Gravitational Fields
Every mass in the universe pulls on every other mass. That pull keeps your feet on the floor, the Moon going round the Earth and the Earth going round the Sun. In this topic we describe gravity as a field, work out how strong it is at any distance from a planet, meet gravitational potential, and then use Newton's laws to explain how satellites, moons and planets orbit. Electric fields follow almost the same rules, so we finish by comparing the two.
Gravitational fields and field strength
Nothing has to touch you for the Earth to pull on you: a skydiver speeds up with nothing attached to them. Physicists describe this "action at a distance" with the idea of a field. Every mass fills the space around it with a gravitational field, and any other mass placed in that field feels a force.
Gravitational field and field strength
A gravitational field is a region in which a mass experiences a force.
The gravitational field strength \(g\) at a point is the gravitational force per unit mass acting on a small test mass placed at that point.
Gravitational field strength
Field strength is a vector. It points the same way as the force on a mass, which is always towards the mass that creates the field. We imagine a small test mass so that its own field doesn't disturb the field we are trying to measure.
Field strength is the acceleration of free fall
The units N kg⁻¹ and m s⁻² are the same thing (1 N = 1 kg m s⁻²). Newton's second law says \(a = F/m\), which is exactly \(g = F/m\). So "the field strength at the Earth's surface is 9.81 N kg⁻¹" and "objects fall with an acceleration of 9.81 m s⁻²" say the same thing. Every object at the same place gets the same acceleration, whatever its mass, because doubling \(m\) doubles the force too.
When you draw gravitational field lines, always put arrows on them pointing towards the mass. Lines never cross, and they meet the surface of a spherical planet at right angles, heading for its centre.
Newton's law of universal gravitation
Newton's big idea (1687) was that the force that makes an apple fall is the same force that holds the Moon in its orbit. His law says that any two point masses attract each other with a force that is
- proportional to the product of their masses: double either mass and the force doubles;
- inversely proportional to the square of the distance between their centres: double the distance and the force drops to a quarter.
Newton's law of universal gravitation
Newton's law of gravitation
Two point masses attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
Why the square of the distance?
Picture the field lines leaving a planet. At a distance \(r\) they are spread over the surface of a sphere of area \(4\pi r^2\). Double the distance and the same number of lines is shared over four times the area, so the field, and the force on a mass, is a quarter as strong. Treble it and you get one ninth. Anything that spreads out evenly in three dimensions without being absorbed follows an inverse-square law. You will meet the same \(4\pi r^2\) in electric fields and in the intensity of light from stars.
\(G\) is tiny, which is why you never notice the pull between everyday objects. Two 70 kg people standing 1 m apart attract each other with about \(3\times10^{-7}\) N, roughly the weight of a grain of sand. Gravity only matters when at least one of the masses is enormous, like a planet or a star.
The force obeys Newton's third law: the Earth pulls the Moon, and the Moon pulls the Earth with a force of exactly the same size in the opposite direction. The effects are very different only because the masses are.
\(r\) is measured between the centres of the masses. For a satellite 400 km above the Earth, \(r = 6.37\times10^6 + 4.0\times10^5 = 6.77\times10^6\ \text{m}\), not \(4.0\times10^5\) m. And don't forget to square it.
The Earth (mass \(5.97\times10^{24}\) kg) and the Moon (mass \(7.35\times10^{22}\) kg) are \(3.84\times10^8\) m apart. Calculate the gravitational force between them and the acceleration each one has because of it.
Show solution
\[ F = \frac{Gm_1m_2}{r^2} = \frac{6.67\times10^{-11}\times5.97\times10^{24}\times7.35\times10^{22}}{(3.84\times10^8)^2} = 1.98\times10^{20}\ \text{N} \]The same size of force acts on each body (Newton's third law), so use \(a = F/m\) for each:
\[ a_\text{Moon} = \frac{1.98\times10^{20}}{7.35\times10^{22}} = 2.70\times10^{-3}\ \text{m s}^{-2}, \qquad a_\text{Earth} = \frac{1.98\times10^{20}}{5.97\times10^{24}} = 3.32\times10^{-5}\ \text{m s}^{-2} \]The Moon's acceleration is the centripetal acceleration that keeps it in orbit. The Earth's is about 80 times smaller: it makes the Earth wobble slightly as the Moon goes round.
The field around a point mass: g = Gm/r²
Now combine the two equations you have met so far. This is a derivation the specification says you must be able to do, and it only takes three lines.
