Gravitation, Orbit, and Escape Velocity

Every object with mass pulls on every other object with mass. Newton described this attraction with a simple inverse-square rule: the force between two masses \(m_1\) and \(m_2\) separated by a distance \(r\) is

\( F = \frac{G m_1 m_2}{r^2} \)

Here \(G\) is the universal gravitational constant, \(6.67 \times 10^{-11}\ \)N\(\cdot\)m\(^2/\)kg\(^2\). Because the force depends on \(1/r^2\), doubling the distance between two objects cuts the gravitational force to one quarter of its original value.

A satellite stays in orbit because gravity supplies exactly the centripetal force needed to keep it moving in a circle. Setting the gravitational force equal to the centripetal force requirement gives

\( \frac{G M m}{r^2} = \frac{m v^2}{r} \)

where \(M\) is the mass of the central body (like Earth), \(m\) is the mass of the orbiting object, and \(r\) is the orbital radius measured from the center of \(M\). Solving for \(v\) gives the orbital velocity formula:

\( v_{orbit} = \sqrt{\frac{G M}{r}} \)

Notice the orbiting mass \(m\) cancels out completely. Every satellite at the same radius, no matter how heavy, needs the same orbital speed to maintain a stable circular orbit. This is closely related to the ideas used in position, velocity, and acceleration problems, since orbital motion is still governed by the same kinematics, just applied to a circular path.

Quick example: Low Earth orbit

A satellite orbits at \(r = 6.7 \times 10^6\ \)m\(\) from Earth's center (about 300 km above the surface), with Earth's mass \(M = 5.97 \times 10^{24}\ \)kg\(\).

\( v_{orbit} = \sqrt{\frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{6.7 \times 10^6}} \approx 7.7 \times 10^3\ \)m/s\( \)

That is roughly 7.7 kilometers per second, which is why low-orbit satellites circle Earth in about 90 minutes.

Escape velocity is the minimum speed an object needs, fired straight up with no further thrust, to break free of a body's gravity permanently and never fall back. It comes from setting the object's kinetic energy equal to the magnitude of its gravitational potential energy, so that total mechanical energy is exactly zero at infinity:

\( \frac{1}{2} m v_{esc}^2 = \frac{G M m}{r} \)

Solving for \(v_{esc}\):

\( v_{esc} = \sqrt{\frac{2 G M}{r}} \)

Just like orbital velocity, the mass \(m\) of the escaping object cancels out. A pebble and a rocket launched from the same point need exactly the same escape speed, since it depends only on the mass \(M\) and radius \(r\) of the body being escaped.

Escape velocity is always \(\sqrt{2}\) times the circular orbital velocity at the same radius, since \(v_{esc} = \sqrt{2}\, v_{orbit}\). This is why a spacecraft that reaches escape speed follows an open, unbound path (a parabolic or hyperbolic trajectory) instead of a closed escape orbit that loops back around.

Worked example: Earth's escape velocity

Using Earth's mass \(M = 5.97 \times 10^{24}\ \)kg\(\) and radius \(r = 6.37 \times 10^6\ \)m\(\):

\( v_{esc} = \sqrt{\frac{2(6.67 \times 10^{-11})(5.97 \times 10^{24})}{6.37 \times 10^6}} \approx 1.12 \times 10^4\ \)m/s\( \)

That is about 11.2 kilometers per second, the well-known value for Earth's escape velocity.

Comparing escape speeds

Because escape velocity depends on both mass and radius, smaller or less massive bodies generally have much lower escape speeds:

• The Moon (\(M \approx 7.35 \times 10^{22}\ \)kg\(\), \(r \approx 1.74 \times 10^6\ \)m\(\)) has an escape velocity of about 2.4 kilometers per second, roughly a fifth of Earth's.
• The Sun (\(M \approx 1.99 \times 10^{30}\ \)kg\(\), \(r \approx 6.96 \times 10^8\ \)m\(\)) has an enormous escape velocity of about 618 kilometers per second, because its mass is so large compared to its radius.
• A black hole is an extreme case where so much mass is packed into such a small radius that the escape velocity reaches the speed of light itself, which is why not even light can escape from within its event horizon.

Gravitation, orbital motion, and escape velocity are three views of the same inverse-square force. The same formula \(F = GMm/r^2\) explains why the Moon circles Earth, why satellites need a specific orbital speed to avoid spiraling in or drifting away, and why a rocket needs a threshold speed to leave a planet for good. Recognizing that both orbital and escape velocity depend only on the central mass and the radius (never on the smaller object's own mass) is the key idea to remember for these problems. As with other motion problems, comparing speeds and directions can also connect back to ideas from the relative velocity formula when analyzing motion between an orbiting body and an observer.

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