Gravitational potential energy describes the energy stored in a gravitational configuration.
With the standard convention that gravitational potential energy is zero at infinite separation,
the potential energy of two masses is negative:
\[
\boxed{
U=-\frac{GMm}{r}
}.
\]
1. Meaning of the negative sign
The negative sign comes from choosing
\[
U(\infty)=0.
\]
Since gravity is attractive, a mass at a finite distance is bound to the central mass.
Energy must be added to move it away to infinity. Therefore,
\[
U(r)<0.
\]
2. Gravitational potential energy around a planet
For a satellite or object of mass \(m\) at distance \(r\) from the center of a planet of mass \(M\),
\[
\boxed{
U(r)=-\frac{GMm}{r}
}.
\]
The distance \(r\) is measured from the center of the planet, not from the surface.
If the object is at altitude \(h\) above a planet of radius \(R\), then
\[
r=R+h.
\]
3. Change in gravitational potential energy
If an object moves from \(r_1\) to \(r_2\), then
\[
U_1=-\frac{GMm}{r_1},
\qquad
U_2=-\frac{GMm}{r_2}.
\]
The change in potential energy is
\[
\boxed{
\Delta U=U_2-U_1
}.
\]
Substituting the formula gives
\[
\boxed{
\Delta U
=
GMm\left(\frac{1}{r_1}-\frac{1}{r_2}\right)
}.
\]
If \(r_2>r_1\), the object moves outward and \(\Delta U>0\). This means energy must be added.
If \(r_2
4. Binding energy preview
The energy needed to move a mass from radius \(r\) to infinity with zero final kinetic energy is
\[
E_{\mathrm{bind}}
=
U(\infty)-U(r).
\]
Since \(U(\infty)=0\),
\[
\boxed{
E_{\mathrm{bind}}=-U(r)=\frac{GMm}{r}
}.
\]
This is a useful preview of how strongly the object is gravitationally bound.
5. Gravitational potential per unit mass
Gravitational potential is potential energy per unit mass:
\[
\Phi=\frac{U}{m}.
\]
For a central mass,
\[
\boxed{
\Phi=-\frac{GM}{r}
}.
\]
Its unit is \(\mathrm{J/kg}\).
6. Escape speed connection
Escape speed follows from setting total mechanical energy equal to zero:
\[
\frac12mv_{\mathrm{esc}}^2-\frac{GMm}{r}=0.
\]
Solving gives
\[
\boxed{
v_{\mathrm{esc}}=\sqrt{\frac{2GM}{r}}
}.
\]
This is linked to binding energy because the required kinetic energy is
\[
\frac12mv_{\mathrm{esc}}^2
=
\frac{GMm}{r}
=
E_{\mathrm{bind}}.
\]
7. Worked example: satellite near Earth
Suppose a \(1000\ \mathrm{kg}\) satellite is \(8000\ \mathrm{km}\) from Earth’s center.
\[
M_{\oplus}=5.9722\times10^{24}\ \mathrm{kg},
\qquad
m=1000\ \mathrm{kg},
\qquad
r=8000\ \mathrm{km}=8.000\times10^6\ \mathrm{m}.
\]
Apply the formula:
\[
U=-\frac{GMm}{r}.
\]
\[
U
=
-\frac{(6.67430\times10^{-11})(5.9722\times10^{24})(1000)}
{8.000\times10^6}.
\]
\[
\boxed{
U\approx -4.98\times10^{10}\ \mathrm{J}
}.
\]
If the distance is instead close to Earth’s surface radius, \(r\approx6.37\times10^6\ \mathrm{m}\),
the value is about
\[
U\approx -6.25\times10^{10}\ \mathrm{J}.
\]
This explains why different rounded examples may quote different values.
8. Multiple-mass systems
For a system of several point masses, calculate every distinct pair:
\[
U_{ij}=-\frac{Gm_im_j}{r_{ij}}.
\]
Then add all pair energies:
\[
\boxed{
U_{\mathrm{total}}
=
\sum_{i
For example, with three masses there are three pairs:
\[
U_{\mathrm{total}}
=
U_{12}+U_{13}+U_{23}.
\]
With four masses there are six pairs:
\[
U_{\mathrm{total}}
=
U_{12}+U_{13}+U_{14}+U_{23}+U_{24}+U_{34}.
\]
9. Summary table
| Quantity |
Formula |
Meaning |
| Potential energy |
\(U=-GMm/r\) |
Energy of one mass in the field of another |
| Potential |
\(\Phi=-GM/r\) |
Potential energy per kilogram |
| Energy change |
\(\Delta U=GMm(1/r_1-1/r_2)\) |
Change in gravitational potential energy |
| Binding energy |
\(E_{\mathrm{bind}}=GMm/r\) |
Energy needed to separate the object to infinity |
| Escape speed |
\(v_{\mathrm{esc}}=\sqrt{2GM/r}\) |
Speed needed to escape with zero final speed at infinity |
| Multi-body total |
\(\sum_{i
| Total potential energy of all distinct mass pairs |
|
Key idea: gravitational potential energy is negative for bound systems when zero is chosen at infinity.
