Imagine you have a tiny test mass, like a marble, and you want to bring it from the far‑away point called infinity down to a spot near a planet. The gravitational potential at that spot is the amount of work per unit mass that would be needed to move the marble from infinity to that spot. It’s written in physics as \$V\$.
If the work done to bring a mass \$m\$ from infinity to a point at distance \$r\$ from a mass \$M\$ is \$W\$, then the gravitational potential at that point is:
\$V = \dfrac{W}{m}\$
For a point mass \$M\$, the work done is \$W = -\dfrac{GMm}{r}\$, so the potential becomes:
\$V(r) = -\dfrac{GM}{r}\$
Here, \$G\$ is the gravitational constant (\$6.674\times10^{-11}\,\text{N·m}^2/\text{kg}^2\$).
\$V = -\dfrac{(6.674\times10^{-11})(5.97\times10^{24})}{6.371\times10^6}\$
\$V \approx -6.25\times10^7\,\text{J/kg}\$
Think of gravitational potential like a hill that a ball rolls down. The higher you are (farther from the planet), the less “downhill” you have to go, so the work required is smaller. At infinity, the hill is flat – no work is needed. As you get closer, the hill steepens, and you must do more work to bring the ball down. The potential tells you how steep the hill is at any point.
| Distance from Earth’s centre (m) | Gravitational Potential \$V\$ (J/kg) |
|---|---|
| \$6.371\times10^6\$ (surface) | \$-6.25\times10^7\$ |
| \$1.0\times10^7\$ (above surface) | \$-4.00\times10^7\$ |
| \$1.0\times10^8\$ (far away) | \$-2.50\times10^6\$ |
Gravitational potential lets us compare how “deep” different places are in a planet’s gravity field without worrying about the mass of the object we’re moving. It’s a key concept for understanding orbits, escape velocity, and the energy required for space missions. 🚀