# Newton's Law of Gravitation ## Newton's Law of Gravitation - The gravitational force between two masses, for instance between Mars and its moon Phobos, is defined by Newton's Law of Gravitation - Newton's Law of Gravitation states: **The gravitational force $F$ between two masses $m_1$ and $m_2$ is proportional to the product of their masses and inversely proportional to the square of their separation, $r$** - In equation form, this is written as: $$F = -\frac{Gm_1m_2}{r^2}$$

Newton’s Law of Gravitation: F = -Gm₁m₂/r², where F is the attractive force between two point masses.

- Where: - $F$ = gravitational force between two point masses $m_1$ and $m_2$ (N) - $G$ = Newton's gravitational constant ($6.67 \times 10^{-11}$ Nm² kg⁻²) - $m_1$ and $m_2$ = mass of body 1 and mass of body 2 (kg) - $r$ = distance between the centre of the two masses (m) ### The Inverse Square Law - The $1/r^2$ relation is called the **'inverse square law'** - This means that if the distance between two masses doubles, $r$ becomes $2r$ - Therefore, $1/r^2$ becomes $1/(2r)^2$, which is equal to $1/4r^2$ - Hence, the gravitational force between the two masses reduces by a factor of **four** - Consider what happens when distance triples: - $r$ becomes $3r$ - $1/r^2$ becomes $1/(3r)^2 = 1/9r^2$ - The force reduces by a factor of **nine**

The gravitational force follows an inverse square law: doubling distance quarters the force, tripling distance reduces it to one-ninth

- The graph below shows how gravitational force decreases as separation increases for two 1000 kg masses:

- Notice the characteristic curve shape: the force drops rapidly at first, then more gradually as distance increases - This is fundamentally different from a linear relationship where force would decrease at a constant rate ### The Negative Sign - The **negative sign** indicates that the gravitational force $F$ between the two point masses $m_1$ and $m_2$ is **attractive** - This is fundamental to gravity: unlike electric forces (which can be repulsive), gravitational forces always pull masses together - The direction of force is always along the line connecting the two masses, pointing towards each mass --- ## Worked Example ### Problem A communications satellite with mass 4800 kg is orbiting Earth at 3500 km above the Earth's surface. The magnitude of the gravitational force between them is 22 kN. Calculate the mass of the Earth. (Radius of the Earth = 6400 km) ### Answer **Step 1: Write Newton's Law of Gravitation** $$F = \frac{Gm_1m_2}{r^2}$$ Where: - $m_1$ = mass of the satellite = 4800 kg - $m_2$ = mass of the Earth (to find) - $F$ = 22 kN = 22 × 10³ N **Step 2: Rearrange for $m_2$ (mass of Earth)** $$m_2 = \frac{Fr^2}{Gm_1}$$ **Step 3: Calculate the distance $r$** $r$ is the distance between the **centre of the Earth** and the **centre of the satellite** Therefore: $$r = \text{distance above surface} + \text{radius of Earth}$$ $$r = 3500 + 6400 = 9900 \text{ km} = 9900 \times 10^3 = 9.9 \times 10^6 \text{ m}$$ **Step 4: Substitute values** | Variable | Value | |----------|-------| | $F$ | $22 \times 10^3$ N | | $r$ | $9.9 \times 10^6$ m | | $G$ | $6.67 \times 10^{-11}$ Nm²kg⁻² | | $m_1$ | 4800 kg | $$m_2 = \frac{(22 \times 10^3) \times (9.9 \times 10^6)^2}{(6.67 \times 10^{-11}) \times 4800}$$ $$m_2 = \frac{(22 \times 10^3) \times (9.801 \times 10^{13})}{3.20 \times 10^{-7}}$$ $$m_2 = 6.7 \times 10^{24} \text{ kg}$$

When calculating r, always measure from centre to centre of the masses — not from surface to surface.

--- ## Key Points to Remember ### Common Mistakes - **Forgetting to add distances correctly:** - Remember to add together the distance from the surface of the planet and its radius to obtain the value of $r$ - The distance $r$ is measured between the **centre** of each mass - For a satellite problem: $r$ = distance from centre of planet to centre of satellite - **Forgetting to square the distance:** - The distance between point masses $m_1$ and $m_2$ is **squared** in the denominator - Remember this whenever you use Newton's Law of Gravitation! - It's $r^2$, not $r$ - **Calculating magnitude vs. direction:** - When calculating the **magnitude** of gravitational force $F$, you can ignore the negative sign - The negative sign indicates direction (attractive), not size of the force - Be clear whether the question asks for magnitude or includes direction

Always draw a diagram showing the two masses, their centres, and the distance r between centres to avoid confusion.

--- ## Summary Table | Quantity | Symbol | Unit | Notes | |----------|--------|------|-------| | Gravitational force | $F$ | N (newtons) | Always attractive (negative sign) | | Mass 1 | $m_1$ | kg | Can be either mass | | Mass 2 | $m_2$ | kg | Can be either mass | | Separation | $r$ | m | Centre to centre distance | | Gravitational constant | $G$ | Nm²kg⁻² | $6.67 \times 10^{-11}$ |