In August 2023 the rover Pragyan rolled out of the Chandrayaan-3 lander onto the Moon. Its mass was about \(26\,\text{kg}\), exactly as it had been in Bengaluru. Yet the pull on it was about six times smaller on the Moon. Something about the place had changed, not the rover.
Force per kilogram
Place a small body of mass \(m\) at some point near the Earth. The Earth pulls it with a force \(\vec F\). Place a body of mass \(2m\) at the same point, and the pull doubles. The force depends on the body, but the ratio \(\vec F/m\) does not. It is the same for every body placed there. So the ratio describes the point itself.
This force per unit mass is the gravitational field at the point, written \(\vec E\). The small body used to measure it is a test mass. It must be small, so that its own pull does not move the masses that make the field. A ration shop gives a similar idea. The bill depends on how much rice you buy, but the price per kilogram belongs to the shop.
The gravitational field at a point is the gravitational force per unit mass on a small test mass placed at that point: \(\vec E = \vec F/m\). It is a vector, and it points the way the force on the test mass points.
The unit of the field is the newton per kilogram. Since \(1\,\text{N} = 1\,\text{kg m/s}^2\), one N/kg is one m/s². That is no accident. A body on which only gravity acts has acceleration \(\vec F/m = \vec E\). So the field at a point is the acceleration of free fall there. At the Earth's surface the field is \(g = 9.8\,\text{N/kg}\), pointing down.
At a point P, a \(2\,\text{kg}\) body feels a gravitational force of \(10\,\text{N}\). It is replaced by a \(5\,\text{kg}\) body. What is the gravitational field at P now?
The field of a point mass
Put a test mass \(m\) at a distance \(r\) from a point mass \(M\). The law of gravitation gives the pull \(GMm/r^2\) towards \(M\). Divide by \(m\) and the test mass disappears. The field of a point mass is \(E = GM/r^2\), directed towards \(M\). It falls as \(1/r^2\), like the force, and it points inward everywhere.
At the surface, one radius \(R\) from the centre, the Earth's field is \(9.8\,\text{N/kg}\). At \(2R\) it is \(9.8/4 = 2.45\,\text{N/kg}\), and at \(3R\) it is \(9.8/9 = 1.09\,\text{N/kg}\). A picture of the field uses arrows, or lines that follow them. Such field lines point towards the mass and crowd together where the field is strong.
Fields add as vectors, exactly as forces do, because each is only a force divided by the same test mass. The field at any point is the vector sum of the fields of all the masses. For example, at the Earth the Sun's field is about \(5.9 \times 10^{-3}\,\text{N/kg}\) and the Moon's is about \(3.3 \times 10^{-5}\,\text{N/kg}\). Both add to the Earth's own field.
The Earth's field at its surface is \(9.8\,\text{N/kg}\). What is the field at a height equal to one Earth radius above the surface, in N/kg?
Where two pulls cancel
Between two masses the fields point in opposite directions: each points towards its own mass. Near one mass its field wins; near the other, the other wins. Somewhere between them the two fields are equal in size, and the net field is zero. That place is the null point. A body placed there feels no net gravitational force. Predict where it lies for the Earth and the Moon, then watch.
The Earth has \(81\) times the mass of the Moon. A probe travels from the Earth towards the Moon. Where do the two pulls on it cancel?
In the figure above the Earth's pull wins for most of the journey. The two pulls become equal at \(54\) Earth radii from the Earth's centre, about \(346\,000\,\text{km}\). There each pull on the \(1\,\text{kg}\) probe is only \(3.4\,\text{mN}\). Beyond that point the Moon's pull wins, and the net force turns round.
\[ \dfrac{Gm_1}{x^2} = \dfrac{Gm_2}{(d - x)^2} \]
Why. at a distance \(x\) from \(m_1\), on the line between the masses, the point is \(d - x\) from \(m_2\). The two fields must be equal in size.
\[ \dfrac{d - x}{x} = \sqrt{\dfrac{m_2}{m_1}} \]
Why. cancel \(G\), rearrange, and take the positive square root, because both distances are positive.
\[ x = \dfrac{d}{1 + \sqrt{m_2/m_1}} \]
Read it. solve for \(x\). If \(m_2 < m_1\), the root is less than \(1\) and \(x > d/2\): the null point lies nearer the smaller mass.
