Gravitational potential
Introduced| English | Meaning |
|---|---|
| escape velocity/eˈskeɪp vəˈlɒsɪti/ | The minimum initial speed needed to escape a body's gravitational field without further propulsion, for specified starting conditions. |
| gravitational potential/ˌɡrævɪˈteɪʃənl pəˈtenʃl/ | Potential energy per unit mass at a point in a gravitational field, relative to a specified reference. |
| infinity/ɪnˈfɪnɪti/ | An idealised limit at an unbounded distance, often used as a reference in field models. |
| gravitational potential energy/ˌɡrævɪˈteɪʃənl pəˈtenʃl ˈenədʒi/ | Energy stored because of an object’s position in a gravitational field. |
| tightly bound/ˈtaɪtli baʊnd/ | tightly bound |
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:
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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 ____.
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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.
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$E_{\text{P}} = -\dfrac{GMm}{r}$ — smaller $r$ gives a more negative value, i.e. a more tightly bound pair.
Match each term to the definition the examiner marks.
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Two of these are per unit mass and one is per unit mass of WORK. That single word is what separates potential from field strength.
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:
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It assumes a uniform field. For big changes in $r$, use $-\dfrac{GMm}{r}$ at each radius instead.
Why is gravitational potential always negative? Select all that apply.
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"Because gravity attracts" on its own is not the marked answer. The mark needs the work argument: zero at infinity, and the field doing the work on the way in.
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.)
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$\phi = -\dfrac{GM}{R} = -\dfrac{6.67 \times 10^{-11} \times 7.35 \times 10^{22}}{1.74 \times 10^{6}} = -2.8 \times 10^{6}\ \dfrac{\text{J}}{\text{kg}}$ — 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?
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$v_{\text{esc}} = \sqrt{\dfrac{2GM}{r}} \approx 11\ \dfrac{\text{km}}{\text{s}}$ for Earth.
Escape velocity depends on the mass of the escaping object.
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The object mass cancels: $v_{\text{esc}} = \sqrt{\dfrac{2GM}{r}}$ depends only on the planet and the distance.
A planet has GM/r = 6.3e7 J/kg at its surface. What is its escape speed, in km/s?
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v = sqrt(2GM/r) = sqrt(1.26e8) = 1.12e4 m/s = 11.2 km/s, the Earth's value. Note the factor of 2 and that the mass of the escaping object cancels out entirely.
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.
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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)