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1.2.24 · D5Newton's Laws & Dynamics

Question bank — Orbital velocity for circular orbit — derivation

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Throughout, the symbols mean exactly what the parent note built them to mean:

  • = the constant tangential (sideways) speed for a circular orbit.
  • = distance from the centre of the planet to the satellite, i.e. ( = planet radius, = altitude).
  • = planet's mass, = satellite's mass, = gravitational constant.
  • Master formula: .

True or false — justify

A satellite in a stable circular orbit is not accelerating.
False. Its speed is constant but its direction changes every instant, so it has a real inward (centripetal) acceleration — velocity is a vector, and turning is accelerating.
Two satellites of very different mass at the same radius orbit at the same speed.
True. In the force balance the satellite mass cancels, leaving — only the planet's mass and the radius matter.
A higher orbit means a faster satellite.
False. , so a larger radius gives a smaller speed — low satellites race, distant ones amble.
Gravity does positive work on a satellite in a perfectly circular orbit.
False. Gravity points toward the centre while velocity points tangentially — they are perpendicular, so the work is zero and the speed never changes.
Escape velocity from a given radius equals the orbital velocity there.
False. — escape speed is always about 41% larger than the circular-orbit speed at the same radius.
If you switched off gravity mid-orbit, the satellite would fly straight off tangentially.
True. With no inward force, Newton's first law takes over and the body keeps its current velocity — a straight line tangent to the old circle, not a curve outward.
The formula works for a satellite orbiting far above the surface.
False. That shortcut uses ; it is only valid near the surface. For altitude you must use with .
Doubling the planet's mass doubles the required orbital speed at a fixed radius.
False. , so doubling multiplies the speed by , not by 2.

Spot the error

"To orbit at altitude 400 km, plug km into ."
The error is : it is measured from Earth's centre, so km. Using km gives a wildly wrong (far too large) speed.
"A heavier satellite needs a stronger rocket to stay in the same orbit, so it must move faster."
Both the gravitational pull and the required centripetal force grow in exact proportion to , so cancels — the needed speed is identical for any mass.
", so a bigger orbit needs a bigger speed because is bigger."
The sits in the denominator under the root: larger makes the fraction smaller, so decreases. The reasoning ignored where lives in the formula.
"Gravity pushes the satellite forward along its path to keep it moving."
Gravity points inward (toward the planet's centre), perpendicular to the motion — it bends the path but never speeds it up. In vacuum there is no forward push and none is needed.
"Since decreases with altitude, use the surface value for any orbit."
You may keep the surface only if you use the identity and the true ; writing at high altitude double-counts by ignoring how changed.
"We add gravity and the centripetal force to get the net inward force."
There is no separate "centripetal force" to add — gravity is the centripetal force here. The derivation sets equal to the required ; they are the same single force viewed two ways.
"Escape velocity is because you need to fully overcome gravity."
It is , not . The factor comes from the energy condition , which yields , not .

Why questions

Why does the satellite's own mass disappear from the final formula?
Because both sides of carry a factor ; cancelling it shows the orbit speed depends only on and , mirroring how all masses fall at the same rate.
Why must gravity exactly equal , not more or less?
If gravity were larger the path would spiral inward; if smaller, the satellite would drift outward. Only exact equality keeps the radius constant — the defining condition of a circular orbit.
Why do we measure from the centre, not the surface?
Newton's law treats a spherical planet as if all its mass sits at its centre, so the relevant distance is centre-to-satellite, which is .
Why is a low satellite so fast (≈ 7.9 km/s) yet the Moon so slow (≈ 1 km/s)?
: the Moon is roughly 60 Earth-radii out, so times slower — the great distance weakens the pull and slackens the pace.
Why does the derivation use specifically?
Because turning at constant speed still requires acceleration, and geometry shows an object on a circle of radius at speed curves inward at exactly this rate — that's the force the orbit "demands."
Why does the substitution help in practice?
We often don't know and separately, but we always know a planet's surface gravity and radius ; the identity lets us compute from measurable quantities.

Edge cases

What would predict as ?
The speed tends to zero — infinitely far out, gravity is negligibly weak, so an orbiting body would barely need to move. (In reality it also takes an infinite time.)
What does the formula "want" as (orbit at the very centre)?
It blows up to infinity, which is a signal the model breaks down: you can't orbit inside the planet, and the point-mass law no longer holds there.
Is a circular orbit at the surface (, ) physically real?
Mathematically km/s, but physically air resistance and mountains make it impossible — it's the idealized limiting case, not an achievable orbit.
Does the formula apply to an elliptical orbit?
No. assumes constant speed and constant radius; in an ellipse both change (fast near the planet, slow far away), so this single-speed result doesn't hold.
If a satellite's speed at radius is slightly less than , what happens?
Gravity now exceeds the centripetal requirement, so the path curves inward — the orbit becomes an ellipse dipping closer to the planet, not a stable circle.
What if the speed is slightly more than at that radius?
Gravity is too weak to bend the faster motion into the same circle, so the satellite swings outward into a larger elliptical orbit (and if fast enough, escapes entirely).

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