Gravitational fields

Gravitational fields and field lines

A gravitational field is a region in which a mass experiences a force because of the mass it has. It is an example of a field of force: instead of something the mass has to touch, the field is the space around a mass where its influence is felt, and any other mass placed anywhere in that space is pulled towards the mass producing the field. The gravitational field strength g at a point is defined as the gravitational force per unit mass acting on a small test mass placed at that point, so g = F / m. Force is measured in newtons and mass in kilograms, which makes the unit of field strength the newton per kilogram (N kg-1). Field strength is a vector, and its direction is the direction of the force acting on the test mass.

Field lines are the standard way of picturing a field. A field line is drawn so that its direction at any point is the direction of the force acting on a small mass placed there, which means every gravitational field line points towards the mass producing the field, because gravity is always attractive. Where the lines crowd together the field is strong and where they spread apart it is weak, so the spacing of the lines carries as much information as the lines themselves. Field lines never cross each other, because a small mass placed at a point would otherwise have two possible directions of force at once.

Two shapes of field matter for this course. Around a point mass, or around a planet that may be treated as one, the field is radial: the lines point inwards towards the centre and are drawn closer together near the mass, so the field is strongest at the surface of the planet and weakens with distance. Over a small region near the surface of the Earth the lines are almost parallel and equally spaced, and a field of that kind is a uniform field, in which the field strength has the same size and the same direction everywhere. The uniform picture is the one to use for an object dropped or thrown a few metres, and the radial picture is the one to use for a satellite.

• Field strength g = F / m, the gravitational force per unit mass, measured in N kg-1; it is a vector pointing towards the mass producing the field.
• Gravitational field lines always point towards the mass, and the spacing of the lines shows how strong the field is at that place.
• A point mass or a planet gives a radial field whose strength falls with distance; a small region near a planet's surface is a uniform field.

Newton's law of gravitation

Newton's law of gravitation states that 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, so F = G m1 m2 / r2. The distance r is measured between the centres of the two masses, and each force acts along the line joining them. The constant G is the gravitational constant, 6.67 x 10-11 N m2 kg-2, and it is universal: the same value describes the attraction between two lead spheres in a laboratory and the attraction between the Earth and the Sun. The two forces are equal in size and opposite in direction, which is Newton's third law, so the Sun pulls the Earth with a force of exactly the same size as the force with which the Earth pulls the Sun.

The inverse square part of the law is what makes gravity a long-range but very weak force. Doubling the separation quarters the force, and trebling it cuts the force to one ninth. The size of G shows how weak the attraction is. Two 5.0 kg masses 0.20 m apart attract each other with a force of (6.67 x 10-11 x 5.0 x 5.0) / (0.20)2 = 4.2 x 10-8 N, which is about four millionths of the 9.8 x 10-3 N weight of a 1 g paperclip. Gravity only dominates the motion of very large masses, which is why the attraction between ordinary laboratory objects cannot be felt and why the constant G was not measured until 1798.

The point-mass approximation. For a point outside a uniform sphere the mass of the sphere may be considered to be a point mass at its centre, so r in F = G m1 m2 / r2 is measured from the centre of the planet and never from its surface. The reason is that a uniform sphere can be divided into thin spherical shells, and a shell attracts a point outside it as though the whole mass of that shell were concentrated at its centre; adding the shells of a solid sphere gives the same force as a single point mass of the same total mass. This is why the radius of a planet is added to the height of a satellite before anything is calculated, and why the weight of an object at the Earth's surface has the finite value it does. The result holds outside the sphere: inside a hollow uniform shell the field is zero, because the pulls from the different parts of the shell cancel.

The law also gives the force on a satellite directly, and it is worth seeing the size of the numbers. A 1200 kg satellite at a height of 1.0 x 10^6 m above the Earth's surface is 7.37 x 10^6 m from the centre, where the gravitational force on it is (6.67 x 10-11 x 5.97 x 10^24 x 1200) / (7.37 x 10^6)2 = 8.8 x 10^3 N. That force is the whole of the force acting on the satellite, so it is both the gravitational force and the resultant force, and it is what keeps the satellite turning in its circular path.

Deriving and using g = GM / r2

The field strength due to a point mass follows from the two equations you already have. The definition of field strength gives g = F / m for a small mass m placed at a distance r from a mass M, and Newton's law of gravitation gives the force on that small mass as F = G M m / r2. Substituting the force into the definition gives g = (G M m / r2) / m, and the test mass m cancels, leaving g = G M / r2. The cancellation is the important step: the field strength at a point depends only on the mass M producing the field and on the distance r from the centre of that mass, and not at all on the mass placed there to test it. A small satellite and a large one at the same height experience exactly the same field strength.

Because g = G M / r2, the field of a planet follows the same inverse square law as the force. At the surface of the Earth, M = 5.97 x 10^24 kg and r = 6.37 x 10^6 m, so g = (6.67 x 10-11 x 5.97 x 10^24) / (6.37 x 10^6)2 = 9.81 N kg-1, the value used in every free-fall calculation near the surface. On the Moon, with M = 7.35 x 10^22 kg and r = 1.74 x 10^6 m, the same equation gives 1.62 N kg-1, about a sixth of the Earth's value, which is why an astronaut can jump about six times higher on the Moon. Farther out the field falls away quickly: at the distance of the Moon's orbit from the Earth's centre, 3.84 x 10^8 m, the field strength of the Earth is only 2.70 x 10-3 N kg-1.

