Gravitational fields · ⁨สนามโน้มถ่วง⁩

Definition

A gravitational field 重力场 is a region where a mass 质量 feels a force 力 from other masses. The gravitational field strength 重力场强度 $g$ at a point is the gravitational force per unit mass on a small test mass 检验质量 placed there:

Unit: $\text{N kg}^{-1}$ (the same as $\text{m s}^{-2}$ — the acceleration of free fall in the field). $g$ is a vector 矢量, pointing the way the force acts — towards the source mass.

The examiner's wording. A gravitational field is a region of space in which a mass experiences a force. Gravitational field strength at a point is the gravitational force per unit mass acting on a small test mass placed at the point. The direction of a field line at a point is the direction of the force on a (small test) mass placed there. All three are one-mark definitions; "force per unit mass" is the phrase that scores, "the force on 1 kg" is not.

Field lines

A gravitational field is drawn with field lines 场线 that point the way the force acts on a test mass:

• around a point mass 质点 or a uniform sphere (treated as a point mass from outside), the field lines are radial 径向, pointing inwards.
• near the Earth's surface over a small area, the field lines are nearly parallel and equally spaced, pointing straight down — a uniform field 匀强场.

Closer lines mean a stronger field.

Why $g$ is constant near the surface, in terms of field lines (a two-mark explanation): over a region whose size is small compared with the Earth's radius, the radial field lines are almost parallel and their spacing hardly changes with height, so the field strength — which the spacing represents — is almost the same at the top of the region as at the bottom. Over the whole planet the lines spread out with distance and the field falls.

Worked example. A point P represents a point mass. Draw lines to represent the gravitational field around P, and state what the direction of a line shows.

Straight radial lines, evenly spaced around P, with arrows pointing inwards towards P. The direction of a line is the direction of the force on a small test mass placed there — always towards the mass that produces the field, because gravity only attracts.

For two point masses $m_{1}, m_{2}$ a distance $r$ apart, the force on each is

pulling them together along the line joining them. This is Newton's law of gravitation 万有引力定律. The constant $G = 6.67 \times 10^{-11}\ \text{N m}^{2}\ \text{kg}^{-2}$ is the universal gravitational constant 万有引力常量.

In words, as the mark scheme wants it: the gravitational force between two point masses is proportional to the product of their masses and inversely proportional to the square of their separation — both proportionalities, and "point masses". Asked to "state the equation and the meaning of any other symbols", give $F = Gm_{1}m_{2}/r^{2}$ with $G$ the gravitational constant and $r$ the distance between the centres.

Worked example. Two isolated uniform spheres, each of mass $1.0 \times 10^{3}\ \text{kg}$, have their centres $2.0\ \text{m}$ apart. Find the gravitational force between them and comment on its size.

$F = \dfrac{6.67 \times 10^{-11} \times 1.0 \times 10^{3} \times 1.0 \times 10^{3}}{2.0^{2}} = 1.7 \times 10^{-5}\ \text{N}$. This is about $10^{-9}$ of each sphere's weight ($9.8 \times 10^{3}\ \text{N}$): gravity between everyday objects is negligible, and only a planet-sized mass produces a noticeable field. "Isolated" in a question means that no other masses need be considered.

Spheres treated as point masses

For a uniform sphere (such as a planet or star), the field at any point outside is the same as that of a point mass equal to the total mass at the centre. So from above the surface, you can treat the Earth as a point mass at its centre. (Points inside a sphere are different, and are not in the syllabus.)

Between two masses the fields subtract and the potentials add. Field strength is a vector: on the line joining two spheres the two fields point in opposite directions, so there is a point where they cancel and the resultant field is zero. Potential is a scalar: the two potentials simply add, and both are negative, so the potential is least negative (a maximum) at the point where the field is zero.

Worked example. Two identical isolated uniform spheres X and Y, each of mass $M$ and radius $R$, have their centres a distance $L$ apart. Point P lies on the line joining the centres. State where on the line the resultant field strength is zero, and find the gravitational potential there.

