Gravitational Fields

Every object with mass attracts every other object with mass. Newton's law of gravitation makes this quantitative: two point masses $M$ and $m$ separated by a distance $r$ attract each other with a force given by:

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

Here $G = 6.67 \times 10^{-11}\ \text{N m}^2\ \text{kg}^{-2}$ is the universal gravitational constant. The force acts along the line joining the masses and is always attractive.

Two features matter most. First, this is an inverse-square law: doubling the separation cuts the force to a quarter, and tripling it cuts the force to a ninth.

Second, the two forces form a Newton's third law pair: the Earth pulls on you with a force equal to your weight, and you pull back on the Earth with an equal and opposite force; the Earth simply does not accelerate noticeably because its mass is enormous.

Although the law is stated for point masses, a uniform sphere behaves as if all its mass were concentrated at its centre, so $r$ is always measured centre to centre, not surface to surface.

You should be able to calculate the force between two bodies and reason proportionally about how $F$ changes when the masses or the separation change.

A field is how physics describes action at a distance: a mass modifies the space around it, and any other mass placed in that region experiences a force. The gravitational field strength $g$ at a point is the force per unit mass on a small test mass placed there, $g = \frac{F}{m}$, with units $\text{N kg}^{-1}$ (equivalent to $\text{m s}^{-2}$). It is a vector pointing towards the mass creating the field.

For a point or spherical mass $M$, combining $F = \frac{GMm}{r^2}$ with $g = \frac{F}{m}$ gives:

$$g = \frac{GM}{r^2}$$

At the Earth's surface this evaluates to about $9.8\ \text{N kg}^{-1}$; at a distance of two Earth radii from the centre it falls to a quarter of that, roughly $2.45\ \text{N kg}^{-1}$.

Field lines make the pattern visible. For a spherical mass the lines:

• are radial
• point inwards
• crowd together where the field is stronger

Close to a planet's surface, over distances small compared with its radius, the lines are effectively parallel and evenly spaced, which is why $g$ is treated as uniform in projectile problems.

You should be able to calculate $g$ at any distance, compare surface values for different planets, and sketch both radial and uniform field patterns.

A satellite in a circular orbit is in constant free fall: the gravitational pull supplies exactly the centripetal force needed to keep it turning. (HL) Setting $\frac{GMm}{r^2} = \frac{mv^2}{r}$ gives the orbital speed:

$$v_{\text{orbital}} = \sqrt{\frac{GM}{r}}$$

This depends only on the central mass and the orbital radius, never on the satellite's own mass.

Substituting $v_{\text{orbital}} = \frac{2\pi r}{T}$ produces Kepler's third law for circular orbits:

$$T^2 = \frac{4\pi^2}{GM} r^3$$

Wider orbits therefore have much longer periods; this is why a geostationary satellite must sit at one particular radius for its period to match the Earth's 24-hour rotation.

Kepler's laws summarise orbital motion in general:

• First law: each planet moves in an ellipse with the Sun at one focus (a circle is the special case).
• Second law: the line from the Sun to a planet sweeps out equal areas in equal times, so a planet moves fastest at its closest approach.
• Third law: $T^2 \propto r^3$ for every body orbiting the same central mass.

You should be able to derive the third law from Newton's law of gravitation and use ratios such as $\frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}$ to compare two orbits without knowing $G$ or $M$.

Gravitational potential $V_g$ at a point is the work done per unit mass in bringing a small test mass from infinity to that point, measured in $\text{J kg}^{-1}$:

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

Because the zero of potential is set at infinity and gravity is attractive, $V_g$ is negative everywhere; the field does the work as the mass falls inwards. The potential energy of a specific mass $m$ follows directly: $E_p = m V_g = -\frac{GMm}{r}$.

Moving away from a planet makes the potential less negative, so it increases. For a change of orbit the useful form is:

$$\Delta E_p = GMm\left(\frac{1}{r_i} - \frac{1}{r_f}\right)$$

This is positive whenever $r_f > r_i$, matching the intuition that raising a satellite costs energy.

Equipotential surfaces around a spherical mass are concentric spheres. They are always perpendicular to field lines, and no work is done moving a mass along one. The field strength is the negative gradient of the potential, $g = -\frac{\Delta V_g}{\Delta r}$, so on a graph of $V_g$ against $r$ a steep slope means a strong field.

You should be able to calculate $V_g$ and $E_p$, find energy changes between orbits, and read field strengths from potential-distance graphs.

Escape speed is the minimum launch speed that lets an unpowered projectile leave a body's gravitational field completely. It comes from energy conservation: escaping 'just barely' means arriving at infinity with nothing to spare, so the total energy must be at least zero. Setting $\frac{1}{2}mv_{esc}^2 - \frac{GMm}{r} = 0$ gives:

$$v_{esc} = \sqrt{\frac{2GM}{r}}$$

The projectile's own mass cancels: from the Earth's surface, $v_{esc} \approx 11.2\ \text{km s}^{-1}$ for a pebble or a spacecraft alike.

For a satellite in a circular orbit of radius $r$:

• kinetic energy: $E_k = \frac{GMm}{2r}$
• potential energy: $E_p = -\frac{GMm}{r}$
• total mechanical energy: $E_{total} = E_k + E_p = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}$

This negative value represents a bound state: it indicates that the system lacks the energy to escape to infinity.

For a satellite to move to a wider orbit with a larger radius, its potential energy must increase (becoming less negative) while its kinetic energy must decrease. Because the potential energy increase is twice as large as the kinetic energy decrease, the satellite's overall total energy must increase, requiring positive work to be done on the satellite by an external source such as its thrusters.

The same accounting explains atmospheric drag: a satellite in low orbit loses total energy, spirals inwards to a smaller $r$, and, counter-intuitively, ends up moving faster. You should be able to derive $v_{esc}$, compute each energy term, and sketch how $E_k$, $E_p$ and $E_{total}$ vary with $r$.