on Earth
Use:
\[
M_{\oplus}=5.97\times10^{24}\ \mathrm{kg},
\qquad
m=70\ \mathrm{kg},
\qquad
r=6.37\times10^6\ \mathrm{m}.
\]
Apply Newton’s law:
\[
F
=
G\frac{M_{\oplus}m}{r^2}.
\]
\[
F
=
(6.67430\times10^{-11})
\frac{(5.97\times10^{24})(70)}{(6.37\times10^6)^2}.
\]
\[
F\approx6.87\times10^2\ \mathrm{N}.
\]
\[
\boxed{
F\approx686.7\ \mathrm{N}
}.
\]
This is the approximate weight of the \(70\ \mathrm{kg}\) person near Earth’s surface.
9. Common gravitational data
| Body or quantity |
Approximate value |
Use |
| Earth mass |
\(5.97\times10^{24}\ \mathrm{kg}\) |
Earth-surface and Earth-orbit problems |
| Earth radius |
\(6.37\times10^6\ \mathrm{m}\) |
Distance from Earth’s center to surface |
| Moon mass |
\(7.35\times10^{22}\ \mathrm{kg}\) |
Earth–Moon force |
| Sun mass |
\(1.99\times10^{30}\ \mathrm{kg}\) |
Solar-system gravity |
| Astronomical unit |
\(1.496\times10^{11}\ \mathrm{m}\) |
Approximate Earth–Sun distance |
Key idea: gravity is universal, attractive, and follows an inverse-square dependence on center-to-center distance.
