Circular orbit — velocity, period, energy: Question bank

The four core facts everything below leans on, so you don't have to scroll: $v = \frac{GM}{r}, T = 2 π \frac{r ^{3}}{GM}, E = - \frac{GM m}{2 r}, K = + \frac{GM m}{2 r}, U = - \frac{GM m}{r}$ Here $G$ is Newton's gravitational constant, $M$ the big central mass, $m$ the orbiting mass, $r$ the orbit radius (measured from the center, not the surface), $v$ the speed, $T$ the period, $K$ / $U$ / $E$ the kinetic/potential/total energy.

True or false — justify

A heavier satellite at the same altitude orbits slower than a light one.

False. In

\frac{GM m}{r ^{2}} = \frac{m v ^{2}}{r}

the satellite mass

m

cancels on both sides, so

v = GM / r

has no

m

in it — a feather and a bus at the same

r

share the same speed and period.

A higher circular orbit has more total energy but less kinetic energy.

True.

E = - \frac{GM m}{2 r}

rises toward zero as

r

grows (more total energy), yet

K = \frac{GM m}{2 r}

shrinks — the extra energy all went into raising the (negative) potential energy

U

If total energy

E

is exactly zero, the orbit is a very large circle.

False.

E = 0

is the escape boundary: the path is a parabola, not a closed circle. A circle needs

E = - \frac{GM m}{2 r} < 0

; any circle has strictly negative energy.

Doubling the orbit radius doubles the period.

False.

T \propto r^{3/2}

, so doubling

r

multiplies

T

2^{3/2} \approx 2.83

, not by 2. Only quantities linear in

r

would double.

For a circular orbit, kinetic energy equals minus the total energy.

True.

K = + \frac{GM m}{2 r}

and

E = - \frac{GM m}{2 r}

, so

K = - E

. This

K = - E

U = 2 E

pattern is the virial theorem for a

1/ r

force — see Virial Theorem.

The Moon and a low satellite feel the same gravity per kilogram, so they orbit at the same speed.

False. Gravity per kilogram is

GM / r^{2}

, which depends on

r

. The Moon is far out (large

r

), so its

g

is much weaker and its orbital speed

v = GM / r

is much smaller than a low satellite's.

E = 0

the object is "just barely escaped," meaning it arrives at infinity at rest.

True.

E = K + U

; as

r \to \infty

U \to 0

, so

E = 0

forces

K \to 0

, i.e. it coasts to infinity with vanishing speed. See Escape Velocity (the

E = 0

boundary,

v_{esc} = 2 v_{c i r c}

Escape speed at a given radius is exactly twice the circular speed there.

False. It is

2

times, not

2

times.

v_{esc} = 2 GM / r = 2 GM / r = 2 v_{c i r c}

— the factor lives under the square root.

Spot the error

"Use

U = m g h

to find the satellite's potential energy at 400 km."

The error is using the flat-Earth constant-

g

formula. Over orbital distances

g = GM / r^{2}

changes a lot, so you must use

U = - \frac{GM m}{r}

;

m g h

is only the tangent-line approximation near the surface — see Gravitational Potential Energy.

"Since a higher orbit has more energy, the rocket must speed the satellite up to get there."

Deceptive. The final orbit is slower (

v = GM / r

drops with

r

). The added energy raises

U

, not

K

. Firing forward does start the transfer, but the destination circle has lower kinetic energy than the start.

"Gravity is the centripetal force plus an extra outward centrifugal force that balances it."

There is no real outward force in the inertial frame. Gravity alone is the centripetal force; nothing "balances" it — an unbalanced inward force is exactly what bends a straight path into a circle. See Centripetal Force and Uniform Circular Motion.

"Because

T^{2} \propto r^{3}

depends on

GM

, it also depends on the satellite's mass

m

The constant is

\frac{4 π ^{2}}{GM}

— only the central mass

M

appears, not the orbiting

m

. Newton's law gives

\frac{GM m}{r ^{2}} = \frac{m v ^{2}}{r}

and the

m

's cancel before

T

is ever computed.

"The satellite stays up because gravity is too weak to reach it out there."

Gravity absolutely reaches it — that inward pull is precisely what curves the orbit. The satellite stays up because its sideways speed makes the ground curve away as fast as it falls, not because gravity is absent.

K = \frac{1}{2} m v^{2}

and

U = - \frac{GM m}{r}

, so

E = K + U = + \frac{GM m}{2 r}

Sign slip.