Why g is nearly constant near the surface. Field strength is a force per unit mass, so the weight of an object of mass m near the surface is W = m g, and the acceleration of free fall is numerically the same as g, measured in m s-2. The value 9.81 N kg-1 is a good approximation only because we live so close to the centre compared with the radius of the Earth. From g = G M / r2, a small change in height delta h changes g by a fraction of about 2 delta h / r. A few metres changes g by about one part in a million, which is far below the precision of any school measurement, so g is treated as constant in projectile and free-fall problems. Over much larger distances the approximation fails: at a height of 100 km the field strength has fallen to 9.51 N kg-1, a change of about 3 per cent, and at 1000 km it is only 7.33 N kg-1.

Circular orbits: the gravitational force is the centripetal force

A satellite in a circular orbit travels at a constant speed, but its velocity is changing all the time because the direction in which it moves keeps turning. The only force acting on it is the gravitational attraction towards the centre of the planet, and a force acting at right angles to the velocity changes the direction of motion without changing the speed. That is the whole physics of an orbit: the gravitational force is the centripetal force. Equating the two expressions gives G M m / r2 = m v2 / r.

Cancelling the mass m of the satellite and one power of r gives v2 = G M / r, so the orbital speed is v = sqrt(G M / r). Notice what is not in that equation. The mass of the satellite has gone, so satellites of very different masses at the same orbital radius move at the same speed, and only the mass of the central body and the radius of the orbit matter. The speed falls as the radius grows, so a satellite close to the Earth must travel faster than one far out: the International Space Station, at about 410 km above the surface, moves at 7.66 x 10^3 m s-1, while a geostationary satellite manages only 3.07 x 10^3 m s-1.

The orbital period follows from the speed. In one period T the satellite travels a distance 2 pi r, so T = 2 pi r / v. Substituting v = sqrt(G M / r) and squaring gives T2 = (4 pi2 / G M) r3, so the square of the period is proportional to the cube of the radius. This is why a satellite in a low orbit circles the Earth many times a day while a geostationary satellite takes exactly a day: 93 minutes at the radius of the space station against 24 hours at the geostationary radius. A graph of T2 against r3 is a straight line through the origin with gradient 4 pi2 / G M, and measuring that gradient is a standard way of finding the mass of a planet from the motion of its satellites.

One result of all this surprises most students: moving a satellite to a higher orbit needs energy to be supplied, even though the satellite ends up moving more slowly. The engines have to drive the satellite away from the planet against the gravitational force, and the gravitational potential energy gained is larger than the kinetic energy lost, so the total energy of the satellite is increased. The reverse is also true, and it matters in practice: a satellite that loses energy through atmospheric drag spirals into a lower orbit and speeds up as it does so.

Geostationary orbits

A geostationary orbit keeps a satellite above the same point on the Earth's surface at all times, so a dish on the ground can be fixed in place and pointed at the satellite once and for all. Three features make this possible, and all three are needed. The orbit must be a circle lying in the plane of the Equator, so that the satellite stays directly above the Equator. The satellite must travel in the same direction as the Earth rotates, from west to east. And its period must be 24 hours, so that it completes exactly one orbit while the Earth turns once on its axis.

The equatorial requirement follows from the geometry of circular orbits. A satellite in a circular orbit moves around the centre of the Earth, so a satellite that is permanently above one point on the ground can only be above a point on the Equator; an orbit tilted to the Equator would carry the satellite alternately north and south of the observer and it would trace out a figure of eight in the sky. The direction of travel is fixed in the same way: a satellite travelling from east to west would complete a circuit relative to the ground in less than 24 hours and would drift steadily westwards across the sky.

Once the period is fixed at 24 hours the radius is fixed too, and it follows from T2 = (4 pi2 / G M) r3. Putting T = 24 x 60 x 60 = 8.64 x 10^4 s and M = 5.97 x 10^24 kg gives r3 = (6.67 x 10-11 x 5.97 x 10^24 x (8.64 x 10^4)2) / (4 pi2) = 7.53 x 10^22 m3, so r = 4.22 x 10^7 m. The radius of the Earth is 6.37 x 10^6 m, so the satellite is 4.22 x 10^7 - 6.37 x 10^6 = 3.6 x 10^7 m above the surface, about 35800 km up, and its orbital speed is v = 2 pi r / T = (2 x pi x 4.22 x 10^7) / (8.64 x 10^4) = 3.07 x 10^3 m s-1. There is only one circle that satisfies all three conditions, so the number of satellites that can share the orbit is limited.

