Gravitation • Topic 1 of 3

Universal Law of Gravitation & g

Why do things fall? Long before Newton, people knew apples drop and rivers flow downhill, but nobody could explain why the Moon does not fall to Earth the way an apple does. Newton's brilliant leap was to realise that the same force pulls the apple down and keeps the Moon in orbit. He called it gravitation, and he stated it as a single rule that works for any two objects in the universe.

Newton's Universal Law of Gravitation: Every object in the universe attracts every other object with a force that is (a) directly proportional to the product of their masses, and (b) inversely proportional to the square of the distance between their centres. In symbols:

  • $F \propto m_1 m_2$ — double either mass and the force doubles.
  • $F \propto \dfrac{1}{r^2}$ — double the distance and the force falls to one-quarter (the inverse-square law).

Combining these gives the famous equation $F = \dfrac{G m_1 m_2}{r^2}$, where $G$ is the universal gravitational constant. Its value, measured by Henry Cavendish, is $G = 6.674 \times 10^{-11}\ \text{N m}^2/\text{kg}^2$. Because $G$ is so tiny, the pull between everyday objects is far too weak to feel — gravitation only becomes obvious when at least one body (like a planet) is enormous.

Acceleration due to gravity ($g$): When Earth pulls a freely falling object, the object accelerates. This acceleration is given the special symbol $g$, and near Earth's surface $g \approx 9.8\ \text{m/s}^2$ (often rounded to $10\ \text{m/s}^2$ for quick calculations). Crucially, $g$ does not depend on the falling object's mass — a feather and a coin fall with the same $g$ in a vacuum.

How $g$ is linked to the law: Put the falling object of mass $m$ at Earth's surface. The gravitational pull is $F = \dfrac{G M m}{R^2}$, where $M$ is Earth's mass and $R$ its radius. But this force also equals $mg$ (force = mass × acceleration). Setting them equal, the $m$ cancels and we get $g = \dfrac{G M}{R^2}$. This single result explains a lot: $g$ depends only on the planet's mass and radius, not on the object dropped.

Why $g$ varies:

  • Altitude: higher up, $r$ increases, so $g$ decreases.
  • Depth: deep inside Earth only the mass below pulls, so $g$ decreases, reaching zero at the centre.
  • Shape: Earth is flattened at the poles, so $g$ is largest at the poles ($\approx 9.83$) and smallest at the equator ($\approx 9.78\ \text{m/s}^2$).

The same law explains the orbits of planets and why a satellite stays up: it keeps ‘falling’ toward Earth but moves sideways fast enough to miss it.

Two masses attracting each other with gravitational force F over distance rUniversal Law of Gravitationm1m2FFr (centre to centre)F = G m1 m2 / r squaredThe two forces are equal in size and opposite in direction (action-reaction).
1
Worked Example
State Newton's universal law of gravitation and write its mathematical form, naming every symbol.
Solution
  1. The law: every object attracts every other object with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.
  2. In symbols, $F \propto m_1 m_2$ and $F \propto \dfrac{1}{r^2}$.
  3. Combining, $F = \dfrac{G m_1 m_2}{r^2}$.
  4. Here $F$ = gravitational force, $m_1, m_2$ = the two masses, $r$ = distance between their centres, $G$ = universal gravitational constant $= 6.674 \times 10^{-11}\ \text{N m}^2/\text{kg}^2$.

Answer: $F = \dfrac{G m_1 m_2}{r^2}$, the inverse-square law of gravitation.

2
Worked Example
Two bodies of masses $50\ \text{kg}$ and $80\ \text{kg}$ are placed $2\ \text{m}$ apart. Find the gravitational force between them. (Take $G = 6.67 \times 10^{-11}\ \text{N m}^2/\text{kg}^2$.)
Solution
  1. Write the formula: $F = \dfrac{G m_1 m_2}{r^2}$.
  2. Substitute: $F = \dfrac{(6.67 \times 10^{-11})(50)(80)}{(2)^2}$.
  3. Numerator: $6.67 \times 10^{-11} \times 4000 = 2.668 \times 10^{-7}$.
  4. Divide by $r^2 = 4$: $F = \dfrac{2.668 \times 10^{-7}}{4} = 6.67 \times 10^{-8}\ \text{N}$.

Answer: $F \approx 6.67 \times 10^{-8}\ \text{N}$ — an extremely small force, which is why we never feel the pull between ordinary objects.