By symmetry the fields of X and Y are equal and opposite at the midpoint, $L/2$ from each centre, so the resultant field is zero there. The potential at P is the sum $\phi = -\dfrac{GM}{L/2} - \dfrac{GM}{L/2} = -\dfrac{4GM}{L}$. A mass released at P would stay there; a mass released anywhere else on the line falls towards the nearer sphere.

Field strength from a point mass

Put the gravitational force on a test mass $m$ at distance $r$ from a point mass $M$ into $g = F/m$:

So $g$ falls off as $1/r^{2}$ as you move away from the source.

The two-mark derivation. The force on a test mass $m$ at distance $r$ from a point mass $M$ is $F = GMm/r^{2}$ (Newton's law). Field strength is force per unit mass, $g = F/m$, so $g = GM/r^{2}$. Write both lines: the law and the definition are the two marks. Because $g \propto 1/r^{2}$, halving the distance makes the field four times stronger: at distance $x/2$ the field is $4g$, in the same direction, towards $M$.

Worked example. Find the gravitational field strength at the Earth's surface. (Earth's mass $M = 6.0 \times 10^{24}\ \text{kg}$, radius $R = 6.4 \times 10^{6}\ \text{m}$, $G = 6.67 \times 10^{-11}\ \text{N m}^{2}\ \text{kg}^{-2}$.)

Why $g$ is nearly constant near the Earth's surface

The Earth's radius is $R \approx 6.4 \times 10^{6}\ \text{m}$. Rising to height $h$ changes the distance from the centre from $R$ to $R + h$. For $h \ll R$ (any building or mountain), $(R + h)/R \approx 1$, so $g$ barely changes — going from $5\ \text{m}$ to $10\ \text{m}$ high changes $r$ by about one part in a million. In the laboratory, $g$ is effectively constant.

For a satellite 卫星 of mass $m$ in a circular orbit 轨道 of radius $r$ around a body of mass $M$, gravity provides the centripetal force 向心力:

Cancel $m$ (the orbital speed does not depend on the satellite's mass):

Worked example. A satellite orbits the Earth in a circular orbit of radius $r = 7.0 \times 10^{6}\ \text{m}$. Find its orbital speed. (For the Earth, $GM = 4.0 \times 10^{14}\ \text{m}^{3}\ \text{s}^{-2}$.)

The period 周期 follows from $T = 2\pi r / v$:

This is Kepler's third law 开普勒第三定律 for circular orbits: $T^{2} \propto r^{3}$. A plot of $T^{2}$ against $r^{3}$ is a straight line through the origin with gradient $4\pi^{2}/(GM)$, so orbital data gives the central mass.

The two-mark derivation, as the exam sets it. For a satellite of mass $m$ in a circular orbit of radius $R$ and period $T$ around a planet of mass $M$: the gravitational force provides the centripetal force, $\dfrac{GMm}{R^{2}} = mR\omega^{2}$, and $\omega = 2\pi/T$; so $\dfrac{GM}{R^{2}} = \dfrac{4\pi^{2}R}{T^{2}}$, giving $T^{2} = \dfrac{4\pi^{2}R^{3}}{GM}$. For an orbit at height $h$ above a planet of radius $R_{\text{p}}$ the orbital radius is $R_{\text{p}} + h$, so $T^{2} = 4\pi^{2}(R_{\text{p}} + h)^{3}/GM$: a graph of $T^{2}$ against $(R_{\text{p}} + h)^{3}$ is a straight line through the origin whose gradient gives $M$.

Worked example. Satellite X, of mass $M_{\text{s}}$, orbits a planet at a distance $4R$ from its centre; satellite Y, of mass $2M_{\text{s}}$, orbits at $3R$. Compare their speeds, periods and kinetic energies.

From $v = \sqrt{GM/r}$, $\dfrac{v_{\text{Y}}}{v_{\text{X}}} = \sqrt{\dfrac{4R}{3R}} = 1.15$: Y moves faster, and the satellite masses do not enter. From $T \propto r^{3/2}$, $\dfrac{T_{\text{Y}}}{T_{\text{X}}} = \left(\dfrac{3}{4}\right)^{3/2} = 0.65$. Kinetic energy $= \tfrac{1}{2}mv^{2} = \dfrac{GMm}{2r}$, so $\dfrac{E_{\text{Y}}}{E_{\text{X}}} = \dfrac{2M_{\text{s}}/3R}{M_{\text{s}}/4R} = \dfrac{8}{3}$. In ratio questions cancel $G$, $M$ and the satellite mass first; only the radii and masses that differ survive.