Place a small test mass \(m\) a distance \(r\) from a point mass \(M\). Newton's law of gravitation gives the force on it:
\[ F = \frac{GMm}{r^2} \]Gravitational field strength is defined as force per unit mass, so divide by \(m\):
\[ g = \frac{F}{m} = \frac{GMm}{r^2}\cdot\frac{1}{m} \]The test mass cancels:
\[ \boxed{g = \frac{GM}{r^2}} \]The field depends only on the mass that creates it and how far away you are, not on what you put there. That is why a hammer and a feather dropped together on the Moon land together.
The field points towards \(M\). Outside a spherical planet or star, the field is the same as if all its mass were concentrated at its centre, so the equation works for any \(r\) from the surface outwards, with \(r\) measured from the centre.
Newton proved this "shell theorem" with some clever geometry. The proof is not in the spec; you just need to know you can treat a sphere as a point mass.
Radial gravitational field
Because \(g \propto 1/r^2\), going from the surface (\(r = R\)) to \(r = 2R\) divides \(g\) by 4, and \(r = 3R\) divides it by 9. One Earth radius above the ground, \(g\) is only about 2.45 N kg⁻¹.
The Earth's radius is \(6.37\times10^6\) m and \(g\) at its surface is 9.81 N kg⁻¹. (a) Calculate the mass of the Earth. (b) The International Space Station orbits 408 km above the surface. Calculate \(g\) at that height.
Show solution
(a) Rearrange \(g = GM/R^2\):
\[ M = \frac{gR^2}{G} = \frac{9.81\times(6.37\times10^6)^2}{6.67\times10^{-11}} = 5.97\times10^{24}\ \text{kg} \](b) Measure from the centre: \(r = 6.37\times10^6 + 0.408\times10^6 = 6.778\times10^6\ \text{m}\).
\[ g = \frac{GM}{r^2} = \frac{6.67\times10^{-11}\times5.97\times10^{24}}{(6.778\times10^6)^2} = 8.67\ \text{N kg}^{-1} \]That is still 88% of the value on the ground. Gravity has certainly not "switched off" up there, which matters when we explain why astronauts float.
Two masses: where do the fields cancel?
Field strengths are vectors, so where two fields overlap you add them as vectors. On the line between the Earth and the Moon the two pulls point in opposite directions. Close to the Earth its pull wins easily, but somewhere near the Moon the two are equal and opposite and the resultant field is zero. This is the neutral point.
Using the data from the first worked example, find how far from the centre of the Earth the resultant gravitational field on the Earth–Moon line is zero.
Show solution
Let the point be a distance \(x\) from the Earth, so \(d - x\) from the Moon, with \(d = 3.84\times10^8\ \text{m}\). The two field strengths are equal there:
\[ \frac{GM_E}{x^2} = \frac{GM_M}{(d-x)^2} \quad\Rightarrow\quad \frac{d-x}{x} = \sqrt{\frac{M_M}{M_E}} = \sqrt{\frac{7.35\times10^{22}}{5.97\times10^{24}}} = 0.111 \]Taking the square root of both sides keeps the algebra linear. Use the positive root: the point is between the two bodies.
\[ d = 1.111\,x \quad\Rightarrow\quad x = \frac{3.84\times10^8}{1.111} = 3.46\times10^8\ \text{m} \]That is 90% of the way to the Moon, about \(3.8\times10^7\) m from the Moon's centre. \(G\) cancelled, so you didn't need its value.
Gravitational potential
Lift a book and you do work against gravity; let go and the field does work on it as it falls. Gravitational potential tells you how much work per kilogram is involved in getting to each point in a field. It works like height on a map: the field pulls masses "downhill" towards lower potential.
Gravitational potential
The gravitational potential at a point is the work done per unit mass in bringing a small test mass from infinity to that point. It is a scalar, measured in J kg⁻¹.
Gravitational potential in a radial field
Why is the potential negative?
We choose the potential to be zero at infinity, far from every mass. Gravity is attractive, so as a mass falls in from infinity the field does the work; to bring it in slowly you would have to hold it back, which means you do negative work. So at every finite distance \(V < 0\), and the closer you get to the mass the more negative it becomes. A mass near a planet sits at the bottom of a "potential well", and you must supply energy to lift it out.