Outside the pair there is no null point. On either side beyond the masses, both fields point the same way, towards the pair, so they cannot cancel. For the Earth and the Moon, \(\sqrt{m_2/m_1} = 1/9\). So \(x = d/(1 + 1/9) = 0.9d = 0.9 \times 3.84 \times 10^8 = 3.46 \times 10^8\,\text{m}\) from the Earth.
Masses of \(4\,\text{kg}\) and \(16\,\text{kg}\) are \(6\,\text{m}\) apart. How far from the \(4\,\text{kg}\) mass is the null point, in metres?
Two equal masses sit \(d\) apart. Where is the net gravitational field zero?
A uniform field near the ground
Look again at the right of Fig 3.1. Over a small patch of ground the Earth's field arrows are parallel and equal. Strictly they point to the Earth's centre, so they meet far below. But over a hundred metres they are parallel for every practical purpose. A field that has the same size and direction everywhere in a region is a uniform field.
How good is this? At the top of the Qutub Minar, \(73\,\text{m}\) high, the distance from the Earth's centre has grown by a fraction of about \(10^{-5}\). The field falls by twice that fraction, \(2.3 \times 10^{-5}\), or about \(0.0002\,\text{m/s}^2\). That is why every problem of the chapters on Kinematics and Dynamics could use one constant \(g\).
The Earth's field points towards its centre everywhere. Why may we still treat it as uniform over a cricket ground?
A cricket ground is about \(150\,\text{m}\) across, which is tiny compared with the Earth's radius, \(6.4 \times 10^6\,\text{m}\). Field arrows at the two ends of the ground point to the centre, so they differ in direction by only about \(150/6.4 \times 10^6\) radian, about \(0.001°\). Over a few metres of height, the distance from the centre changes by a fraction of about \(10^{-6}\). So the size of the field hardly changes either. Both the size and the direction are constant to far better than we can measure, so the field is uniform for every practical purpose.
A rover's weight follows from the field at once. On the Moon's surface the field is about \(1.63\,\text{N/kg}\), as the last lesson of this chapter shows. So Pragyan's \(26\,\text{kg}\) felt a pull of \(26 \times 1.63 = 42\,\text{N}\). On Earth the same rover felt \(26 \times 9.8 = 255\,\text{N}\). The rover did not change; the field of the place did.
From 2002 a pair of satellites called GRACE flew one behind the other around the Earth. The gap between them changed very slightly as they passed over places where the field was a little stronger or weaker. From these changes scientists mapped tiny changes in the Earth's field over the years. In 2009 they found that water under Rajasthan, Punjab and Haryana was being pumped out faster than rain could refill it. The field there was growing weaker, year by year, as the water was removed.
Newton described a pull that acts across empty space at once, and he admitted he could not explain how. Michael Faraday, working with magnets in London, pictured the space around a magnet filled with lines of force. The lines, he said, belonged to the space itself, whether or not a body was placed there to feel them. In a paper of 1857 he asked whether gravitation might work in the same way. The field of this lesson is his idea: a property of each point in space.
- Thinking the field at a point depends on the test mass placed there.\(E = F/m\), and \(F\) grows in proportion to \(m\). The ratio belongs to the point.
- Putting the null point halfway between two unequal masses.The null point lies nearer the smaller mass, at \(x = d/(1 + \sqrt{m_2/m_1})\).
- Looking for a null point outside a pair of masses.Outside the pair both fields point the same way; the only null point is between them.