Gravitation

Task 1 - The universal law

Hook

If every mass attracts every other mass, then you are pulling on the person sitting next to you right now, and they are pulling on you. Why do you not slowly drift together?

Scientific notation refresher

Gravity involves numbers that are awkward to write out. Scientific notation writes a number as a value between 1 and 10 multiplied by a power of ten. The mass of the Earth, 5 972 000 000 000 000 000 000 000 kg, becomes $5.972 \times 10^{24}\text{ kg}$. A very small number such as 0.000 000 33 becomes $3.3 \times 10^{-7}$. To multiply, multiply the front numbers and add the powers. Calculators usually show the power after a letter E, so $3.3 \times 10^{-7}$ appears as 3.3E-7.

Intuition

The gravitational force between two objects depends on three things. More mass on either side means a stronger pull. More distance between them means a weaker pull, and it weakens fast. The force is shared, so both objects feel exactly the same size of pull toward each other, as Newton's third law requires.

Drag body B away from body A below and watch the force fall along the curve. Notice that the two red arrows always stay the same length as each other, even when the masses are very different.

Formalise

Newton's law of universal gravitation, where $r$ is the distance between the centres of the two masses and $G$ is the gravitational constant.

Worked example

Two 70 kg people stand 1 m apart. How strongly do they attract each other?

1. Substitute. $F = 6.674 \times 10^{-11} \times \dfrac{70 \times 70}{1^2}$.
2. Calculate. $F = 6.674 \times 10^{-11} \times 4900 = 3.3 \times 10^{-7}\text{ N}$.

That is about the weight of a few grains of dust, far smaller than the friction between your shoes and the floor. Gravity is only noticeable when at least one of the masses is enormous, like a planet.

Your turn

Task 2 - Where does g come from?

Hook

Everyone learns that $g = 9.81\text{ N/kg}$ on Earth. Where does that number actually come from, and what would it be on another world?

Intuition

An object of mass $m$ on the surface of a planet of mass $M$ and radius $R$ is pulled by the universal law, with the distance equal to the planet's radius. The pull on each kilogram, the force divided by $m$, is the field strength $g$. The object's own mass cancels out, so $g$ depends only on the planet.

Formalise

Worked example

The Earth has mass $M = 5.972 \times 10^{24}\text{ kg}$ and radius $R = 6.371 \times 10^{6}\text{ m}$.

1. Top. $GM = 6.674 \times 10^{-11} \times 5.972 \times 10^{24} = 3.986 \times 10^{14}$.
2. Bottom. $R^2 = (6.371 \times 10^6)^2 = 4.059 \times 10^{13}$.
3. Divide. $g = 3.986 \times 10^{14} / 4.059 \times 10^{13} = 9.82\text{ N/kg}$.

The same calculation for the Moon gives about 1.62 N/kg, roughly one sixth of Earth's value. A 70 kg astronaut weighs 687 N on Earth and only 113 N on the Moon.

Your turn

Mars has a mass of $6.42 \times 10^{23}\text{ kg}$ and a radius of $3.39 \times 10^{6}\text{ m}$.

Task 3 - Falling

Hook

Drop a bowling ball and a tennis ball from the same height at the same moment. Which one hits the ground first?

Intuition

The bowling ball is pulled down by a much larger force, but it also has much more mass to accelerate. The two effects cancel exactly, so without air both balls accelerate at $g$ and land together. Astronauts on Apollo 15 tested this on the Moon with a hammer and a feather.

Air changes things. As an object speeds up, air resistance grows. Eventually it becomes as large as the weight, the net force becomes zero, and the object stops speeding up. It falls at a steady speed called terminal velocity. A heavier object needs a larger air resistance to balance its weight, so it reaches a higher terminal velocity.

Try the diagram with the air off first, then on. Make the two masses very different and watch the speed curves separate.

Formalise

With no air resistance, the only force is the weight.

Worked example

Describe a skydiver's fall using free body diagrams.

1. Just after jumping. Speed is low, so air resistance is tiny. The net force is nearly the full weight and the acceleration is close to $g$.
2. Speeding up. Air resistance grows with speed. The net force and the acceleration shrink.
3. Terminal velocity. Air resistance equals weight. Net force is zero, so speed stays constant, often around 55 m/s for a skydiver lying flat.
4. Parachute opens. Air resistance suddenly becomes much larger than the weight. The net force points upward, so the skydiver slows down until a new, much lower terminal velocity is reached.

Your turn

Task 4 - Orbits

Hook

Newton imagined a cannon on top of an impossibly tall mountain, firing cannonballs horizontally. Each shot is faster than the last. What happens to the cannonballs?

Intuition

A slow cannonball curves down and lands near the mountain. A faster one lands further away, and because the Earth is round, the ground has curved away a little beneath it. At one particular speed, the cannonball falls toward the Earth at exactly the rate the surface curves away. It never lands. That is an orbit.

Fire the cannon below at different speeds. Try below orbital speed, exactly orbital speed, a little above it, and above 1.41 times orbital speed.

Formalise

Moving in a circle needs a force toward the centre. The acceleration toward the centre needed to keep turning in a circle of radius $r$ at speed $v$ is $v^2 / r$. In orbit, gravity supplies that force. Setting gravity equal to mass times this acceleration gives the orbital speed. The time for one full orbit, the period $T$, is the circumference divided by the speed.

Worked example

The International Space Station orbits about 408 km above the Earth's surface. Find its speed and period.

1. Orbit radius. Measure from the Earth's centre. $r = 6.371 \times 10^6 + 0.408 \times 10^6 = 6.779 \times 10^6\text{ m}$.
2. Speed. $v = \sqrt{3.986 \times 10^{14} / 6.779 \times 10^6} = 7670\text{ m/s}$, about 7.67 km/s.
3. Period. $T = 2\pi \times 6.779 \times 10^6 / 7670 = 5550\text{ s}$, about 92.6 minutes.

The ISS circles the planet about 15 times a day. Notice that the station's own mass cancelled out. Every object at that height orbits at the same speed, whether it is a space station or a spanner.

Your turn

Task 5 - Practice set and game connection

Game connection

Most platform games fake gravity as a constant downward acceleration added to every object each frame. That is the near-surface case from Task 2, and it works because levels are tiny compared with a planet. Designers often pick a value much larger than 9.81, sometimes three or four times larger, because a strong gravity makes jumps feel quick and responsive. That is a design decision, not a physics mistake.

Space games with planets need the full inverse square law. Each frame, for every ship, they add up the pull from every nearby planet, each using $GM/r^2$ pointed toward that planet's centre. That is exactly what the cannon diagram does every frame.

Practice set