The Gravitational Force and the Gravitational Field Revision Notes for HSC SSCE Physics

The Gravitational Force and the Gravitational Field

Introduction to universal gravitation

Understanding forces is fundamental to physics. Newton's development of the Law of Universal Gravitation was a groundbreaking achievement because it was the first mathematical model to describe a fundamental force that applies everywhere in the universe. Before Newton, there was no single equation that could explain both how objects fall on Earth and how planets move through space.

infoNote

Newton's Law of Universal Gravitation works for all objects with mass, from tiny atoms to massive galaxies. This universality was revolutionary - it showed that the same mathematical principles govern both "heavenly" motion (planets and stars) and "earthly" motion (falling objects). This was shocking to many people at the time, who believed celestial and terrestrial physics were fundamentally different.

Newton's Law of Universal Gravitation

The gravitational force equation

Newton's law states that every object with mass attracts every other object with mass. The strength of this attractive force depends on:

The masses of both objects
The distance between them

The mathematical formula is:

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

Where:

$F$ is the gravitational force (in newtons, N)
$G$ is the universal gravitational constant = $6.67 \times 10^{-11} \text{ N kg}^{-2} \text{ m}^2$
$M$ and $m$ are the masses of the two objects (in kg)
$r$ is the distance between the centres of the two objects (in m)

This equation shows that gravitational force follows an inverse square law - if you double the distance between two objects, the force becomes one-quarter as strong.

The gravitational force always points toward the object creating it, making gravity an attractive force. Each object pulls on the other - the small mass is pulled toward the large mass, and the large mass is pulled toward the small mass (although the effect on the large mass may be harder to observe).

lightbulbExample

Worked Example: Gravitational force between Earth and the Sun

Let's calculate the gravitational force that the Sun exerts on Earth.

Given data:

Mass of Earth, $M_E = 5.97 \times 10^{24}$ kg
Mass of the Sun, $M_S = 2.00 \times 10^{30}$ kg
Mean radius of Earth's orbit, $r = 1.50 \times 10^{11}$ m

Solution:

Using Newton's Law of Universal Gravitation:

$F = \frac{GM_S M_E}{r^2}$

Substituting the values:

$F = \frac{(6.67 \times 10^{-11} \text{ N kg}^{-2} \text{ m}^2)(2.00 \times 10^{30} \text{ kg})(5.97 \times 10^{24} \text{ kg})}{(1.50 \times 10^{11} \text{ m})^2}$

$F = 3.54 \times 10^{22} \text{ N}$

The force is directed towards the Sun.

chatImportant

Exam tip: Always include the direction when stating a force as your final answer. For gravitational forces, this is always toward the object creating the force.

Gravity and Newton's third law

Newton's third law tells us that forces always come in pairs. If object A exerts a force on object B, then object B exerts an equal and opposite force on object A. We write this as:

$F_{a \text{ on } b} = -F_{b \text{ on } a}$

This applies to gravitational forces too. The force the Sun exerts on Earth is matched by an equal and opposite force that Earth exerts on the Sun.

For the Sun and Earth:

Force of Sun on Earth: $F_{S \text{ on } E} = -\frac{GM_S M_E}{r^2}$ (pointing toward the Sun)
Force of Earth on Sun: $F_{E \text{ on } S} = \frac{GM_E M_S}{r^2}$ (pointing toward Earth)

Both forces have the same magnitude but opposite directions, so:

$F_{S \text{ on } E} = -F_{E \text{ on } S}$

lightbulbExample

Worked Example: Earth's force on the Sun

What is the gravitational force that Earth exerts on the Sun?

Solution:

From Newton's third law, we know that:

$F_{S \text{ on } E} = -F_{E \text{ on } S}$

From our previous calculation, $F_{S \text{ on } E} = 3.54 \times 10^{22}$ N toward the Sun.

Therefore:

$F_{E \text{ on } S} = -F_{S \text{ on } E} = -3.54 \times 10^{22} \text{ N}$

The negative sign indicates the opposite direction. So the force Earth exerts on the Sun is $3.54 \times 10^{22}$ N directed toward Earth.

infoNote

Key point: The forces are equal in magnitude but opposite in direction. Even though the Sun is much more massive than Earth, both objects experience the same size force - they just respond differently due to their different masses.

The gravitational field

Definition and concept

Like electric and magnetic forces, gravity acts at a distance without direct contact. We explain this using the concept of a gravitational field. Just as we defined the electric field as force per unit charge, we define the gravitational field as the force per unit mass acting on a small test mass:

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

The units of gravitational field are newtons per kilogram (N kg⁻¹) or, equivalently, metres per second squared (m s⁻²).

