Gravitation • Topic 1 of 3

Universal Gravitation & g

Every object in the universe attracts every other object. This single idea, stated by Isaac Newton in 1687, explains why an apple falls and why the Moon stays in orbit. Newton's law of universal gravitation says the force of attraction between two point masses $m_1$ and $m_2$ separated by a distance $r$ is directly proportional to the product of the masses and inversely proportional to the square of the distance between them:

  • $F=\frac{Gm_1m_2}{r^2}$, where $G$ is the universal gravitational constant.
  • $G=6.67\times10^{-11}\ \text{N}\,\text{m}^2\,\text{kg}^{-2}$ — the same value everywhere in the universe.
  • The force is always attractive and acts along the line joining the two masses (a central force).

Gravitation is the weakest of the four fundamental forces, but because it is always attractive and has unlimited range, it dominates on the scale of planets, stars and galaxies. The force is an action–reaction pair: the Earth pulls a stone with the same magnitude of force with which the stone pulls the Earth.

Acceleration due to gravity ($g$) is the acceleration a body experiences because of the Earth's pull. Treating the Earth as a sphere of mass $M$ and radius $R$, the weight $mg$ of a body of mass $m$ at the surface equals the gravitational force, so $mg=\frac{GMm}{R^2}$ giving $g=\frac{GM}{R^2}$. Notice that $g$ is independent of the mass of the falling body — that is why all objects fall with the same acceleration in vacuum. At the Earth's surface $g\approx9.8\ \text{m/s}^2$.

Variation of g. The value of $g$ is not truly constant — it changes with position:

  • With altitude (height $h$): $g_h=g\left(1-\frac{2h}{R}\right)$ for $h\ll R$. The value of $g$ decreases as we go up.
  • With depth $d$: $g_d=g\left(1-\frac{d}{R}\right)$. The value of $g$ also decreases going down, becoming zero at the centre of the Earth.
  • With rotation and latitude: the spin of the Earth reduces effective $g$: $g_{\lambda}=g-R\omega^2\cos^2\lambda$. The reduction is maximum at the equator ($\lambda=0$) and zero at the poles ($\lambda=90^\circ$), so $g$ is greatest at the poles.

This is why a body weighs slightly more at the poles than at the equator, and why $g$ falls off both as you climb a mountain and as you descend into a mine.

Newton's law of gravitation between two massesm1m2rF (on m1)F (on m2)F = G m1 m2 / r^2
Solved Examples
1
Worked Example
Two masses of 5 kg and 10 kg are placed 2 m apart. Find the gravitational force between them. ($G=6.67\times10^{-11}\ \text{N}\,\text{m}^2\,\text{kg}^{-2}$)
Solution
  1. Step 1: Use $F=\frac{Gm_1m_2}{r^2}$.
  2. Step 2: Substitute: $F=\frac{6.67\times10^{-11}\times5\times10}{2^2}=\frac{6.67\times10^{-11}\times50}{4}$.
  3. Step 3: Compute: $F=\frac{3.335\times10^{-9}}{4}=8.34\times10^{-10}\ \text{N}$.

Answer: $F\approx8.34\times10^{-10}\ \text{N}$ (attractive).

2
Worked Example
If the distance between two masses is doubled, how does the gravitational force change?
Solution
  1. Step 1: Since $F\propto\frac{1}{r^2}$, replacing $r$ by $2r$ gives $F'\propto\frac{1}{(2r)^2}=\frac{1}{4r^2}$.
  2. Step 2: Therefore $F'=\frac{F}{4}$.
  3. Step 3: The force becomes one-fourth of the original.

Answer: The force reduces to one-fourth, $\frac{F}{4}$.

3
Worked Example
Calculate the value of $g$ on the Earth's surface given $M=6\times10^{24}\ \text{kg}$, $R=6.4\times10^{6}\ \text{m}$ and $G=6.67\times10^{-11}$ SI units.
Solution
  1. Step 1: Use $g=\frac{GM}{R^2}$.
  2. Step 2: Substitute: $g=\frac{6.67\times10^{-11}\times6\times10^{24}}{(6.4\times10^{6})^2}=\frac{4.0\times10^{14}}{4.096\times10^{13}}$.
  3. Step 3: Compute: $g\approx9.77\ \text{m/s}^2$.

