Gravity: Mass and Distance described gravity in words: more mass, more pull; more distance, less pull. Newton made it exact, and the same law describes a falling apple and the Moon's orbit.
Key idea
Every two masses attract each other with $F = \dfrac{Gm_1m_2}{r^2}$, where $r$ is the distance between their centers and $G = 6.674 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$. Each object feels the same size force (a third-law pair).
Why it works
The pull grows with each mass, so it is proportional to the product $m_1m_2$. It spreads out in all directions from a mass, over spheres whose areas grow as $r^2$, so it weakens as $1/r^2$: twice as far, a quarter of the pull. Newton showed that a uniform sphere pulls on things outside it as if all its mass were at its center, which is why $r$ is measured center to center.
Worked examples
Worked example: Two people
Two $70\ \text{kg}$ students stand $1\ \text{m}$ apart. Find the gravitational force between them.
- $F = \dfrac{(6.674 \times 10^{-11})(70)(70)}{1^2}$.
- $F \approx 3.27 \times 10^{-7}\ \text{N}$, far too small to feel. Gravity is weak unless a mass is enormous.
Answer: about $3.27 \times 10^{-7}\ \text{N}$
Worked example: A satellite
A $1{,}000\ \text{kg}$ satellite orbits $400\ \text{km}$ above Earth ($M = 5.972 \times 10^{24}\ \text{kg}$, $R = 6.371 \times 10^{6}\ \text{m}$). Find Earth's pull on it.
- Center to center: $$r = 6.371 \times 10^{6} + 4.00 \times 10^{5} = 6.771 \times 10^{6}\ \text{m}.$$
- $$F = \dfrac{(6.674 \times 10^{-11})(5.972 \times 10^{24})(1000)}{(6.771 \times 10^{6})^2} \approx 8.69 \times 10^{3}\ \text{N}.$$
- That is almost 90% of its $9{,}800\ \text{N}$ weight on the ground.
Answer: about $8{,}690\ \text{N}$
Worked example: Scaling
If the distance between two objects triples, what happens to the force?
- $F \propto \dfrac{1}{r^2}$, so the force is multiplied by $\dfrac{1}{3^2} = \dfrac19$.
Answer: It becomes $\tfrac19$ as large.
Common mistake
Measuring $r$ from the planet's surface instead of its center. A satellite $400\ \text{km}$ up is $6{,}771\ \text{km}$ from Earth's center, and that is the $r$ to square. Check: for an object near the ground, your $r$ should be close to Earth's radius, never close to zero.
Try it in the Science Lab's Orbit Lab.