Binary stars. Two stars of masses $M_{\text{A}}$ and $M_{\text{B}}$ a distance $d$ apart orbit their common centre of mass 质心, always on opposite sides of it, with the same period. The centre of mass divides $d$ in the inverse ratio of the masses, $M_{\text{A}}r_{\text{A}} = M_{\text{B}}r_{\text{B}}$ with $r_{\text{A}} + r_{\text{B}} = d$, and the same gravitational force $GM_{\text{A}}M_{\text{B}}/d^{2}$ is the centripetal force on each star.

Worked example. A binary star consists of star A, mass $4.0 \times 10^{30}\ \text{kg}$, and star B, mass $2.0 \times 10^{30}\ \text{kg}$, with centres $3.3 \times 10^{11}\ \text{m}$ apart. Find the distance of A from the centre of mass and the period of the orbit.

$M_{\text{A}}r_{\text{A}} = M_{\text{B}}r_{\text{B}}$ with $r_{\text{A}} + r_{\text{B}} = 3.3 \times 10^{11}$ gives $r_{\text{A}} = \dfrac{2.0}{6.0} \times 3.3 \times 10^{11} = 1.1 \times 10^{11}\ \text{m}$ (and $r_{\text{B}} = 2.2 \times 10^{11}\ \text{m}$). For star A: $\dfrac{GM_{\text{A}}M_{\text{B}}}{d^{2}} = M_{\text{A}}r_{\text{A}}\omega^{2}$, so $\omega^{2} = \dfrac{GM_{\text{B}}}{d^{2}r_{\text{A}}} = \dfrac{6.67 \times 10^{-11} \times 2.0 \times 10^{30}}{(3.3 \times 10^{11})^{2} \times 1.1 \times 10^{11}} = 1.1 \times 10^{-14}\ \text{s}^{-2}$, $\omega = 1.06 \times 10^{-7}\ \text{rad s}^{-1}$ and $T = 2\pi/\omega = 6.0 \times 10^{7}\ \text{s}$, about $1.9$ years. Using $d$ in the force but $r_{\text{A}}$ in the centripetal term is the whole point of the question.

Geostationary orbit

A geostationary 地球同步 satellite:

• stays directly above the same point on the Earth (so a fixed dish always points at it),
• has a period of 24 hours (the same as the Earth's rotation),
• orbits west to east (the same way the Earth turns),
• must be directly above the equator 赤道.

It must have the same angular speed 角速度 as the Earth, in the same direction, in the equatorial plane (or it would drift north–south during the day). From $T = 24\ \text{h}$ and $T^{2} = 4\pi^{2} r^{3}/(GM)$, the radius is $r \approx 4.2 \times 10^{7}\ \text{m}$ (about $3.6 \times 10^{7}\ \text{m}$ above the surface).

Worked example. Calculate the radius of a geostationary orbit around the Earth ($M = 5.98 \times 10^{24}\ \text{kg}$), and explain why a satellite with the same orbital radius and period may still not be geostationary.

$T = 24 \times 3600 = 8.64 \times 10^{4}\ \text{s}$; $r^{3} = \dfrac{GMT^{2}}{4\pi^{2}} = \dfrac{6.67 \times 10^{-11} \times 5.98 \times 10^{24} \times (8.64 \times 10^{4})^{2}}{4\pi^{2}} = 7.5 \times 10^{22}\ \text{m}^{3}$, so $r = 4.2 \times 10^{7}\ \text{m}$ — a height of about $3.6 \times 10^{7}\ \text{m}$ above the surface. A satellite with this period is geostationary only if its orbit is in the plane of the equator and it travels from west to east; with the same period in a tilted orbit (over the poles, say) it returns to the same point each day but is not always above it. Mars turns once in about $25$ hours, so a satellite that stays above one point on Mars has a $25$-hour period and the same two other features.