K = + \frac{GM m}{2 r}

and

U = - \frac{GM m}{r}

, so

E = \frac{GM m}{2 r} - \frac{GM m}{r} = - \frac{GM m}{2 r}

— negative, because

∣ U ∣ = 2 K

Why questions

Why does the satellite's mass drop out of the speed but not out of the energy?

Speed comes from a force balance where

m

appears on both sides and cancels. Energies

K, U, E

are the actual joules stored, and doubling the mass genuinely doubles each of them — there's no second

m

to cancel against.

Why is the total energy of every bound orbit negative?

We set

U = 0

at infinity; a bound object sits below that reference because pulling it in from infinity released energy. Negative

E

literally means "you'd have to add energy to set it free."

Why does

T^{2} \propto r^{3}

come out of Newton for free, when Kepler had to discover it from data?

Substituting

v = GM / r

into

T = 2 π r / v

gives

T = 2 π r^{3} / GM

automatically. Kepler saw the pattern in planetary data; Newton showed it must follow from an inverse-square force — see Kepler's Three Laws and Newton's Law of Universal Gravitation.

Why must gravity supply exactly

\frac{m v ^{2}}{r}

— no more, no less — for a circle?

Too much inward force curves the path tighter than a circle (it spirals in); too little lets the body drift outward (an ellipse or escape). Only the exact match keeps

r

constant, which is the definition of a circle.

Why is a higher orbit slower even though it has more energy?

Both

v

and

E

change with

r

but in opposite senses:

v \sim r^{- 1/2}

falls, while

E \sim - r^{- 1}

rises toward zero. The rising energy is stored as potential energy, so speed can drop while total energy grows.

Why can we treat gravity as pointing exactly at the center for the whole orbit?

In a circle every point is the same distance

r

from the center, and gravity from a spherical mass always points radially inward. So the force keeps constant magnitude

GM m / r^{2}

and always aims at the center — perfect for supplying a constant centripetal pull.

Edge cases

What happens to

v

r \to \infty

(orbit infinitely large)?

v = GM / r \to 0

. An infinitely distant circular orbit has vanishing speed and, by

T \sim r^{3/2}

, an infinite period — it barely moves.

What happens to

v

r \to 0

(skimming the center)?

v = GM / r \to \infty

. Mathematically the required speed blows up; physically you'd hit the surface first, so real orbits have a minimum radius (the body's radius plus atmosphere).

If you set the orbital speed but let

r

be the actual surface radius, is that a valid orbit?

Only in the idealized airless, mountain-free limit. On a real planet the surface, terrain, and atmosphere make grazing orbits impossible; the formula stays valid but the geometry doesn't.

What does

E > 0

(positive total energy) correspond to?

An unbound trajectory — a hyperbola. The body has more than enough energy to reach infinity with speed to spare, so it never returns; this is beyond the circular regime and links to Elliptical Orbits and the Vis-viva Equation.

What is special about the orbit where

v

equals the local escape speed?

No such circular orbit exists: reaching

v_{esc} = 2 v_{c i r c}

at that radius gives

E = 0

, which is a parabola, not a circle. Any speed above circular at that

r

opens the path into an ellipse, parabola, or hyperbola.

If two satellites share the same period, what must be true of their radii?

They must have the same radius.

T = 2 π r^{3} / GM

is one-to-one in

r

for a fixed

M

, so equal periods force equal

r

(assuming they orbit the same central body).

Recall One-line self-test before you close this page

Say aloud: "Why is a higher orbit slower yet higher-energy?" Answer ::: Because $v \sim r^{- 1/2}$ falls while $E \sim - r^{- 1}$ rises toward zero — the extra energy is stored as potential energy, not speed.

Connections

Parent: Circular orbit — velocity, period, energy
Newton's Law of Universal Gravitation — the inward force we balance.
Centripetal Force and Uniform Circular Motion — supplies the $m v^{2} / r$ requirement.
Kepler's Three Laws — why $T^{2} \propto r^{3}$ is no coincidence.
Gravitational Potential Energy — origin of $U = - GM m / r$ and why $m g h$ fails.
Escape Velocity — the $E = 0$ boundary, $v_{esc} = 2 v_{c i r c}$ .
Elliptical Orbits and the Vis-viva Equation — where $E > 0$ and $E = 0$ paths live.
Virial Theorem — explains $K = - E$ , $U = 2 E$ .