Notice that \(V\) goes as \(1/r\), not \(1/r^2\): the potential fades away more slowly than the field strength. At the Earth's surface \(V = -6.25\times10^7\ \text{J kg}^{-1}\) and at twice that distance it is half as big, while \(g\) has dropped to a quarter.
Potential difference and potential energy
To move a mass \(m\) between two points you must do work equal to the mass times the change in potential: \(\Delta W = m\,\Delta V\). It follows that a mass \(m\) at a point where the potential is \(V_\text{grav}\) has gravitational potential energy
\[ E_\text{grav} = mV_\text{grav} = -\frac{GMm}{r} \]This is also negative, and zero at infinity. Close to the Earth, over a height change that is small compared with the radius, \(\Delta V \approx g\,\Delta h\), which gives back the AS result \(\Delta E = mg\Delta h\). Over large distances \(g\) changes, so you must use \(\Delta V\) instead.
A 1000 kg satellite is to be placed in a circular orbit 400 km above the Earth (radius \(6.37\times10^6\) m, mass \(5.97\times10^{24}\) kg). (a) Calculate the increase in its gravitational potential energy. (b) Compare your answer with \(mg\Delta h\). (c) Calculate the kinetic energy it needs in orbit.
Show solution
(a) \(GM = 6.67\times10^{-11}\times5.97\times10^{24} = 3.982\times10^{14}\ \text{N m}^2\,\text{kg}^{-1}\). Find the potential at each radius:
\[ V_1 = -\frac{3.982\times10^{14}}{6.37\times10^6} = -6.251\times10^7\ \text{J kg}^{-1}, \qquad V_2 = -\frac{3.982\times10^{14}}{6.77\times10^6} = -5.882\times10^7\ \text{J kg}^{-1} \] \[ \Delta E = m\,\Delta V = 1000\times(-5.882\times10^7 - (-6.251\times10^7)) = 3.69\times10^9\ \text{J} \](b) \(mg\Delta h = 1000\times9.81\times4.0\times10^5 = 3.92\times10^9\ \text{J}\). This is about 6% too big, because \(g\) gets smaller as the satellite rises (it is 8.69 N kg⁻¹ at 400 km).
(c) In a circular orbit \(v^2 = GM/r\) (see the orbits section below), so
\[ E_k = \tfrac{1}{2}mv^2 = \frac{GMm}{2r} = \frac{3.98\times10^{14}\times1000}{2\times6.77\times10^6} = 2.94\times10^{10}\ \text{J} \]Eight times the potential energy. Getting up to orbit is the easy part; getting fast enough to stay there is what needs the huge rockets.
Equipotentials and field lines
An equipotential is a surface (a line on a flat diagram) on which the potential is the same everywhere. Around a single planet they are spheres. Three things to know:
- No work is done moving a mass along an equipotential, because \(\Delta V = 0\).
- Equipotentials are always perpendicular to field lines. If they weren't, the field would have a component along the equipotential and would do work there.
- If you draw them at equal steps of \(\Delta V\), they crowd together where the field is strong, like contour lines on a steep hill.
Field strength is the potential gradient
The field strength tells you how steeply the potential changes with distance:
\[ g = -\frac{\Delta V}{\Delta r} \]On a graph of \(V\) against \(r\), the gradient at any point equals the size of \(g\) there. The minus sign says the field points the way the potential decreases, which is towards the mass, just as a ball rolls downhill. You can check the units: J kg⁻¹ m⁻¹ = N m kg⁻¹ m⁻¹ = N kg⁻¹. The same link between field and potential holds for electric fields.
Orbital motion
Newton imagined a cannon on top of a very high mountain, firing balls horizontally. A slow ball curves down and hits the ground nearby. A faster one lands further away. Fire it fast enough and, as it falls, the curved Earth drops away beneath it at the same rate: it keeps falling but never lands. That is an orbit. An orbiting satellite is in continuous free fall.
Newton's three laws in an orbit
- First law: without a force the satellite would fly off in a straight line along the tangent. Something must keep pulling it round.
- Second law: that something is gravity, the only force acting. It is the resultant force, so it causes an acceleration towards the centre: it provides the centripetal force \(mv^2/r\) (see Circular Motion).
- Third law: the satellite pulls the Earth with an equal and opposite force. Because the Earth is so massive, its acceleration is negligible.
