Gravitational potential
How fast to escape forever?
- Throw a ball up and it comes back. Throw it fast enough and it never returns.
- That "never returns" speed is the escape velocity.
- To find it we need gravitational potential.
Gravitational potential
- $\phi$ is the work done per unit mass to bring a test mass from infinity to a point: $\phi = \dfrac{W}{m}$.
- Taken as zero at infinity, so $\phi = -\dfrac{GM}{r}$ is negative everywhere else.
Gravitational potential is:
Gravity does the work as a mass falls in, so $\phi = -\dfrac{GM}{r}$ is negative, reaching zero only at infinity.
Gravitational potential is taken to be zero at ____.
That choice makes the potential negative everywhere a mass actually is.
Gravitational potential energy
- A mass $m$ at potential $\phi$ has $E_{\text{P}} = m\phi = -\dfrac{GMm}{r}$.
- It is negative; closer masses are more negative — more tightly bound.
Two masses closer together have a more negative (lower) gravitational potential energy.
$E_{\text{P}} = -\dfrac{GMm}{r}$ — smaller $r$ gives a more negative value, i.e. a more tightly bound pair.
Link with $mg\Delta h$
- For small height changes near the surface, $r$ barely changes, so $\Delta E_{\text{P}} \approx mg\Delta h$.
- For large changes (raising a satellite), use $-\dfrac{GMm}{r}$ at each radius and subtract.
The simple formula $\Delta E_{\text{P}} = mg\Delta h$ is valid when:
It assumes a uniform field. For big changes in $r$, use $-\dfrac{GMm}{r}$ at each radius instead.
Escape velocity
- To reach infinity, kinetic energy must match the depth of the well: $\tfrac{1}{2}mv_{\text{esc}}^{2} = \dfrac{GMm}{r}$.
- $v_{\text{esc}} = \sqrt{\dfrac{2GM}{r}} \approx 11\ \dfrac{\text{km}}{\text{s}}$ at Earth's surface — independent of the object's mass.
Roughly, what is the escape velocity from the Earth's surface, in km/s?
$v_{\text{esc}} = \sqrt{\dfrac{2GM}{r}} \approx 11\ \dfrac{\text{km}}{\text{s}}$ for Earth.
Escape velocity depends on the mass of the escaping object.
The object mass cancels: $v_{\text{esc}} = \sqrt{\dfrac{2GM}{r}}$ depends only on the planet and the distance.
You've got it
- gravitational potential $\phi = -\dfrac{GM}{r}$ (negative, zero at infinity)
- gravitational PE $E_{\text{P}} = -\dfrac{GMm}{r}$ — more negative = more tightly bound
- escape velocity $v_{\text{esc}} = \sqrt{\dfrac{2GM}{r}}$ (mass-independent, $\approx 11\ \dfrac{\text{km}}{\text{s}}$ for Earth)