A trekking map of the Kedarkantha trail marks the height of every point with contour lines. The map does not care who walks. Yet from it any trekker can find two things. The steepness of the slope tells how hard the ground pulls a walker back. The difference in height between two camps tells how much work the climb will take, which is weight times height gained.
A property of a place
In the last lesson, when some masses were fixed, only the terms of the moving body mattered. For a small body of mass \(m\) near fixed masses, those terms add up to a potential energy \(U\) that is proportional to \(m\). Double the body, and \(U\) doubles. So the ratio \(U/m\) depends only on where the body is. It describes the place, not the body.
The gravitational potential at a point is this ratio, \(V = U/m\), the potential energy per kilogram of a small body placed there. Its unit is J/kg. It is a scalar, a single number at each point, like the height on a trekking map. Once \(V\) is known, any body of mass \(m\) placed at that point has \(U = mV\).
The gravitational potential \(V\) at a point is the potential energy per unit mass of a small test body placed there: \(V = U/m\). Equivalently, it is the work that an outside agent does per unit mass to bring the body slowly from infinity to that point. \(V = 0\) at infinity.
For a single point mass \(M\), divide \(U = -GMm/r\) by \(m\). The potential at a distance \(r\) is \(V = -GM/r\). At the Earth's surface this is \(-GM/R = -gR\). With \(g = 9.8\,\text{m/s}^2\) and \(R = 6.4 \times 10^6\,\text{m}\), that is \(-6.27 \times 10^7\,\text{J/kg}\).
Several masses give a potential that is the plain sum of their separate potentials. No angles or components are needed, because \(V\) is a scalar. This is the great convenience of \(V\) over the field, whose contributions must be added as vectors.
Four equal masses \(m\) sit at the corners of a square of side \(a\). Find the potential at the centre, and the field there.
(a) Each corner is \(a/\sqrt2\) from the centre, so each mass gives \(-Gm/(a/\sqrt2) = -\sqrt2\,Gm/a\).
(b) The four add as numbers: \(V = -4\sqrt2\,Gm/a = -5.66\,Gm/a\).
(c) The four pulls at the centre are equal and point to the four corners, so they cancel in pairs. The field is zero there, although the potential is not.
The gravitational potential at a point P is \(V\). A \(2\,\text{kg}\) body at P is replaced by a \(4\,\text{kg}\) body. What happens?
Field from potential
In the chapter on Potential Energy and Conservation of Energy you found that a force is minus the slope of its potential energy, \(F = -dU/dx\). Divide both sides by the test mass \(m\). Force per unit mass is the gravitational field \(E\), and potential energy per unit mass is \(V\). So the field is minus the slope of the potential: \(E = -dV/dr\) along a line.
Check it on a point mass. With \(V = -GM/r\), the slope is \(dV/dr = GM/r^2\), so \(E_r = -GM/r^2\). The minus sign says that the field points inward, towards the mass. Its size is \(GM/r^2\), as the chapter on Gravitational Force and Field found. At the Earth's surface its size is \(GM/R^2 = g\).
The field always points from higher potential to lower potential, down the slope. A body let go from rest moves that way, just as a ball released on a hillside rolls downhill. Where \(V\) is the same over a whole region, the slope is zero, and so is the field. That is exactly the case inside a hollow shell, as the next topic shows.
Along the \(x\) axis the potential is \(V = 5x^2\), with \(V\) in J/kg and \(x\) in metres. What is the \(x\)-component of the field at \(x = 3\,\text{m}\), in N/kg?
A ring on its axis
Take a thin ring of mass \(M\) and radius \(R\), and a point P on its axis at a distance \(x\) from the centre. Cut the ring into small pieces. Every piece is the same distance \(\sqrt{R^2 + x^2}\) from P. So every piece contributes the same share, and the pieces add as numbers. The potential at P is \(V = -GM/\sqrt{R^2 + x^2}\).
At the centre of the ring, \(x = 0\) and \(V = -GM/R\). Yet the field there is zero, because pulls from opposite sides of the ring cancel. The potential is lowest at the centre, and the curve \(V(x)\) is flat there. A zero field needs a flat potential, not a zero potential.
