Gravity Beyond the Ground - Physics for Pilot Assessments

So far gravity has been a constant downward pull of about 10 newtons per kilogram, which is a perfectly good description near the Earth's surface. This lesson steps back to the general picture: gravity acts between any two masses anywhere, it weakens with distance in a specific way, and understanding that turns orbits and weightlessness from mysteries into consequences.

Gravity is universal and mutual

Every mass attracts every other mass. The Earth pulls on you, and you pull on the Earth with exactly the same size of force in the opposite direction — which is Newton's third law, applied to gravity. You do not notice your half of it because the Earth's enormous mass means that force produces an utterly negligible acceleration, while the same-sized force on you produces a very noticeable one. The reason gravity between everyday objects is undetectable is simply that the effect is extremely weak unless at least one of the masses is astronomically large.

The inverse-square law

The strength of the pull depends on both masses and on the distance between their centres — and the distance enters as a square, which makes the fall-off steep.

How gravity weakens with distance

F \propto \frac{1}{r^2}

$F$: the gravitational force, in newtons (N)
$r$: distance between the centres of the two masses, in metres (m)
$\propto$: "is proportional to" — the force scales this way as r changes

Squaring the distance means doubling it does not halve the force — it quarters it. Three times the distance leaves a ninth. Ten times leaves a hundredth. The measurement is always taken from the centre of the mass, not its surface, which matters for a body the size of a planet: standing on the ground you are already 6,400 kilometres from the Earth's centre, so climbing a mountain or flying at cruising altitude changes r by a fraction of a per cent. That is why g feels perfectly constant in everyday life despite obeying this law.

The same two masses at one, two and three times the separation. The arrows are in the ratio 9 : 2.25 : 1 — a quarter and then a ninth of the original pull.

An orbit is a fall that keeps missing

Fire something horizontally and, as the projectiles lesson showed, it falls while it travels forwards. Fire it faster and it travels further before landing. Newton's insight was to ask what happens if you fire it fast enough — and the answer is that the Earth's surface curves away beneath it at the same rate as it falls, so it never reaches the ground. It is still falling, continuously and exactly as any dropped object falls; it simply keeps missing the planet.

Faster and faster horizontal launches from the same point. Each falls the same way, but travels further before meeting the ground — until the path closes and never meets it at all.

That balance also fixes how fast the satellite has to travel — orbiting is not possible at just any speed. The satellite is in free fall, so gravity gives it an acceleration of g towards the centre. The circular-motion lesson showed that following a circle of radius r at speed v requires a centripetal acceleration of $\frac{v^2}{r}$ . An orbit is exactly the case where the one supplies the other, nothing more and nothing less: $g = \frac{v^2}{r}$ . Rearranged, the orbital speed is the square root of g times r.

Orbital speed for a circular orbit

v = \sqrt{g \, r}

$v$: orbital speed, in metres per second (m/s)
$g$: gravitational acceleration at the orbit's height, in m/s² (about 10 close to the Earth)
$r$: radius of the orbit, measured from the Earth's CENTRE, in metres (m)

Two things in that formula deserve a second look. The radius is measured from the Earth's centre, not from the ground — for a satellite skimming just above the atmosphere, r is essentially the Earth's radius itself. And the g that appears is the g at the orbit's height: climb to a higher orbit and the inverse-square law from earlier in this lesson makes g smaller, so a higher orbit is a slower orbit. That is why nothing about orbiting is a matter of engine power once the speed is reached — the orbit's height fixes the speed, and the speed fixes the height.

Example 1 — how fast is fast enough?

How fast must a satellite in a low circular orbit, just above the atmosphere, be travelling? Take g = 10 m/s² and the Earth's radius as 6,400 km.

