Gravitational potential
| English | Chinese | Pinyin |
|---|---|---|
| escape velocity | 逃逸速度 | táo yì sù dù |
| gravitational potential | 引力势 | yǐn lì shì |
| infinity | 无穷远 | wú qióng yuǎn |
| gravitational potential energy | 重力势能 | zhòng lì shì néng |
| tightly bound | 紧密束缚 | jǐn mì shù fù |
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 逃逸速度.
- To find it we need gravitational potential 引力势.
Gravitational potential
- $\phi$ is the work done per unit mass to bring a small test mass from infinity 无穷远 to a point: $\phi = \dfrac{W}{m}$. Unit: $\dfrac{\text{J}}{\text{kg}}$.
- Taken as zero at infinity, so $\phi = -\dfrac{GM}{r}$ is negative everywhere else.
Gravitational potential
V = −GM / r
Potential ∝ −1/r — deep near the mass, flattening with distance.
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 重力势能
- 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 紧密束缚.
Gravity provides the centripetal force that keeps a planet in a circular orbit
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.
Worked example: the potential at the surface of Mars
Mars has mass $6.4 \times 10^{23}\ \text{kg}$ and radius $3.4 \times 10^{6}\ \text{m}$. Find the gravitational potential at its surface, and the energy needed to remove $1\ \text{kg}$ from the surface to infinity.
- Potential: $\phi = -\dfrac{GM}{R} = -\dfrac{6.67 \times 10^{-11} \times 6.4 \times 10^{23}}{3.4 \times 10^{6}} = -1.3 \times 10^{7}\ \dfrac{\text{J}}{\text{kg}}$.
- The unit is asked for: joules per kilogram, because potential is energy per unit mass.
- Energy to escape per kilogram: from $\phi$ up to zero at infinity is $+1.3 \times 10^{7}\ \text{J}$ for each kilogram.
- Check: the minus sign is not optional. It says work must be done on the mass to remove it — the mass is in a "well" $1.3 \times 10^{7}\ \dfrac{\text{J}}{\text{kg}}$ deep.
The Moon has mass $7.35 \times 10^{22}\ \text{kg}$ and radius $1.74 \times 10^{6}\ \text{m}$. What is the gravitational potential at its surface, as a multiple of $10^{6}\ \dfrac{\text{J}}{\text{kg}}$? (Include the sign.)
$\phi = -\dfrac{GM}{R} = -\dfrac{6.67 \times 10^{-11} \times 7.35 \times 10^{22}}{1.74 \times 10^{6}} = -2.8 \times 10^{6}\ \dfrac{\text{J}}{\text{kg}}$ — about a twentieth of the Earth's, which is why leaving the Moon is easy.
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.
Worked example: lifting a satellite
A $500\ \text{kg}$ satellite is raised from the Earth's surface ($R = 6.37 \times 10^{6}\ \text{m}$) to an orbit of radius $7.0 \times 10^{6}\ \text{m}$. Earth's mass is $5.97 \times 10^{24}\ \text{kg}$. Find the gain in potential energy, and compare it with $mg\Delta h$.
- Two potentials, subtract: $\Delta E_{\text{P}} = GMm\left(\dfrac{1}{R} - \dfrac{1}{r}\right)$.
- Substitute: $6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 500 \times \left(\dfrac{1}{6.37 \times 10^{6}} - \dfrac{1}{7.0 \times 10^{6}}\right) = 2.8 \times 10^{9}\ \text{J}$.
- The rough way: $mg\Delta h = 500 \times 9.81 \times 6.3 \times 10^{5} = 3.1 \times 10^{9}\ \text{J}$ — about $10\%$ too big, because $g$ falls with height and $mg\Delta h$ assumes it does not.
- Check: the gain is positive — the satellite ends less negative, less tightly bound — even though both potentials are negative numbers.
Potential is negative and increases towards zero as you move away. "Energy gained" is final minus initial with the signs kept: $-1.3 \times 10^{7} \to -0.9 \times 10^{7}$ is a gain of $0.4 \times 10^{7}$ per kilogram. $mg\Delta h$ is only for $\Delta h \ll R$. And the unit of $\phi$ is $\dfrac{\text{J}}{\text{kg}}$ — writing joules loses the mark.
Which statements about gravitational potential are correct? Select all that apply.
Potential is energy per unit mass, so its unit is joules per kilogram. It is negative, rises towards zero with distance, and its gradient gives the field: $g = -\dfrac{\Delta\phi}{\Delta r}$.
Field strength from the potential graph
- The field strength is the gradient of the potential–distance graph: $g = -\dfrac{\Delta\phi}{\Delta r}$.
- Where the $\phi$ graph is steep (close to the mass) the field is strong; far away the graph flattens and the field fades.
- The minus sign says the field points down the potential slope — towards the mass, where $\phi$ is most negative.
You've got it
- gravitational potential $\phi = -\dfrac{GM}{r}$ in $\dfrac{\text{J}}{\text{kg}}$ (negative, zero at infinity); $g = -\dfrac{\Delta\phi}{\Delta r}$
- gravitational PE $E_{\text{P}} = -\dfrac{GMm}{r}$ — more negative = more tightly bound; large changes use $GMm\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right)$
- escape velocity $v_{\text{esc}} = \sqrt{\dfrac{2GM}{r}}$ (mass-independent, $\approx 11\ \dfrac{\text{km}}{\text{s}}$ for Earth)