A weather satellite high above India is pulled by the rock under Delhi and Chennai, and by the floor of the Pacific, all at once. Each piece pulls from a different distance and in a different direction. Yet the engineers at ISRO compute its path with one formula, as if the whole Earth were a single point at its centre.
Cutting a body into point masses
The law of gravitation is written for point masses. A rod, a ring or a planet is not a point. The way forward is the one used for the centre of mass in the chapter on Centre of Mass. Cut the body into small elements, each of mass \(dm\). Each element is a point mass and makes a field \(G\,dm/r^2\) towards itself. By superposition the field of the whole body is the vector sum, and for tiny elements the sum becomes an integral.
For a thin uniform rod of mass \(M\) and length \(L\), the mass per unit length is the linear mass density \(\lambda = M/L\). A piece of length \(dx\) then has mass \(dm = \lambda\,dx\). Two points are worth solving: a point on the line of the rod, and a point on its perpendicular bisector.
\[ dE = \dfrac{G\,\lambda\,dx}{x^2} \]
Why. an element of length \(dx\) at a distance \(x\) from \(P\) has mass \(\lambda\,dx\). Its field at \(P\) points along the rod, towards the element.
\[ E = G\lambda \int_a^{a+L} \dfrac{dx}{x^2} = G\lambda \left[-\dfrac{1}{x}\right]_a^{a+L} \]
Why. every element's field points the same way, so the sizes simply add. The rod runs from \(x = a\) to \(x = a + L\).
\[ E = G\lambda\left(\dfrac1a - \dfrac{1}{a+L}\right) = \dfrac{G\lambda L}{a(a+L)} \]
Why. substitute the two limits and combine the fractions.
\[ E = \dfrac{GM}{a(a+L)} \]
Read it. \(\lambda L = M\), the mass of the whole rod.
Let \(d\) be the distance from \(P\) to the middle of the rod. Then \(a = d - L/2\) and \(a + L = d + L/2\), so \(E = GM/(d^2 - L^2/4)\). Far from the rod, where \(d\) is much larger than \(L\), this becomes \(GM/d^2\). From far away a rod looks like a point mass at its centre, as it should.
On the perpendicular bisector the elements pull in different directions. For every element above the middle there is one below, at the same distance. Their components along the rod cancel, and only the components towards the rod add. The integral of those components gives the result below. For a very long rod, \(L\) much larger than \(r\), it becomes \(2G\lambda/r\), a field that falls as \(1/r\), not \(1/r^2\).
A uniform rod of mass \(12\,\text{kg}\) and length \(2\,\text{m}\) lies along the \(x\) axis. A point \(P\) on the same line is \(2\,\text{m}\) from the near end. (a) Find the field at \(P\) as a multiple of \(G\), and in N/kg. (b) Compare it with the field of a \(12\,\text{kg}\) point mass at the rod's middle.
(a) \(E = GM/(a(a + L)) = 12G/(2 \times 4) = 1.5G = 1.0 \times 10^{-10}\,\text{N/kg}\), along the rod, towards it.
(b) The middle is \(3\,\text{m}\) from \(P\), so a point mass would give \(12G/9 = 1.33G\). The rod's field is larger, because its near half is closer to \(P\) and \(1/x^2\) favours the near elements.
A uniform rod of mass \(10\,\text{kg}\) and length \(1.2\,\text{m}\) is fixed in space. A point lies on its perpendicular bisector, \(0.8\,\text{m}\) from its middle. Take \(G = 6.67 \times 10^{-11}\,\text{N m}^2/\text{kg}^2\). Find the field there, in units of \(10^{-10}\,\text{N/kg}\).
The ring on its axis
Take a thin uniform ring of mass \(M\) and radius \(R\). Find the field at a point \(P\) on its axis, a distance \(z\) from its centre. Every element of the ring is the same distance from \(P\), \(\sqrt{R^2 + z^2}\). Each makes a field of the same size, pointing from \(P\) to the element. For every element there is one opposite it on the ring, and their sideways parts cancel.
The part of each element's field along the axis is its field times \(z/\sqrt{R^2 + z^2}\), the cosine of its angle with the axis. That factor is the same for every element. So the sum is easy: replace the element's mass by the whole mass \(M\). The field on the axis is \(E = GMz/(R^2 + z^2)^{3/2}\), directed along the axis towards the ring's centre.