For a satellite of mass \(m\) in a circular orbit of radius \(r\) round a planet of mass \(M\), gravity provides the centripetal force:
\[ \frac{GMm}{r^2} = \frac{mv^2}{r} \]Cancel \(m\) and one factor of \(r\):
\[ v^2 = \frac{GM}{r} \]Take the square root:
\[ \boxed{v = \sqrt{\frac{GM}{r}}} \]The satellite's mass has cancelled, so a spanner and a space station in the same orbit move at the same speed. A bigger \(r\) means a smaller \(v\): higher orbits are slower.
Why do astronauts float?
Not because there is no gravity: \(g\) at the ISS is 8.7 N kg⁻¹. The astronauts and the station are falling together with the same acceleration, so the floor doesn't need to push on them, and they feel weightless. Examiners like the phrase "in free fall, with the same acceleration as the spacecraft".
Period and radius: T² ∝ r³
A satellite goes once round its orbit, a distance \(2\pi r\), in one period \(T\), so \(v = 2\pi r/T\). Put that into the orbital speed equation and you get a relationship between the period and the radius of any orbit.
Start from \(v^2 = GM/r\) and substitute \(v = 2\pi r/T\):
\[ \frac{4\pi^2 r^2}{T^2} = \frac{GM}{r} \]Rearrange to make \(T^2\) the subject:
\[ \boxed{T^2 = \frac{4\pi^2}{GM}\,r^3} \]Kepler discovered \(T^2 \propto r^3\) from planetary observations in 1619 ("Kepler's third law"), almost 70 years before Newton's law of gravitation explained it.
So for everything orbiting the same central body, a graph of \(T^2\) against \(r^3\) is a straight line through the origin with
\[ \text{gradient} = \frac{4\pi^2}{GM} \quad\Rightarrow\quad M = \frac{4\pi^2}{G\times\text{gradient}} \]This is how astronomers "weigh" planets and stars: time how long a moon takes to go round and measure the size of its orbit. You don't need to know anything about the moon itself.
Io orbits Jupiter with a period of 1.77 days at a radius of \(4.22\times10^8\) m. Calculate the mass of Jupiter.
Show solution
Convert the period to seconds: \(T = 1.77\times24\times3600 = 1.529\times10^5\ \text{s}\). Rearrange \(T^2 = 4\pi^2r^3/GM\):
\[ M = \frac{4\pi^2 r^3}{GT^2} = \frac{4\pi^2\times(4.22\times10^8)^3}{6.67\times10^{-11}\times(1.529\times10^5)^2} = 1.90\times10^{27}\ \text{kg} \]About 320 times the mass of the Earth.
Geostationary orbits
A geostationary satellite stays above the same point on the Earth's surface. For that it must
- have a period of 24 hours, the same as the Earth's rotation;
- orbit above the equator;
- move in the same direction as the Earth rotates (west to east).
Because \(T\) is fixed, \(T^2 \propto r^3\) fixes the radius too: there is only one geostationary orbit. Communications and TV satellites use it, so a dish on your roof can point at one spot in the sky and never move. Low orbits like the ISS (about 90 minutes per orbit) are better for photographing the ground in detail, and GPS satellites take about 12 hours.
Calculate the height above the Earth's surface of a geostationary orbit, and the speed of a satellite in it. Mass of Earth \(= 5.97\times10^{24}\) kg, radius of Earth \(= 6.37\times10^6\) m.
Show solution
\(T = 24\times3600 = 86\,400\ \text{s}\). Rearrange \(T^2 = 4\pi^2 r^3/GM\) for \(r\):
\[ r^3 = \frac{GMT^2}{4\pi^2} = \frac{6.67\times10^{-11}\times5.97\times10^{24}\times(86\,400)^2}{4\pi^2} = 7.53\times10^{22}\ \text{m}^3 \] \[ r = \sqrt[3]{7.53\times10^{22}} = 4.22\times10^7\ \text{m} \]This is measured from the centre, so subtract the Earth's radius:
\[ h = 4.22\times10^7 - 6.37\times10^6 = 3.59\times10^7\ \text{m} \quad(\text{about } 36\,000\ \text{km}) \] \[ v = \frac{2\pi r}{T} = \frac{2\pi\times4.22\times10^7}{86\,400} = 3.07\times10^3\ \text{m s}^{-1} \]Two classic slips in orbit questions: leaving \(T\) in hours or days (it must be in seconds), and giving \(r\) as the answer when the question asks for the height above the surface.