Geostationary satellites are used for communications and for weather monitoring: television broadcasts, telephone links and the images from weather satellites such as the Meteosat series all come from this orbit, and one satellite covers roughly a third of the globe. The drawbacks all follow from the height. The signal travels 3.6 x 10^7 m up to the satellite and the same distance back down, so even at the speed of light a two-way link has a delay of about 0.24 s, which is long enough to be noticed in a telephone conversation, and the signal at that distance is weak, so large dishes are needed. A low Earth orbit gives a much stronger signal with almost no delay, but a satellite 410 km up sweeps round the whole Earth in 93 minutes, so a constellation of many satellites and dishes that track them is needed instead.

Gravitational potential

Gravitational potential at a point is defined as the work done per unit mass in bringing a small test mass from infinity to that point. Infinity is used because two masses at an infinite separation are completely out of each other's influence, and the potential there is taken as zero, so every other potential is measured from that starting value. Potential is a scalar and is measured in joules per kilogram (J kg-1). Field strength is a vector in N kg-1 while potential is a scalar in J kg-1, and keeping the two quantities apart is essential in an exam.

Every gravitational potential is negative. Gravity is attractive, so a test mass released at a finite distance from the Earth falls towards the Earth: the test mass has less potential energy at the point than it has at infinity, and the work done per unit mass in bringing it in from infinity is negative. The potential is therefore negative everywhere and it increases towards zero as the distance increases. Two consequences follow immediately: work must be done by an outside agent to lift a mass away from a planet, and the potential has its most negative value at the surface of the planet.

For the field of a point mass the potential is phi = -G M / r, with r measured from the centre of the mass. At the surface of the Earth, phi = -(6.67 x 10-11 x 5.97 x 10^24) / (6.37 x 10^6) = -6.25 x 10^7 J kg-1, and at a height of 1.0 x 10^6 m, where r = 7.37 x 10^6 m, it has risen to -5.40 x 10^7 J kg-1. Notice that the potential is a property of the field and not of any object placed in it, which is exactly why it is quoted per unit mass.

A potential difference between two points is the work done per unit mass in moving between them, so the work done on a mass m is W = m x delta phi. Lifting 50 kg from the Earth's surface to a height of 1.0 x 10^6 m raises the potential by (-5.40 x 10^7) - (-6.25 x 10^7) = 8.5 x 10^6 J kg-1, so the work done is 50 x 8.5 x 10^6 = 4.2 x 10^8 J. A surface on which every point has the same potential is an equipotential surface; around a point mass the equipotentials are spheres centred on the mass, and since no work is done when a mass moves along an equipotential, moving a satellite within one circular orbit costs no energy. The field strength is linked to the potential through the gradient of a potential against distance graph: g is the negative of the gradient, because the field points in the direction in which the potential falls.

Gravitational potential energy and escape speed

Gravitational potential energy of two point masses. Multiplying the potential by the mass m of the object gives its gravitational potential energy, EP = m x phi = -G M m / r. The energy belongs to the pair of masses and it is negative, which is the sign that the object is bound: energy has to be supplied to separate the two masses completely. As the separation increases, EP rises towards zero, and a bound object can never have positive gravitational potential energy.

The equation EP = -G M m / r looks nothing like m g h, but m g h is the small-height approximation of it. For a rise delta h that is small compared with r, the change in potential is delta phi = G M delta h / r2, so the change in potential energy is m delta phi = (G M / r2) m delta h = m g delta h, using g = G M / r2 near the surface. Lifting 50 kg through 100 m on the Earth gives 50 x 9.81 x 100 = 4.9 x 10^4 J from m g h and 4.9 x 10^4 J from the potential difference, so over distances of this size the two methods agree. Use m g h only while g can be treated as constant.

The full equation matters for spacecraft. A 1200 kg satellite sitting on the launch pad has a gravitational potential energy of -G M m / r = -(6.67 x 10-11 x 5.97 x 10^24 x 1200) / (6.37 x 10^6) = -7.5 x 10^10 J, and in geostationary orbit at r = 4.22 x 10^7 m its potential energy is -1.1 x 10^10 J. The satellite must therefore gain 6.4 x 10^10 J of potential energy simply to reach the orbit, and it must be given kinetic energy as well, which is why launch to geostationary orbit requires so much fuel and such a large rocket.

Escape speed. An object escapes from a planet altogether only if it can be carried out to infinity, where the potential energy is zero. Leaving the surface of a planet of mass M and radius r with speed v, an object of mass m has kinetic energy (1/2) m v2 and potential energy -G M m / r, so the minimum condition is (1/2) m v2 = G M m / r. The mass m cancels once more, giving the escape speed v = sqrt(2 G M / r), which depends only on the mass and the radius of the planet. For the Earth, v = sqrt((2 x 6.67 x 10-11 x 5.97 x 10^24) / (6.37 x 10^6)) = 1.12 x 10^4 m s-1, about 11.2 km s-1, and a rocket must reach that speed to leave the Earth for good. The escape speed of the Moon is only 2.4 x 10^3 m s-1, which is of the same order as the speeds of gas molecules at the Moon's surface temperature, so the faster molecules of any gas that formed there have escaped over billions of years and the Moon has no atmosphere, while the Earth's 11.2 km s-1 is far too high for that to have happened.