3
Worked Example
The gravitational force between two objects is $F$. What happens to the force if the distance between them is doubled, all masses unchanged?
Solution
  1. Force depends on distance as $F \propto \dfrac{1}{r^2}$.
  2. New distance $= 2r$, so new force $F' \propto \dfrac{1}{(2r)^2} = \dfrac{1}{4r^2}$.
  3. Therefore $F' = \dfrac{1}{4} F$.

Answer: The force becomes one-quarter of the original (the inverse-square law).

4
Worked Example
Show that the acceleration due to gravity at Earth's surface is $g = \dfrac{GM}{R^2}$, and hence find $g$. (Take $M = 6 \times 10^{24}\ \text{kg}$, $R = 6.4 \times 10^{6}\ \text{m}$, $G = 6.67 \times 10^{-11}$.)
Solution
  1. An object of mass $m$ at the surface feels $F = \dfrac{GMm}{R^2}$.
  2. This same force gives it acceleration $g$, so $F = mg$.
  3. Equate: $mg = \dfrac{GMm}{R^2}$; cancel $m$ to get $g = \dfrac{GM}{R^2}$.
  4. Substitute: $g = \dfrac{(6.67 \times 10^{-11})(6 \times 10^{24})}{(6.4 \times 10^{6})^2}$.
  5. Numerator $= 4.0 \times 10^{14}$; denominator $= 4.096 \times 10^{13}$.
  6. $g = \dfrac{4.0 \times 10^{14}}{4.096 \times 10^{13}} \approx 9.8\ \text{m/s}^2$.

Answer: $g = \dfrac{GM}{R^2} \approx 9.8\ \text{m/s}^2$, independent of the falling object's mass.

5
Worked Example
If both masses are tripled and the distance between them is also tripled, by what factor does the gravitational force change?
Solution
  1. Original: $F = \dfrac{G m_1 m_2}{r^2}$.
  2. New masses $3m_1$ and $3m_2$ give a numerator factor of $3 \times 3 = 9$.
  3. New distance $3r$ gives a denominator factor of $(3)^2 = 9$.
  4. New force $F' = \dfrac{G(3m_1)(3m_2)}{(3r)^2} = \dfrac{9}{9} \cdot \dfrac{G m_1 m_2}{r^2} = F$.

Answer: The force is unchanged (the factor of 9 on top cancels the factor of 9 on the bottom).

6
Worked Example
A planet has twice Earth's mass and twice Earth's radius. Find the value of $g$ on this planet compared with Earth's $g$.
Solution
  1. Surface gravity: $g = \dfrac{GM}{R^2}$.
  2. For the planet, $M' = 2M$ and $R' = 2R$.
  3. $g' = \dfrac{G(2M)}{(2R)^2} = \dfrac{2GM}{4R^2} = \dfrac{1}{2} \cdot \dfrac{GM}{R^2} = \dfrac{g}{2}$.
  4. With Earth's $g = 9.8\ \text{m/s}^2$, the planet's $g' = 4.9\ \text{m/s}^2$.

Answer: $g' = \dfrac{g}{2} = 4.9\ \text{m/s}^2$.

Key Points

  • Newton's law: $F = \dfrac{G m_1 m_2}{r^2}$ — force grows with the product of masses and falls with the square of the distance.
  • The universal gravitational constant is $G = 6.674 \times 10^{-11}\ \text{N m}^2/\text{kg}^2$ and is the same everywhere in the universe.
  • Acceleration due to gravity near Earth is $g \approx 9.8\ \text{m/s}^2$ and does not depend on the falling object's mass.
  • $g = \dfrac{GM}{R^2}$ links $g$ to a planet's mass $M$ and radius $R$, not to the dropped object.
  • $g$ varies: it decreases with altitude and with depth, and is largest at the poles, smallest at the equator.
Tap an option to check your answer0 / 4
Q1.In $F = \dfrac{G m_1 m_2}{r^2}$, if the distance $r$ is halved, the force becomes:
Explanation: $F \propto 1/r^2$; halving $r$ multiplies the force by $1/(1/2)^2 = 4$.
Q2.The value of the universal gravitational constant $G$ is about:
Explanation: $G = 6.674 \times 10^{-11}\ \text{N m}^2/\text{kg}^2$ (measured by Cavendish).
Q3.The acceleration due to gravity $g$ on a falling stone depends on:
Explanation: $g = GM/R^2$ depends on the planet, not on the falling object.
Q4.Where is the value of $g$ greatest on Earth's surface?
Explanation: Earth is flattened at the poles (smaller $R$), so $g = GM/R^2$ is largest there.