Comparing gravitational and electric fields
Gravitational fields act on mass; electric fields act on charge. Apart from that, the maths is almost identical, so learning one helps you with the other (see Electric Fields).
| Gravitational field | Electric field | |
|---|---|---|
| Acts on | mass | charge |
| Field strength | \(g = F/m\), N kg⁻¹ | \(E = F/Q\), N C⁻¹ |
| Force between two point objects | \(F = \dfrac{Gm_1m_2}{r^2}\) | \(F = \dfrac{Q_1Q_2}{4\pi\varepsilon_0 r^2}\) |
| Field of a point object | \(g = \dfrac{Gm}{r^2}\) | \(E = \dfrac{Q}{4\pi\varepsilon_0 r^2}\) |
| Potential in a radial field | \(V = -\dfrac{Gm}{r}\): always negative | \(V = \dfrac{Q}{4\pi\varepsilon_0 r}\): positive or negative |
| Attract or repel? | always attracts | like charges repel, unlike attract |
| Constant | \(G = 6.67\times10^{-11}\ \text{N m}^2\,\text{kg}^{-2}\) | \(\dfrac{1}{4\pi\varepsilon_0} = 8.99\times10^{9}\ \text{N m}^2\,\text{C}^{-2}\) |
| Uniform field | near a planet's surface (approximately) | between parallel plates, \(E = V/d\) |
Similarities and differences
Similar: both follow an inverse-square law for a point source; both radial fields look the same on a diagram; both have equipotentials perpendicular to the field lines; in both, the potential is zero at infinity and the field strength is the potential gradient.
Different: gravity only ever attracts, because there is no negative mass, so gravitational potential is never positive. Electric forces can attract or repel. And gravity is extraordinarily weak by comparison: it only dominates on large scales because big objects are almost exactly neutral, so their electric forces cancel while their gravity adds up.
In a hydrogen atom an electron and a proton attract each other electrically and gravitationally. Show that the electric force is about \(10^{39}\) times bigger. (\(e = 1.60\times10^{-19}\) C, \(m_p = 1.67\times10^{-27}\) kg, \(m_e = 9.11\times10^{-31}\) kg)
Show solution
Both forces have \(r^2\) underneath, so it cancels when you divide one by the other:
\[ \frac{F_E}{F_G} = \frac{e^2/4\pi\varepsilon_0 r^2}{Gm_pm_e/r^2} = \frac{8.99\times10^9\times(1.60\times10^{-19})^2}{6.67\times10^{-11}\times1.67\times10^{-27}\times9.11\times10^{-31}} = 2.27\times10^{39} \]This ratio doesn't depend on how far apart they are, because both are inverse-square laws.
Quick checks
The gravitational field strength at the Earth's surface is 9.81 N kg⁻¹. What is it at a height above the surface equal to the Earth's radius?
The distance from the centre doubles (from \(R\) to \(2R\)), so \(g\) falls by a factor of \(2^2 = 4\): \(9.81/4 = 2.45\ \text{N kg}^{-1}\). 4.91 forgets the square; 1.09 uses \(3R\).
A satellite is moved from a low circular orbit to a higher circular orbit around the Earth. What happens to its speed and its period?
\(v = \sqrt{GM/r}\), so a bigger \(r\) means a lower speed. It also has further to go, so \(T = 2\pi r/v\) increases on both counts (\(T^2 \propto r^3\)).
Why do astronauts on the International Space Station feel weightless?
\(g\) at the ISS is about 8.7 N kg⁻¹, nearly the surface value. Everything on board is falling round the Earth together, so nothing presses on anything else.
At a point in the Earth's field the gravitational potential is −50 MJ kg⁻¹. What is the minimum energy needed to move a 2.0 kg mass from there to infinity?
The potential at infinity is zero, so \(\Delta V = 0 - (-50) = +50\ \text{MJ kg}^{-1}\). Work done \(= m\Delta V = 2.0\times50 = 100\ \text{MJ}\). You have to put energy in, so it is positive.
At the neutral point between the Earth and the Moon, which statement is correct?
Field strengths are vectors and cancel there. Potentials are scalars and simply add: both are negative, so the total is negative. The potential has a maximum at that point, so its gradient (the field) is zero.
Moon X orbits a planet at four times the orbital radius of moon Y. How many times longer is the period of X than the period of Y?
\(T^2 \propto r^3\), so \(T \propto r^{3/2}\). Multiplying \(r\) by 4 multiplies \(T\) by \(4^{3/2} = \sqrt{64} = 8\).