Gravitational potential 引力势 $\phi$ at a point is the work done per unit mass in bringing a small test mass from infinity 无穷远 to that point:

Unit: $\text{J kg}^{-1}$.

The potential is taken as zero at infinity. As the test mass falls in towards the source, gravity does the work for you, so $\phi$ is negative everywhere except at infinity. For a point mass $M$ at distance $r$:

$\phi$ is a scalar 标量. For several masses, add the potentials.

The two-mark definition, and why it is negative. Gravitational potential at a point is the work done per unit mass in bringing a small test mass from infinity to the point. Potential is defined as zero at infinity; gravity is attractive, so as the test mass comes in from infinity the field does work on it and the mass gives out energy — the work that an external agent must do is negative. Hence the potential at every finite distance is below zero. (Electric potential near a positive charge is positive for the opposite reason: that field repels, so work must be done to bring a positive test charge in.)

Similarities and differences with electric potential. Both are defined as work done per unit mass or per unit positive charge from infinity, both are scalars, both vary as $1/r$ from a point source and both are zero at infinity. The difference: gravitational potential is always negative, because gravity only attracts, whereas electric potential can be positive or negative depending on the sign of the charge.

Reading a potential graph. Two graphs appear in questions. On $\phi$ against $r$ the curve is $-GM/r$: the surface value $\phi_{\text{s}} = -GM/R$ gives the mass of the planet, and the gradient at any point gives the field strength there, since $g = -\dfrac{\Delta\phi}{\Delta r}$ — field strength is the negative of the potential gradient (this is the "relationship between potential and field strength" a question may ask for). On $\phi$ against $1/r$ the graph is a straight line through the origin with gradient $-GM$, which is the neater way to extract $M$.

Worked example. The Moon is an isolated uniform sphere of mass $7.3 \times 10^{22}\ \text{kg}$ and radius $1.7 \times 10^{6}\ \text{m}$. Calculate the gravitational potential at its surface, and the minimum speed with which a particle must leave the surface to escape.

$\phi = -\dfrac{GM}{R} = -\dfrac{6.67 \times 10^{-11} \times 7.3 \times 10^{22}}{1.7 \times 10^{6}} = -2.9 \times 10^{6}\ \text{J kg}^{-1}$. To escape, the particle's kinetic energy per unit mass must equal the depth of the potential well: $\tfrac{1}{2}v^{2} = 2.9 \times 10^{6}$, so $v = \sqrt{2 \times 2.9 \times 10^{6}} = 2.4 \times 10^{3}\ \text{m s}^{-1}$. The minus sign carries the meaning: energy of $2.9\ \text{MJ}$ per kilogram must be supplied to lift the particle out.

Gravitational potential energy of two point masses

If a test mass $m$ sits where the potential is $\phi$, the gravitational potential energy 重力势能 of the pair is

Like the potential, $E_{\text{P}}$ is negative and reaches zero only at infinite separation. Closer masses have more negative potential energy (more tightly bound).

Link with $\Delta E_{\text{P}} = mg\Delta h$

For small height changes near the surface, $r$ barely changes, so $\Delta E_{\text{P}} \approx mg\Delta h$. For large changes (a satellite moving to a higher orbit) use $-GMm/r$ at each radius and take the difference:

which is positive (energy must be supplied to raise the satellite).

Worked example. A satellite of mass $1200\ \text{kg}$ is moved from a circular orbit of radius $7.0 \times 10^{6}\ \text{m}$ to one of radius $8.0 \times 10^{6}\ \text{m}$ around the Earth ($GM = 4.0 \times 10^{14}\ \text{m}^{3}\ \text{s}^{-2}$). Find the changes in its gravitational potential energy, its kinetic energy and its total energy.

