Revise Gravitational field of a point mass for Cambridge 9702 (13.3)

A gravitational field is essentially a region of space surrounding a mass where another mass would experience an attractive force. For most external calculations, especially with large, uniform objects like planets or stars, we simplify things by treating them as if all their mass is concentrated at a single point – known as a point mass at their centre. This makes calculations much more manageable.

Sir Isaac Newton's groundbreaking law describes the fundamental attraction between any two masses. It states that the attractive force between them is directly proportional to the product of their masses and inversely proportional to the square of the distance separating their centres. This inverse square law is crucial for understanding how gravity weakens with distance.

$F = G \frac{m_1 m_2}{r^2}$

Gravitational field strength ($g$) quantifies how strong the gravitational pull is at a particular point. It's defined as the gravitational force per unit mass acting on a small test mass placed at that point. Importantly, $g$ also represents the acceleration experienced by a free-falling object in that field. Field lines are radial, point inwards, and their density indicates the field's strength.

We can derive the formula for gravitational field strength by combining Newton's Law of Gravitation with the definition of a field.

Derivation:

1. Start with the definition of gravitational field strength: the force per unit test mass, $m$. $g = \frac{F}{m}$
2. Recall Newton's Law of Gravitation for the force $F$ between the source mass $M$ and the test mass $m$. $F = \frac{GMm}{r^2}$
3. Substitute the expression for $F$ into the equation for $g$. $g = \frac{\left( \frac{GMm}{r^2} \right)}{m}$
4. The test mass $m$ cancels out, leaving the formula for the field strength created by the source mass $M$. $g = \frac{GM}{r^2}$ This shows that the field strength at a point depends only on the source mass and the distance from it, not on the test mass placed there.

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

Gravitational potential ($\phi$) is a scalar quantity representing the work done per unit mass to move a test mass from infinity to a specific point within the field. By convention, the gravitational potential at infinity is defined as zero. Since gravity is always attractive and work is done by the field as a mass approaches the source, gravitational potential is always a negative value.

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

There is a direct mathematical relationship between gravitational field strength ($g$) and gravitational potential ($\phi$). The field strength at a point is equal to the negative of the potential gradient at that point. The potential gradient is the rate of change of potential with distance. This means that a region with a rapidly changing potential (a steep slope on a potential-distance graph) will have a strong gravitational field.

$g = -\frac{\Delta\phi}{\Delta r}$

The negative sign is important: it signifies that the gravitational field vector ($g$) always points in the direction of decreasing potential. Think of it like a ball rolling downhill – it moves from a higher potential to a lower potential, and the direction of its acceleration is down the steepest slope. For a point mass, potential becomes less negative (increases) as you move away from the mass, and the field $g$ points inwards, towards the mass, in the direction of decreasing potential.

This relationship is visualized in graphs. A graph of potential ($\phi$) against distance ($r$) for a point mass is a curve in the negative quadrant, approaching zero as $r$ approaches infinity. The gradient of this graph at any point $r$ is equal to $-g$. A graph of field strength ($g$) against distance ($r$) shows an inverse square relationship, always positive, and also approaching zero as $r$ increases.

Gravitational potential energy ($E_p$) is the energy an object possesses due to its position within a gravitational field. For a system of two masses, it represents the work done to assemble them from infinite separation to their current distance. Like gravitational potential, $E_p$ is also a negative value. To completely remove an object from a field, you'd need to supply energy equal to the negative of its current potential energy.

$E_p = -\frac{GMm}{r}$