Answer: $g\approx9.8\ \text{m/s}^2$.

4
Worked Example
At what height above the Earth's surface will the value of $g$ become one-fourth of its value at the surface? (Take $R=6400\ \text{km}$.)
Solution
  1. Step 1: At height $h$, $g_h=\frac{GM}{(R+h)^2}=g\left(\frac{R}{R+h}\right)^2$.
  2. Step 2: Set $g_h=\frac{g}{4}$, so $\left(\frac{R}{R+h}\right)^2=\frac{1}{4}$, giving $\frac{R}{R+h}=\frac{1}{2}$.
  3. Step 3: Thus $R+h=2R$, so $h=R=6400\ \text{km}$.

Answer: $h=R=6400\ \text{km}$.

5
Worked Example
Find the value of $g$ at a depth equal to half the radius of the Earth. Take $g=9.8\ \text{m/s}^2$ at the surface.
Solution
  1. Step 1: Use $g_d=g\left(1-\frac{d}{R}\right)$ with $d=\frac{R}{2}$.
  2. Step 2: Substitute: $g_d=9.8\left(1-\frac{R/2}{R}\right)=9.8\left(1-\frac{1}{2}\right)$.
  3. Step 3: Compute: $g_d=9.8\times0.5=4.9\ \text{m/s}^2$.

Answer: $g_d=4.9\ \text{m/s}^2$.

6
Worked Example
The mass of a body on Earth is 60 kg. What is its mass and weight on the Moon, where $g_{moon}=1.6\ \text{m/s}^2$? (Take $g_{earth}=9.8\ \text{m/s}^2$.)
Solution
  1. Step 1: Mass is a fixed amount of matter and does not change with location: mass on Moon $=60\ \text{kg}$.
  2. Step 2: Weight on Moon $=m\,g_{moon}=60\times1.6$.
  3. Step 3: Compute: weight $=96\ \text{N}$ (for comparison, weight on Earth $=60\times9.8=588\ \text{N}$).

Answer: Mass $=60\ \text{kg}$; weight on Moon $=96\ \text{N}$.

Key Points

  • Newton's law of gravitation: $F=\frac{Gm_1m_2}{r^2}$; the force is always attractive and acts along the line joining the masses.
  • $G=6.67\times10^{-11}\ \text{N}\,\text{m}^2\,\text{kg}^{-2}$ is universal; $g=\frac{GM}{R^2}\approx9.8\ \text{m/s}^2$ and is independent of the falling body's mass.
  • With altitude: $g_h=g\left(1-\frac{2h}{R}\right)$ — $g$ decreases as you go up.
  • With depth: $g_d=g\left(1-\frac{d}{R}\right)$ — $g$ decreases with depth and is zero at the Earth's centre.
  • Rotation makes $g$ greatest at the poles and least at the equator: $g_{\lambda}=g-R\omega^2\cos^2\lambda$.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.The gravitational force between two bodies depends on the distance $r$ between them as:
Explanation: Newton's law gives $F=\frac{Gm_1m_2}{r^2}$, so $F\propto\frac{1}{r^2}$.
Q2.The acceleration due to gravity $g$ at the Earth's surface is given by:
Explanation: From $mg=\frac{GMm}{R^2}$ we get $g=\frac{GM}{R^2}$.
Q3.As we go to greater depths inside the Earth, the value of $g$:
Explanation: $g_d=g\left(1-\frac{d}{R}\right)$, which falls to zero when $d=R$ (the centre).
Q4.The value of $g$ is maximum at the:
Explanation: Rotation reduces effective $g$ by $R\omega^2\cos^2\lambda$; at the poles ($\lambda=90^\circ$) this term is zero, so $g$ is greatest there.
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