Move \(P\) to the centre, \(z = 0\). The pulls of opposite elements are now exactly opposite, and the field is zero. Move it far away, and the field becomes \(GM/z^2\), the field of a point mass. Between these, the field must rise and then fall. In the sim it is greatest at \(z = 0.71R\), where \(E = 0.385\,GM/R^2\).
Calculus finds the peak exactly. Set \(dE/dz = 0\). The derivative of \(z(R^2 + z^2)^{-3/2}\) is \((R^2 + z^2)^{-5/2}(R^2 + z^2 - 3z^2)\). It is zero when \(R^2 = 2z^2\), that is \(z = R/\sqrt2\). Putting this back gives \(E_{\max} = 2GM/(3\sqrt3\,R^2)\).
A uniform ring has mass \(M\) and radius \(R\). Find the field on its axis at a distance \(z = R\) from the centre, as a multiple of \(GM/R^2\).
A test mass moves along the axis of a uniform ring, from the centre outwards. Where is the field on it greatest?
Shells and spheres
A thin hollow sphere of uniform thickness is a spherical shell. It can be cut into rings, each centred on the line from the shell's centre to the point. Adding the fields of all the rings is a longer integral, which Newton completed in the Principia of 1687. The result is two remarkably simple statements, the shell theorem.
A uniform spherical shell of mass \(M\) attracts a point outside it as if all its mass were at its centre: \(E = GM/r^2\). A uniform spherical shell exerts no net gravitational force at any point inside it: \(E = 0\).
The second theorem can be seen without the integral. Stand at a point \(P\) inside the shell. Draw a thin cone from \(P\) in some direction, and the same cone in the opposite direction. Each cone cuts a small patch from the shell. The patch at distance \(r\) has an area that grows as \(r^2\), so its mass grows as \(r^2\). Its pull on \(P\) falls as \(1/r^2\). So the near patch and the far patch pull equally, in opposite directions.
Every direction from \(P\) pairs with its opposite, so all the pulls cancel. That works only because the force falls exactly as \(1/r^2\). With any other power, the near and far patches would not balance.
A uniform solid sphere of mass \(M\) and radius \(R\) is a set of nested shells. Outside it, every shell acts from the centre, so \(E = GM/r^2\). Inside, at a distance \(r\) from the centre, the shells outside \(r\) pull nothing. Only the sphere of radius \(r\) within counts. Its mass is \(M r^3/R^3\), so its field is \(G(Mr^3/R^3)/r^2 = GMr/R^3\). The field grows in proportion to \(r\), from zero at the centre to \(GM/R^2\) at the surface.
The shell theorem is why a planet, a star or a cricket ball pulls any outside body like a point mass at its centre. Suppose the Earth is made of uniform spherical layers. Then its pull on the Moon is exactly the pull of a point of mass \(6 \times 10^{24}\,\text{kg}\) at its centre.
This answers the question about the weather satellite. The Earth is very nearly a set of uniform spherical layers, denser towards the centre but the same in every direction. Each layer is a shell, and each shell pulls the satellite as if its mass sat at the centre. So the whole Earth does too, and one formula, \(GM/r^2\), gives the pull of every piece of rock at once.
(a) A uniform spherical shell has mass \(100\,\text{kg}\) and radius \(1\,\text{m}\). Find the field \(2\,\text{m}\) from its centre and \(0.5\,\text{m}\) from its centre. (b) Treat the Earth as a uniform sphere with \(g = 9.8\,\text{m/s}^2\) at its surface. Find the field halfway to its centre.
(a) Outside, the shell acts as a point: \(E = G \times 100/2^2 = 25G = 1.67 \times 10^{-9}\,\text{N/kg}\). Inside, at \(0.5\,\text{m}\), the field is zero.
(b) Inside a solid sphere \(E\) grows in proportion to \(r\). At \(r = R/2\) the field is half the surface value, \(9.8/2 = 4.9\,\text{N/kg}\).
A small ball is placed inside a uniform hollow sphere, closer to one side than the other. What is the net gravitational force of the sphere on it?
Treat the Earth as a uniform solid sphere with a surface field of \(9.8\,\text{N/kg}\). What is the field at a point one quarter of the way from the centre to the surface, in N/kg?
- Adding the sizes of element fields that point in different directions.Resolve first. On a ring's axis or a rod's bisector, the sideways parts cancel by symmetry.
- Using \(GM/r^2\) inside a solid sphere.Inside, only the mass within \(r\) counts: \(E = GMr/R^3\).
- Thinking the field inside a shell is zero only at its centre.It is zero at every point inside a uniform shell.