Gravitational fields and Newton's law of gravitationEdexcel International A Level Physics: Revision notes

A gravitational field is a region of space in which a mass experiences a force because of its mass. The field is described by the gravitational field strength $g$, defined as the force per unit mass:

$g = \frac{F}{m}$

It is a vector, measured in N kg⁻¹ (equivalent to m s⁻²), and points in the direction of the force on a mass, towards the source. Near the Earth's surface $g \approx 9.81$ N kg⁻¹, and the weight of a mass is $W = mg$. For a point mass or a uniform sphere the field is radial: field lines point towards the centre, getting further apart with distance, which shows the field becoming weaker.

Any two point masses attract each other with a force that is proportional to the product of the masses and inversely proportional to the square of their separation:

$F = \frac{Gm_1m_2}{r^2}$

where $G = 6.67 \times 10^{-11}$ N m² kg⁻² and $r$ is the distance between the centres. The forces on the two masses are equal and opposite (Newton's third law), always attractive. Doubling $r$ reduces the force to a quarter.

Worked example. Earth (5.97 × 10²⁴ kg) and Moon (7.35 × 10²² kg), $r = 3.84 \times 10^8$ m: $F = 6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 7.35 \times 10^{22} / (3.84 \times 10^8)^2 = 2.0 \times 10^{20}$ N.

Combining $g = F/m$ with Newton's law for a mass $m$ at distance $r$ from a point mass $M$:

$g = \frac{F}{m} = \frac{GMm/r^2}{m} = \frac{GM}{r^2}$

The field strength is independent of the small mass $m$, and is proportional to $1/r^2$. This works outside any spherical mass. At the Moon's surface ($M = 7.35 \times 10^{22}$ kg, $r = 1.74 \times 10^6$ m), $g = 1.62$ N kg⁻¹, so an 85 kg astronaut weighs $85 \times 1.62 = 1.4 \times 10^2$ N.

Where fields from two bodies add, add them as vectors. Between two masses there is a neutral point where the fields are equal and opposite: $GM_1/x^2 = GM_2/(d - x)^2$.

The gravitational potential $V$ at a point is the work done per unit mass in bringing a small mass from infinity to that point. At infinity $V = 0$. In the radial field of a point mass $M$:

$V = -\frac{GM}{r}$

It is a scalar, measured in J kg⁻¹, and is always negative because the field does the work as the mass moves in, so energy must be supplied to move it back to infinity. The potential is more negative closer to the mass. A mass $m$ at that point has potential energy $mV$. For the satellite 4.23 × 10⁷ m from the centre of the Earth, $V = -(6.67 \times 10^{-11} \times 5.97 \times 10^{24}) / (4.23 \times 10^7) = -9.4 \times 10^6$ J kg⁻¹.

Similarities. Both are radial fields around a point source. Both obey an inverse square law ($F \propto 1/r^2$). Field strength is force per unit mass (gravitational) or per unit positive charge (electric). Both have a potential that varies as $1/r$.

Differences. Gravitational forces are always attractive; electric forces can be attractive or repulsive. Gravity depends on mass, electric forces on charge, which can be positive or negative. Gravity is far weaker: for an electron and proton in hydrogen, the electric force is about $2 \times 10^{39}$ times the gravitational force. Gravity dominates on astronomical scales because large bodies are neutral overall.