Gravity and Orbits

Section 4 of 6

Orbital and Escape Velocity

Part 4: Orbital Velocity and Escape Velocity

How Fast Must a Planet Move?

For a stable circular orbit at radius $r$ around a mass $M$ , one specific speed satisfies force balance: gravity equals the centripetal requirement,

$\frac{GMm}{r^2} = \frac{mv_{orb}^2}{r}.$

The orbiting mass $m$ appears on both sides and cancels — all objects orbit at the same speed at a given radius. (This puzzled physicists until Einstein explained that the mass which resists acceleration equals the mass gravity pulls on.) Multiplying by $r$ gives $GM/r = v_{orb}^2$ , so the orbital velocity is:

Orbital velocity

The speed required for a circular orbit at radius $r$ around mass $M$ , $v_{orb} = \sqrt{GM/r}$ . It depends only on $M$ and $r$ — not on the orbiting object’s mass.

Larger $M$ means stronger gravity, so you must orbit faster to avoid falling in; larger $r$ means weaker gravity, so you can orbit more slowly. The orbiting object’s mass doesn’t matter.

Worked Example 1Earth's Orbital Velocity Around the Sun

Problem

Find Earth’s circular orbital speed, given $M_\odot = 2 \times 10^{33}$ g, $r = 1$ AU $= 1.5 \times 10^{13}$ cm, and $G = 6.67 \times 10^{-8}\ \mathrm{cm^3\,g^{-1}\,s^{-2}}$ .

StepSubstitute into the formula

$v_{orb} = \sqrt{\frac{GM_\odot}{r}} = \sqrt{\frac{(6.67 \times 10^{-8})(2 \times 10^{33})}{1.5 \times 10^{13}}}\ \mathrm{cm/s}$

StepCombine powers of ten

$= \sqrt{\frac{6.67 \times 2}{1.5} \times 10^{-8+33-13}} = \sqrt{8.9 \times 10^{12}}\ \mathrm{cm/s} \approx 3.0 \times 10^6\ \mathrm{cm/s}$

Dimensional check

$\sqrt{(\mathrm{cm^3\,g^{-1}\,s^{-2}})(\mathrm{g})/\mathrm{cm}} = \sqrt{\mathrm{cm^2/s^2}} = \mathrm{cm/s}$ ✓ — the grams cancel and the result is a speed.

Result

$v_{orb} \approx 3.0 \times 10^6\ \mathrm{cm/s} = 30$ km/s. Earth travels at about 30 km/s just to stay in orbit. Sanity check: that’s about 0.01% of the speed of light — fast, but not relativistic.

Pause & Predict #4

Use v_orb = sqrt(GM/r).

Jupiter orbits at about 5 AU. Without calculating, is its orbital velocity greater than, less than, or equal to Earth's — and by what factor?

A satellite at r = 6800 km from Earth's center vs a GPS satellite at r = 26,000 km: which moves faster, and by roughly what factor?

Check below.

Problem

Check the two predictions above. (a) Jupiter at 5 AU vs Earth: is its orbital velocity greater, less, or equal — and by what factor? (b) A low satellite at $r = 6800$ km vs a GPS satellite at $r = 26{,}000$ km: which moves faster, and by roughly what factor?

Escape Velocity: Breaking Free of Gravity

What if you want to leave a gravitational system entirely? You need enough kinetic energy to overcome the gravitational binding. The minimum speed is the escape velocity, found by setting the total mechanical energy to zero (just barely reaching infinity):

$E = K + U = \frac{1}{2}mv_{esc}^2 - \frac{GMm}{r} = 0 \quad\Rightarrow\quad v_{esc}^2 = \frac{2GM}{r}.$

Escape velocity

The minimum launch speed needed to coast from radius $r$ to infinity, $v_{esc} = \sqrt{2GM/r} = \sqrt{2}\,v_{orb}$ . The escaping object’s own mass cancels.

Escape velocity is exactly $\sqrt{2}$ times the orbital velocity at the same radius, $v_{esc} = \sqrt{2}\,v_{orb} \approx 1.41\,v_{orb}$ . Larger $M$ or smaller $r$ means a deeper gravitational well and a harder escape. The escaping object’s mass cancels — a rocket and a baseball need the same speed.

Earth with multiple trajectory curves showing different launch velocities: suborbital (falls back), orbital (circular/elliptical), and escape trajectories (parabolic and hyperbolic paths). — Figure 6Escape Velocity: launch speed determines trajectory. Below v_esc, the object falls back; at v_esc, parabolic escape; above v_esc, hyperbolic escape.cococubed.com

Worked Example 2Earth's Escape Velocity

Problem

Find Earth’s escape velocity, given $M_\oplus = 6 \times 10^{27}$ g, $R_\oplus = 6.4 \times 10^{8}$ cm, and $G = 6.67 \times 10^{-8}\ \mathrm{cm^3\,g^{-1}\,s^{-2}}$ .

StepSubstitute into the formula

$v_{esc} = \sqrt{\frac{2GM_\oplus}{R_\oplus}} = \sqrt{\frac{2(6.67 \times 10^{-8})(6 \times 10^{27})}{6.4 \times 10^{8}}}\ \mathrm{cm/s}$

StepCombine powers of ten

$= \sqrt{\frac{2 \times 6.67 \times 6}{6.4} \times 10^{-8+27-8}} = \sqrt{1.25 \times 10^{12}}\ \mathrm{cm/s}$

Dimensional check

The factor of 2 is dimensionless, so the units are identical to $v_{orb}$ : $\sqrt{\mathrm{cm^2/s^2}} = \mathrm{cm/s}$ ✓.

Result

$v_{esc} = 1.12 \times 10^6\ \mathrm{cm/s} = 11.2$ km/s. Sanity check: this is larger than the near-surface orbital velocity ( $\sim 7.9$ km/s) by $\sqrt{2} \approx 1.41$ ✓.

Comparison of key velocities:

Location	$v_{orb}$ (km/s)	$v_{esc}$ (km/s)
Earth at $r = R_\oplus$ (hypothetical near-surface orbit)	7.9	11.2
Moon surface	1.7	2.4
Sun surface	434	618

A physical orbit at Earth’s surface is impossible due to atmospheric drag; the values are for the field strength at that radius. This is why Apollo astronauts needed a smaller rocket to leave the Moon: its escape velocity (2.4 km/s) is only about one-fifth of Earth’s, so less fuel is required.