Combining this definition with Newton's Law of Universal Gravitation gives us:

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

This equation tells us that:

Every object with mass creates a gravitational field around it
The field strength depends on the mass $M$ of the object creating the field
The field strength decreases with the square of the distance from the object
The field is a vector quantity pointing toward the mass creating it

chatImportant

The distance $r$ is measured from the centre of the object, not from its surface. This equation works for point masses and spherically symmetric objects (like planets and stars).

Gravitational field lines

We can visualise gravitational fields using field lines, similar to electric field lines. These lines show:

The direction of the gravitational field (and therefore the direction of force on a test mass)
The strength of the field (indicated by the density of field lines - closer lines mean stronger field)

infoNote

Key difference from electric fields: Gravitational field lines always point toward the mass creating the field, never away from it. This is because gravity is always attractive, whereas electric charges can be positive or negative.

Gravitational field at Earth's surface

Let's find Earth's gravitational field at its surface using:

Mass of Earth, $M_{\text{Earth}} = 5.97 \times 10^{24}$ kg
Radius of Earth, $r_{\text{Earth}} = 6.37 \times 10^6$ m

$g_{\text{Earth surface}} = \frac{GM_{\text{Earth}}}{r_{\text{Earth}}^2} = \frac{(6.67 \times 10^{-11} \text{ N kg}^{-2} \text{ m}^2)(5.97 \times 10^{24} \text{ kg})}{(6.37 \times 10^6 \text{ m})^2}$

$g_{\text{Earth surface}} = 9.8 \text{ N kg}^{-1} = 9.8 \text{ m s}^{-2}$

This is the familiar acceleration due to gravity that we use in many physics problems. The gravitational field strength equals the acceleration that gravity gives to falling objects.

lightbulbExample

Worked Example: Gravitational field on the Moon's surface

Calculate the acceleration due to gravity on the surface of the Moon.

Given data:

Mass of Moon, $M_M = 7.3 \times 10^{22}$ kg
Radius of Moon, $r = 1740$ km $= 1.74 \times 10^6$ m

Solution:

Using the gravitational field equation:

$g_{\text{Moon}} = \frac{GM_M}{r^2}$

Substituting the values:

$g_{\text{Moon}} = \frac{(6.67 \times 10^{-11} \text{ N kg}^{-2} \text{ m}^2)(7.3 \times 10^{22} \text{ kg})}{(1.74 \times 10^6 \text{ m})^2}$

$g_{\text{Moon}} = 1.61 \text{ N kg}^{-1} = 1.6 \text{ m s}^{-2}$

This shows that gravity on the Moon is about one-sixth the strength of gravity on Earth (1.6/9.8 ≈ 1/6), which is why astronauts can jump much higher on the Moon!

The near-Earth approximation

When we're close to Earth's surface (or any large object's surface), and our movements are small compared to Earth's radius, the gravitational field doesn't change much. In this situation, we can treat the gravitational field as constant.

In this near-Earth approximation:

The gravitational field lines are parallel to each other
The field is perpendicular to Earth's surface
We can use $F = mg$ with constant $g = 9.8$ m s⁻²

infoNote

This approximation is what we use for most everyday physics problems involving projectiles, falling objects, or objects on Earth's surface. It's only when we consider large changes in altitude (like satellites in orbit) that we need to use the full equation $g = \frac{GM}{r^2}$ and account for changes in gravitational field strength.

chatImportant

Exam tip: Always check whether the problem involves distances comparable to Earth's radius. If altitude changes are small (a few kilometres), use the near-Earth approximation. If they're large (thousands of kilometres), use the full gravitational field equation.

bookmarkSummary

Key Points to Remember:

Newton's Law of Universal Gravitation: $F = \frac{GMm}{r^2}$ describes the attractive force between any two masses, where $G = 6.67 \times 10^{-11}$ N kg⁻² m²
Gravitational forces are always attractive and always point toward the object creating the force
Newton's third law applies to gravity: If object A pulls on object B with force $F$ , then object B pulls on object A with force $-F$ (equal magnitude, opposite direction)
Gravitational field is defined as force per unit mass: $g = \frac{F}{m} = \frac{GM}{r^2}$ , with units N kg⁻¹ or m s⁻²
At Earth's surface: $g = 9.8$ m s⁻², which is both the gravitational field strength and the acceleration due to gravity
Field lines always point toward masses (unlike electric fields), showing that gravity is purely attractive
Near-Earth approximation: For small altitude changes, treat $g$ as constant; for large changes, use the full inverse square law equation

The Gravitational Force and the Gravitational Field (HSC SSCE Physics): Revision Notes