1.2.21 · D5Newton's Laws & Dynamics
Question bank — Variation of g — with altitude, latitude, depth
The single expression that folds every effect together (pure gravity from altitude/depth, then the spin correction to get effective ) is shown below and unpacked case-by-case in the diagrams that follow.
True or false — justify
Both altitude and depth formulas reduce near the surface, so a 1 km climb and a 1 km dig change by the same amount
False. Altitude uses the inverse-square law giving , but depth is linear giving ; going up changes twice as fast as going down for equal distance (see the s01 figure).
The formula works for any height
False. It is a binomial approximation valid only for ; for large (satellites) you must use the exact .
Going down a mine always weakens
Only for a uniform-density Earth. The clean assumes constant ; the real Earth has a dense iron core, so actually rises slightly in the first ~2900 km of depth before falling to zero at the center.
At Earth's exact center,
True. Every bit of mass is now in shells around you, and by the Shell Theorem a shell exerts zero net force on anything inside it, so the pulls cancel from all directions (see the s02 figure).
is largest at the equator because the equator is closest to the Sun's pull
False. The Sun is irrelevant to local ; effective is actually smallest at the equator because rotation flings you outward there and the equatorial bulge puts you farther from Earth's center.
The rotation term vanishes at the poles
True. At the poles so ; a point on the axis traces a circle of zero radius and needs no centripetal force, so no gravity is "spent" there.
If Earth stopped spinning, effective would instantly be the same everywhere
False. The rotation correction would vanish at once, but the oblate shape (Oblate Spheroid Earth) persists — and it relaxes only over geological timescales (millions of years, if at all), so the shape-driven -variation would linger long after the spin stopped.
A satellite in orbit experiences zero gravity
False. Gravity at orbital height is still large (); the astronauts feel weightless because they are in free fall, not because is zero — see Weight vs Mass.
Mass changes as you move from equator to pole
False. Mass is the amount of matter and never changes with location; only weight () changes because effective changes. This is the core distinction in Weight vs Mass.
Spot the error
" is exact, so I can use it for a satellite at ."
The formula is a binomial approximation valid only for . At you must use the exact , not the linear one which would absurdly give .
"Going deeper, , so at gravity is (points upward)."
The formula only applies for (inside Earth); is measured positive downward and can never exceed . Beyond the center there is no Earth to be "below," so the derivation collapses.
"The latitude formula has because we squared the radius."
Wrong reason. One comes from the circle radius , and the second from projecting the horizontal centripetal vector onto the local vertical — two distinct geometric steps, not a squared radius (see the s03 figure).
"At the equator the full centripetal term reduces ; at latitude the reduction is exactly half."
The reduction is , and , so it happens to be half there — but the dependence is , not linear in , so "half the latitude" does not mean "half the effect" in general.
"Effective gravity always points straight toward Earth's center."
Only at the equator and poles. Elsewhere the centripetal correction has a sideways component, so a plumb line hangs slightly off the true center-direction — the "down" you feel is the effective gravity , tilted toward the equator.
"Since shells above you contribute zero, the deeper you go the faster must drop."
The rate is constant for uniform density: is linear, so falls at a steady per metre, not faster and faster.
Why questions
Why does decrease with altitude at all?
Newton's law is inverse-square in distance from the center; at height you sit at , so the pull is weaker — this is Newton's Law of Universal Gravitation directly.
Why is depth's formula exact while altitude's is only approximate?
Inside a uniform sphere is perfectly linear, so needs no approximation; outside, is nonlinear, so we must binomial-expand (assuming ) to get the tidy .
Why does the spinning Earth make you lighter, not the ground push harder?
Part of pure gravity is redirected to supply the centripetal force (Centripetal Force & Circular Motion) that keeps you moving in a circle; only the leftover — effective — holds you down, so the scale reads less.
Why is the rotation effect strongest at the equator?
The equator sits farthest from the spin axis, so it moves in the largest, fastest circle and needs the most centripetal force ( peaks at ), diverting the most gravity.
Why do both the oblate shape and the rotation make polar the largest?
At the poles you are closest to Earth's center (polar radius is smaller) and the rotation effect is zero — both factors push in the same direction, so effective is maximal there.
Why can't we make effective negative by spinning Earth faster?
The equatorial reduction is capped at ; once that equals , effective gravity is exactly zero and objects begin to orbit/float off rather than press down — you reach , never below it. This is the boundary to Escape Velocity & Orbital Mechanics.
Edge cases
What is at the top of the atmosphere versus deep inside the mantle, roughly?
At high altitude smoothly decreases as ; inside the real (non-uniform) Earth can briefly rise due to the dense core before dropping — the uniform model is only a first approximation.
What happens to the latitude formula precisely at and ?
At (equator) gives the maximum reduction ; at (pole) gives no reduction, — the two extremes of the whole formula.
At (the center) both the depth formula and the shell argument must agree — do they?
Yes. , and the shell theorem independently says all mass is now outside you, giving zero net force — two routes, same answer.
If (same distance up and down), which point has more ?
The downward point: when , since depth loses gravity at half the rate of altitude near the surface.
What is the smallest possible "day length" before equatorial gravity vanishes?
Setting gives , a period of roughly 1.4 hours; spin any faster and the equator can no longer hold objects down.
Does the depth formula work on the Moon or a gas planet?
The linear result needs uniform density; a differentiated Moon or a gas giant with a huge density gradient will deviate strongly, so the formula is a model, not a law.
Recall One-line summary of every trap
UP loses twice as fast as DOWN (and only for ); DOWN only works for uniform density and only for ; SPIN converts pure into effective , steals most at the equator, and can at most drive to zero, never negative; mass never changes, only weight does; and the oblate shape outlives the spin because it relaxes only over geological time.
Connections
- Parent topic — the derivations these traps test.
- Shell Theorem — behind every depth and center-of-Earth question.
- Weight vs Mass — the mass-doesn't-change traps live here.
- Oblate Spheroid Earth — the "shape vs spin" separation.
- Escape Velocity & Orbital Mechanics — where the spin limit leads.