What is the total mechanical energy of a circular orbit?

E = -GMm/(2r). It is negative because the orbit is bound; zero corresponds to escape, and positive means unbound (hyperbolic trajectory).

How do K and U relate in a circular orbit?

K = +GMm/(2r), U = -GMm/r, so |U| = 2K and E = -K. This is the virial theorem for inverse-square gravity.

Why does a higher orbit have more total energy but lower speed?

Moving outward, kinetic energy decreases (the satellite slows) but potential energy increases (becomes less negative) by twice as much, so total energy increases. Counter-intuitively, you must speed the satellite up briefly to slow it down in the long run.

How much energy is needed to escape from a circular orbit?

The change from E = -GMm/(2r) to E = 0 is GMm/(2r). Equivalently, you need to multiply the orbital speed by √2 to reach escape velocity at that radius.

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NSW · HSCModule 5

Orbital energy and orbit transfer calculator

Compute K, U and E for a satellite in two different orbits, and the energy required to move from one to the other.

Inputs

Central body presetCentral mass M (kg)Satellite mass m (kg)Orbit 1 radius r₁ (m)Orbit 2 radius r₂ (m)

Result

Orbit 1

1.472e+10J

-2.944e+10J

-1.472e+10J

7673m/s

Orbit 2

2.361e+9J

-4.722e+9J

-2.361e+9J

3073m/s

Energy to move from r₁ to r₂

1.236e+10J

E = -GMm/(2r), K = +GMm/(2r), U = -GMm/r. Negative E means bound.

How this calculator works

For a circular orbit of radius r, K = GMm/(2r), U = -GMm/r, and E = -GMm/(2r). The calculator evaluates each at two radii r₁ and r₂ and reports the difference ΔE = E₂ - E₁, which is the work the rocket must do (positive if r₂ > r₁).

See the full derivation and worked example in our conservation of energy in orbital motion dot point answer.

Common questions

What is the total mechanical energy of a circular orbit?: E = -GMm/(2r). It is negative because the orbit is bound; zero corresponds to escape, and positive means unbound (hyperbolic trajectory).
How do K and U relate in a circular orbit?: K = +GMm/(2r), U = -GMm/r, so |U| = 2K and E = -K. This is the virial theorem for inverse-square gravity.
Why does a higher orbit have more total energy but lower speed?: Moving outward, kinetic energy decreases (the satellite slows) but potential energy increases (becomes less negative) by twice as much, so total energy increases. Counter-intuitively, you must speed the satellite up briefly to slow it down in the long run.
How much energy is needed to escape from a circular orbit?: The change from E = -GMm/(2r) to E = 0 is GMm/(2r). Equivalently, you need to multiply the orbital speed by √2 to reach escape velocity at that radius.