3.1 - Newton’s Universal Law of Gravitation

Gravitational Force and Newton’s Law

Gravitational force is an attractive force between bodies. It is a universal force because it acts between any two bodies in the universe, including the Sun, Earth and Moon.
Gravitational attraction brings a jumping person back to the ground, holds air molecules around Earth and keeps the Moon orbiting Earth.
Both bodies experience forces of equal magnitude towards each other. A falling leaf and Earth attract each other; Earth’s much greater mass makes its movement imperceptible.
Newton’s Universal Law of Gravitation relates the force to the product of the masses and the square of their centre-to-centre distance:

F \propto m_{1} m_{2}, F \propto \frac{1}{r ^{2}}

F = \frac{G m _{1} m _{2}}{r ^{2}}

Symbol	Meaning and unit
$F$	Gravitational force, $N$
$m_{1}$ , $m_{2}$	Masses of the two bodies, $kg$
$r$	Distance between their centres, $m$
$G$	Gravitational constant: $6.67 \times 1 0^{- 11} N m^{2} k g^{- 2}$

For a 2.0 kg durian, with Earth’s mass $5.97 \times 1 0^{24} kg$ and $r = 6.37 \times 1 0^{6} m$ :

F = \frac{( 6.67 \times 1 0 ^{- 11} ) ( 2.0 ) ( 5.97 \times 1 0 ^{24} )}{( 6.37 \times 1 0 ^{6} ) ^{2}} = 19.63 N

A rocket at the same distance experiences $F = 4.98 \times 1 0^{5} N$ . Its mass is:

m = \frac{F r ^{2}}{GM} = \frac{( 4.98 \times 1 0 ^{5} ) ( 6.37 \times 1 0 ^{6} ) ^{2}}{( 6.67 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} )} = 5.07 \times 1 0^{4} kg

With distance fixed, increasing either mass increases the force.
With masses fixed, increasing the distance decreases the force according to the inverse-square relationship.
Attraction between small masses is very weak. Two people of masses 50 kg and 70 kg, separated by 1 m, experience a force of only $2.3 \times 1 0^{- 7} N$ .
Apply the same law to bodies on Earth’s surface, Earth and a satellite, or Earth and the Sun. Use the distance between centres for $r$ .

Let $M$ be Earth’s mass, $m$ the object’s mass and $r$ the distance between their centres. Equate Newton’s Second Law with gravitational attraction:

m g = \frac{GM m}{r ^{2}} \Rightarrow g = \frac{GM}{r ^{2}}

Cancelling $m$ shows that $g$ does not depend on the object’s mass. It depends on Earth’s mass and the distance from Earth’s centre.
At Earth’s surface, $r = R$ , where $R$ is Earth’s radius:

g = \frac{GM}{R ^{2}}

g = \frac{GM}{( R + h ) ^{2}}

The graph of $g$ against $r$ shows $g \propto r$ below the surface ( $r < R$ ), reaching its maximum at the surface. For $r \geq R$ , $g \propto 1/ r^{2}$ and approaches zero far from Earth.
With $M = 5.97 \times 1 0^{24} kg$ and $R = 6.37 \times 1 0^{6} m$ :

g_{surface} = \frac{( 6.67 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} )}{( 6.37 \times 1 0 ^{6} ) ^{2}} = 9.81 m s^{- 2}

g = \frac{( 6.67 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} )}{( 6.37 \times 1 0 ^{6} + 480 000 ) ^{2}} = 8.49 m s^{- 2}

Use $g = GM / R^{2}$ to compare surface gravitational acceleration on planets, the Moon and the Sun, using each body’s mass and radius.
Knowing $g$ allows the force on an object to be calculated using $F = m g$ . This supports space exploration and continuity of life.
Astronauts travelling far from Earth or near other planets may experience low or high gravity. The human factors considered include density changes, bone fragility, lung size, blood circulation and blood pressure.

In uniform circular motion, speed remains constant but the direction of motion, and therefore velocity, changes continuously.
Centripetal force acts towards the centre of the circle. Its magnitude depends on mass $m$ , linear speed $v$ and radius $r$ :

F = \frac{m v ^{2}}{r}

For a body rotated at uniform speed with the string almost horizontal, the effect of gravitational force on its circular motion can be ignored.
In the Centripetal Force Kit, the hanging slotted weights create tension in the string, which supplies the centripetal force on the rotating rubber stopper.
For fixed $m$ and $r$ , a higher speed requires a larger force; for fixed $m$ and $v$ , a smaller radius requires a larger force.
A 7.2 kg iron ball moving at $20 m s^{- 1}$ in a circle of radius 1.8 m experiences:

F = \frac{7.2 ( 20 ) ^{2}}{1.8} = 1600 N

Newton predicted that an object launched at low speed would return to Earth, whereas sufficient linear speed could keep it moving around Earth.
An orbiting satellite can continue moving without rocket thrust. Earth’s gravitational force provides its centripetal force, directed towards Earth’s centre.
Equating $F = ma$ with $F = m v^{2} / r$ gives centripetal acceleration:

a = \frac{v ^{2}}{r}

For a weather satellite at $h = 480 km$ , with $v = 7.62 \times 1 0^{3} m s^{- 1}$ and $R = 6.37 \times 1 0^{6} m$ :

a = \frac{( 7.62 \times 1 0 ^{3} ) ^{2}}{6.37 \times 1 0 ^{6} + 480 000} = 8.48 m s^{- 2}

v = \frac{2 π r}{T}

Equate gravitational and centripetal forces, cancel the orbiting body’s mass $m$ , substitute for $v$ and rearrange:

\frac{GM m}{r ^{2}} = \frac{m v ^{2}}{r} \Rightarrow GM = v^{2} r

GM = (\frac{2 π r}{T})^{2} r \Rightarrow M = \frac{4 π ^{2} r ^{3}}{G T ^{2}}

To determine Earth’s mass, use the orbital radius and period of the Moon or a satellite.
To determine the Sun’s mass, use the orbital radius and period of a planet.
Use $r$ in metres and $T$ in seconds with the SI value of $G$ .