A shell and a solid sphere
Outside a thin shell or a uniform solid sphere, the field is that of a point mass at the centre. So outside both, \(V = -GM/r\). Inside a thin shell the field is zero, so \(V\) cannot change: it keeps its surface value, \(-GM/R\), all the way to the centre. Inside a solid sphere the field is \(GMr/R^3\), pointing inward. So \(V\) keeps falling as you go deeper.
\[ E_r = -\dfrac{GMr}{R^3} \]
Why. inside a uniform sphere the field grows in proportion to \(r\) and points inward, as the chapter on Gravitational Force and Field showed.
\[ V(r) - V(R) = -\int_R^r E_r\,dr' = \int_R^r \dfrac{GMr'}{R^3}\,dr' \]
Why. the potential difference is minus the integral of the field along the path from the surface inward.
\[ V(r) - V(R) = \dfrac{GM(r^2 - R^2)}{2R^3} \]
Why. the integral of \(r'\) is \(r'^2/2\). The result is negative for \(r < R\): the potential falls inward.
\[ V(r) = -\dfrac{GM}{R} - \dfrac{GM(R^2 - r^2)}{2R^3} = -\dfrac{GM(3R^2 - r^2)}{2R^3} \]
Read it. add the surface value \(V(R) = -GM/R\) and collect the terms over \(2R^3\).
In Fig 3.2 the two curves join at the surface without a jump. The potential never jumps, because a jump would need an infinite field. At the centre of the solid sphere the curve is flat, since the field is zero there. The centre is the lowest point of the potential, \(-3GM/2R\), one and a half times the surface value.
A small body is inside a hollow spherical shell of mass \(M\) and radius \(R\), a distance \(R/2\) from the centre. Which statement is correct?
Work from a difference in potential
Move a body of mass \(m\) slowly from a point A to a point B. Slowly means that it gains no kinetic energy. Then the work of the outside agent all goes into potential energy: \(W = U_B - U_A = m(V_B - V_A)\). The work depends only on the two potentials, never on the route. Where \(V_B > V_A\), the agent must do positive work.
Only differences in \(V\) enter any result, so the zero at infinity could be moved, as the zero of \(U\) could. We keep it at infinity because then \(V\) at a point is the work per kilogram to bring a body there from far away. The same idea explains the fixed masses of the last lesson. Their mutual energy never changes, so it never appears in \(V\) or in any work.
The potential at A is \(-6 \times 10^7\,\text{J/kg}\) and at B is \(-2 \times 10^7\,\text{J/kg}\). How much work must an outside agent do to carry a \(2\,\text{kg}\) probe slowly from A to B?
The probe moves slowly, so its kinetic energy does not change. Only the potentials at the two ends matter.
\(W = m(V_B - V_A)\).
\(V_B - V_A = -2 \times 10^7 - (-6 \times 10^7) = 4 \times 10^7\,\text{J/kg}\). So \(W = 2 \times 4 \times 10^7 = 8 \times 10^7\,\text{J}\).
The agent does \(8 \times 10^7\,\text{J}\) of work. It is positive because the probe is carried to a higher potential, away from the masses.
How much work must be done to carry a \(1\,\text{kg}\) body slowly from the Earth's surface to a distance \(2R\) from the centre, in megajoules? Take \(g = 9.8\,\text{m/s}^2\) and \(R = 6.4 \times 10^6\,\text{m}\).
George Green worked in his father's windmill in Nottingham and taught himself mathematics from library books. In 1828, at his own expense, he printed an essay on electricity and magnetism. In it he used a single function of position whose slope gives the force, and he called it the "potential function". Only about fifty copies were sold, and the essay was nearly lost. Kelvin found it in 1845 and made it famous. The \(V\) of this lesson is Green's potential, applied to gravity.
The trekking map of the hook is a map of potential. Gravity's version around a planet is a set of circles of equal \(V\), closer together near the planet, where \(V = -GM/r\) changes fastest. The field at any point is minus the slope of that map. The work of a journey is the mass times the difference in \(V\) between its ends, whatever route the traveller takes.
- Thinking that a zero field means a zero potential.A zero field means a flat potential. Inside a shell \(V = -GM/R\), not zero.
- Using \(V = -GM/r\) inside a shell or a solid sphere.That form holds only outside. Inside, use \(-GM/R\) for the shell and \(-GM(3R^2 - r^2)/2R^3\) for the sphere.
- Adding the potentials of several masses as vectors.\(V\) is a scalar. Add the numbers, with their minus signs.