How Do You Calculate the Orbital Speed of a Satellite? (HL)

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Why This Equation Works

The satellite's weight provides the centripetal force required for circular motion. Newton's universal law of gravitation gives the gravitational force:

The required centripetal force is:

For a stable circular orbit, these are the same force, so:

Cancel the satellite mass $m$ and one factor of $r$:

Therefore:

The satellite's mass cancels, so orbital speed does not depend on the satellite's mass. The equation also shows that a satellite in a larger circular orbit travels more slowly because $v\propto r^{-1/2}$.

This derivation assumes a circular orbit and neglects atmospheric drag and other bodies. In an elliptical orbit, speed changes with position, so this single-radius result cannot describe the whole path.

The inverse-square gravitational force becomes weaker at larger radius, but the required centripetal acceleration also changes. The final dependence, $v\propto r^{-1/2}$, means doubling the orbital radius reduces speed by a factor of $\sqrt{2}$.

Worked Example

A satellite orbits Earth at an altitude of $400\text{ km}$. Take Earth's radius as $6.37\times10^6\text{ m}$, its mass as $5.97\times10^{24}\text{ kg}$, and $G=6.67\times10^{-11}\text{ N m}^2\text{ kg}^{-2}$.

First find the orbital radius, measured from Earth's centre:

Then:

Thus, the orbital speed is approximately $7.67\text{ km s}^{-1}$.

Notice that the altitude must first be converted to metres and added to Earth's radius; substituting $400\text{ km}$ directly would give the wrong gravitational separation.

Exam Technique

For a Show question, include the force equation, centripetal-force equation, cancellation, and final result. A common misconception is using altitude as $r$; always use the distance from the centre of the planet. Also state that gravity is the resultant inward force. Check that the answer has units of speed and is physically reasonable: low-Earth-orbit speeds are several kilometres per second.