Set up the numbers. For a low orbit, r is essentially the Earth's radius: 6,400 km = 6.4 × 10⁶ m, and g is still close to its surface value of 10 m/s².
Multiply inside the square root first: $g \times r = 10 \times 6.4 \times 10^6 = 6.4 \times 10^7 = 64 \times 10^6$ The last form is the one whose square root can be taken by eye.
Take the root of each factor: the square root of 64 is 8, and the square root of 10⁶ is 10³. So: $v = 8 \times 10^3 = 8{,}000\ \text{m/s}$ That is eight kilometres every second.
Sense-check it with a lap of the planet. The Earth's circumference is about 40,000 km, so one orbit takes: $\frac{40{,}000\ \text{km}}{8\ \text{km/s}} = 5{,}000\ \text{s}$ That is a little over 83 minutes, which is why low-orbiting satellites and space stations circle the Earth in roughly an hour and a half.
Notice what never appeared: the satellite's mass. Like everything else in free fall, a heavy satellite and a light one orbit at exactly the same speed.

Weightlessness

Astronauts float, and it is not because gravity is absent — at the height of a low orbit it is only slightly weaker than at the ground. They float because they and their spacecraft are falling together at the same rate. There is nothing pressing them into a floor, no normal force, and it is that contact force which we actually feel as weight in ordinary life. The correct name for their condition is free fall, and you can sample it briefly in a lift whose cable is released, or at the top of a fairground drop, where the same absence of a supporting force produces the same lurch.

Eclipses: the same geometry, no forces needed

One purely geometric consequence is worth including because it appears in test items. The Moon orbits the Earth while the Earth orbits the Sun, so occasionally the three line up. With the Moon between Earth and Sun, its shadow falls on us — a solar eclipse. With the Earth in the middle, our shadow falls on the Moon — a lunar eclipse. Nothing about forces is needed to answer these; light travels in straight lines, so a shadow falls exactly where the geometry says it must, and the whole question comes down to knowing which body sits in the middle.

Putting it to work

The same mass, different worlds

Because weight is mass multiplied by the local gravitational field strength, the same object weighs different amounts on different bodies while its mass never changes. On Earth g is about 10 newtons per kilogram in round figures; on the Moon roughly 1.6; on Mars about 3.7; on Jupiter around 25. A 60 kilogram astronaut therefore weighs about 600 newtons on Earth, under 100 on the Moon, a little over 200 on Mars, and around 1,500 on Jupiter — and is 60 kilograms of matter throughout. Questions like to give you a mass and an unfamiliar g and ask for the weight, which is nothing more than the multiplication from the mass-and-weight lesson with a different number substituted.

The reason these values differ comes straight from the inverse-square law together with each body's mass. A large mass pulls harder, and a large radius means the surface is further from the centre, so surface gravity depends on both. The Moon has both less mass and a smaller radius than the Earth, and the mass difference wins, leaving it with about a sixth of our surface gravity.

A last point of vocabulary, since questions use both words. Gravitational field strength and gravitational acceleration are the same quantity measured in two ways: newtons per kilogram if you are thinking of it as the pull per unit of mass, and metres per second squared if you are thinking of it as the acceleration it produces on something falling freely. That they are numerically identical is Newton's second law applied to weight — divide the weight by the mass and you get both the field strength and the acceleration, because W = mg and F = ma are the same equation when the force is gravity. So an item quoting g in either unit is telling you the same thing.

Why it matters in aviation

Two consequences of the inverse-square law matter to flying, and they pull in opposite directions. The first is reassuring. Gravity does weaken with distance from the Earth's centre, but a cruising altitude is a very small addition to the Earth's radius, so the reduction over the whole band of heights aircraft occupy amounts to a fraction of one per cent. That is far below the precision of any performance or loading calculation, which is why a single value of g is used throughout aviation and why no weight is ever corrected for altitude.

The second is that the same law which holds the aircraft down is what keeps satellites up. A satellite is in free fall with gravity as the only force acting, travelling sideways fast enough that its curved path keeps pace with the curvature of the Earth beneath it, and its orbital period is set by its distance rather than by anything carried on board. That predictability is what makes satellite navigation work: the orbits of a navigation constellation are known precisely, so a receiver that measures its distance from several satellites at once can solve for its own position, which is how an aircraft fixes where it is without reference to the ground.

What the test asks

The most common claim to reject is that there is no gravity in space — there is, and it is what keeps a satellite in orbit. The second is treating an orbit as a balance between gravity and an outward force, which reintroduces the centrifugal force the previous lesson dismantled: there is only one force, pointing inwards. The third is arithmetic — doubling the distance and halving the force, when squaring makes it a quarter.