$\Delta E_{\text{P}} = GMm\left(\dfrac{1}{r_{1}} - \dfrac{1}{r_{2}}\right) = 4.0 \times 10^{14} \times 1200 \times \left(\dfrac{1}{7.0 \times 10^{6}} - \dfrac{1}{8.0 \times 10^{6}}\right) = +8.6 \times 10^{9}\ \text{J}$. In a circular orbit $E_{\text{K}} = \tfrac{1}{2}mv^{2} = \dfrac{GMm}{2r}$, so $\Delta E_{\text{K}} = \dfrac{GMm}{2}\left(\dfrac{1}{r_{2}} - \dfrac{1}{r_{1}}\right) = -4.3 \times 10^{9}\ \text{J}$: the higher satellite moves more slowly. The total energy $E_{\text{K}} + E_{\text{P}} = -\dfrac{GMm}{2r}$ rises by $+4.3 \times 10^{9}\ \text{J}$, which is the energy the rocket motor must supply. A satellite's total energy is negative — it is bound — and it becomes less negative as the orbit widens.

Escape velocity (from conservation of energy)

To escape from radius $r$ to infinity, an object's kinetic energy 动能 must equal the size of its gravitational potential energy:

At the Earth's surface, the escape velocity 逃逸速度 is $\approx 11\ \text{km s}^{-1}$. It does not depend on the object's mass.

Worked example. Find the escape velocity from the Earth's surface. (For the Earth, $GM = 4.0 \times 10^{14}\ \text{m}^{3}\ \text{s}^{-2}$, $R = 6.4 \times 10^{6}\ \text{m}$.)

Worked example. A particle is projected vertically upwards from the surface of an isolated planet of radius $R$, where the gravitational potential is $\phi_{\text{s}}$. Show that it just reaches a distance $r$ from the centre if its launch speed $v$ satisfies $\tfrac{1}{2}v^{2} = -\phi_{\text{s}}\left(1 - \dfrac{R}{r}\right)$, and deduce the escape speed.

Potential varies as $1/r$, so at distance $r$ it is $\phi_{\text{s}}R/r$. When the particle stops, its kinetic energy per unit mass has all become potential energy per unit mass: $\tfrac{1}{2}v^{2} = \phi(r) - \phi_{\text{s}} = \phi_{\text{s}}\dfrac{R}{r} - \phi_{\text{s}} = -\phi_{\text{s}}\left(1 - \dfrac{R}{r}\right)$, which is positive because $\phi_{\text{s}}$ is negative. Letting $r \to \infty$ gives $\tfrac{1}{2}v_{\text{esc}}^{2} = -\phi_{\text{s}} = GM/R$ — the escape-speed result again, now read straight off the potential.

A definition question is marked against fixed wording. Learn these exactly, and give one answer only.

• Newton's law of gravitation $F = GMm/r^2$ (inverse-square); field strength $g = GM/r^2$ — and quote the derivation as two lines, the law and the definition of $g$.
• Distinguish gravitational potential ($\phi = -GM/r$, always negative, zero at infinity) from field strength; $g$ is the negative gradient of the $\phi$–$r$ graph.
• For an orbit set gravity $=$ centripetal force to get $T^2 \propto r^3$; a geostationary orbit has $T = 24\ \text{h}$ plus two more features: above the equator, west to east.
• Use the orbital radius (centre to centre, $R_{\text{p}} + h$), never the height, and for a binary star the separation $d$ in the force but the orbit radius $r_{\text{A}}$ in the centripetal term.
• Energy changes come from $-GMm/r$ at each radius; for a circular orbit $E_{\text{K}} = GMm/2r$ and the total energy is $-GMm/2r$.

Common mistakes

• Defining potential as "force per unit mass" or field strength as "work done per unit mass". Swap them and both marks go.
• Saying the potential is negative "because gravity attracts" and stopping there. The mark needs the work argument: zero at infinity, and the field does the work as the mass comes in.
• Using the height above the surface as $r$ in $GM/r^{2}$ or $T^{2} \propto r^{3}$. Add the planet's radius first.
• Writing Kepler's law as $T \propto r$ or $T^{2} \propto r^{2}$. It is $T^{2} \propto r^{3}$: a straight line only on $T^{2}$ against $r^{3}$.
• Making "geostationary" mean only "24-hour period". Without the equatorial plane and the west-to-east direction it is not geostationary.
• Forgetting that a satellite's kinetic energy falls when it is raised to a higher orbit, even though energy had to be supplied.