Newton's Law of Gravitation | CAIE 9702 Paper 4 Notes

Paper 4

Physics9702

Newton's Law of Gravitation

Newton found one rule for the size of the pull between any two masses.

It is the starting equation for the whole chapter: field strength, potential and orbits all come from it.

The law

Increasing either mass makes the pull stronger. Increasing the centre-to-centre distance makes the pull weaker.
The distance between the two centres is called their . Its symbol here is $r$ .
This rule is called .
The law treats each object as a : all of its mass is treated as if it is at one position.

The forces have the same size and opposite directions. In Newton's law, r is the distance from centre to centre, not the empty gap between the surfaces.

Definition

2 marks

What is meant by Newton's law of gravitation?

Model answer: The gravitational force between two point masses is directly proportional to the product of their masses and inversely proportional to the square of their separation (9702/41/O/N/24 Q1(a)).

F = \dfrac{GMm}{r^2}

Symbols

$F$ = size of the gravitational force on either mass (N)
$G$ = gravitational constant, 6.67 × 10⁻¹¹ (N m² kg⁻²)
$M,\ m$ = the two masses (kg)
$r$ = separation measured from centre to centre (m)

The two masses feel the same size of force in opposite directions (Newton's third law). The force on the small mass has the same size as the force on the large mass.
means: double the distance and the force becomes a quarter; triple the distance and the force becomes a ninth.
For a point outside a uniform sphere, put all of the sphere's mass at its centre. Therefore, $r$ is measured centre to centre.

Worked example

Calculate the force between two small spheres

Two small spheres, each of mass 100 g, have their centres 1.0 cm apart. Find the gravitational pull between them.

Put the numbers in kilograms and metres.
$F = \dfrac{GMm}{r^2}$
$G=6.67\times10^{-11}$
$Mm=(0.10)(0.10)$
$Mm=0.010$
$GMm=6.67\times10^{-13}$
$r^2=(0.010)^2$
$r^2=1.0\times10^{-4}$
Evaluate the expression.
$F = 6.7 \times 10^{-9} \text{ N}$

Answer

6.7 \times 10^{-9}\ \text{N}

This force is very small. The pull between everyday objects is usually too small to notice.

Your turn9702-style

The separation of the same two spheres becomes 3.0 cm. Without repeating the full calculation, find the new force.

Show worked answer

The distance changes from 1.0 cm to 3.0 cm, so it becomes three times larger.
Use the inverse-square relation.
$F \propto \dfrac{1}{r^2}$
The new force is one ninth of its first value.
$F = \dfrac{6.7 \times 10^{-9}}{9}$
$F = 7.4 \times 10^{-10}\ \text{N}$

Answer: 7.4 × 10⁻¹⁰ N.

Common mistake

r

is the distance centre to centre, not the gap between the surfaces, and it is squared. Using the surface gap, or forgetting the square, are the usual errors.