Understanding Orbital Velocity

Expert reviewed • 22 November 2024 • 8 minute read

What is Orbital Motion?

Orbital motion describes the path of an object around a central body under the influence of gravity. This phenomenon is observed throughout the universe, from planets orbiting stars to moons circling planets and artificial satellites orbiting Earth.

Gravitational Force and Centripetal Force

For an object to maintain a stable circular orbit, two key forces must be in balance:

Gravitational Force: The attractive force between two masses, pulling the orbiting object towards the centre of the larger body.
Centripetal Force: The net force required to keep an object moving in a circular path, directed towards the centre of the circle.

In a stable orbit, the gravitational force provides the necessary centripetal force to maintain circular motion.

Deriving the Orbital Velocity Equation

We can derive the equation for orbital velocity by equating the gravitational force to the centripetal force:

F_c = F_g\\\frac{mv^2}{r} = \frac{GMm}{r^2}\\v^2 = \frac{GM}{r}

Therefore, the orbital velocity is given by:

v_{\text{orbital}} = \sqrt{\frac{GM}{r}}

Where:

$G$ is the Universal Gravitational Constant (6.67 × 10−11 N⋅m²/kg²)
$M$ is the mass of the central body (e.g., Earth) in kg
$r$ is the radius of the orbit in meters

Key Insights from the Equation

Orbital velocity is independent of the mass of the orbiting object.
Orbital velocity decreases as the orbital radius increases.
Objects in lower orbits move faster than those in higher orbits.

Practice Question 1

Calculate the orbital velocity of a satellite in a low Earth orbit, 500 km above the Earth’s surface.

Given:

Mass of Earth $(M_E) = 5.97 × 1024\: kg$
Radius of Earth $(R_E) = 6.371 × 10^6\: m$
Orbital altitude $= 500\times 10^3\: km$

Step 1: Calculate the orbital radius

r = R_E + altitude \\= 6.371 × 10^6  + 500 × 10^3 \\= 6.871 × 10^6

Step 2: Apply the orbital velocity equation

v_{\text{orbital}} = \sqrt{\frac{GM_E}{r}} \\v_{\text{orbital}} = \sqrt{\frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{6.871 \times 10^6}}\\v_{orbital} = 7, 612 m/s \approx 27,400 \:km/h

This high velocity demonstrates why objects in orbit appear to "fall around" the Earth rather than falling towards it.

What is Orbital Period?

The orbital period $(T)$ is the time taken for an object to complete one revolution in its orbit. For circular orbits, it’s related to the orbital velocity and radius:

T = \frac{2\pi r}{v}

Substituting the orbital velocity equation, we get:

T = 2\pi \sqrt{\frac{r^3}{GM}}

This equation shows that objects in higher orbits have longer orbital periods.

What are the Effects of Velocity Changes on Orbit?

The stability of an orbit depends on maintaining the correct velocity for a given altitude. Changes in velocity can significantly alter an object’s orbit:

Velocity Increase: If an object’s velocity exceeds the required orbital velocity, its orbit becomes elliptical or even hyperbolic, potentially causing it to escape the gravitational influence of the central body.
Velocity Decrease: If the velocity falls below the required orbital velocity, the orbit decays, bringing the object closer to the central body. Significant decreases can cause the object to fall into the atmosphere.

Practice Question 2

Earth’s Orbit Around the Sun:

Calculate Earth’s approximate orbital velocity around the Sun, assuming a circular orbit.

Given:

Mass of Sun $(M_S) = 1.99 × 10^{30} \:kg$
Average Earth-Sun distance $= 1.496 × 10^{11} \:m$

Apply the orbital velocity equation:

v_{\text{Earth}} = \sqrt{\frac{GM_S}{r}}\\v_{\text{Earth}} = \sqrt{\frac{(6.67 \times 10^{-11})(1.99 \times 10^{30})}{1.496 \times 10^{11}}}\\v_{Earth} = 29, 784 m/s \approx 107,200 \:km/h

Newton’s Law of Gravitation and Satellite Orbits

Up Next

Kepler’s Laws of Planetary Motion

Return to Module 5: Advanced Mechanics