Orbital Velocity

This topic is part of the HSC Physics syllabus under the section Motion in Gravitational Fields.

HSC Physics Syllabus

• investigate the orbital motion of planets and artificial satellites when applying the relationships between the following quantities:

– gravitational force
– centripetal force
– centripetal acceleration
– mass
– orbital velocity
– orbital period

Orbital Motion: The Basics

Orbital motion refers to the way in which planets, satellites, and other celestial bodies move in orbits due to gravitational forces. This motion is not just a feature of man-made satellites; it's a natural phenomenon that includes the orbits of planets around the sun, moons around their planets, and even the movement of galaxies.

In space, an artificial satellite around Earth is acted by gravitational force exerted by Earth. The direction of this force is directed towards the centre of Earth.

For a satellite in a stable circular orbit around Earth, the centripetal force (net force towards the centre of its orbital motion) is provided by gravitational force.

Therefore, we can make the following derivation for orbital velocity:

F_c=F_g

(mv^2)/r=(GMm)/r^2

v^2=(GM)/r

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

where:

• G is the Universal Gravitational Constant = 6.67 xx 10^{-11}
• M is the mass of the object the satellite is orbiting around e.g. Earth (kg)
• r is the radius of the orbit in metres (m)

The orbital velocity is independent of the mass of the orbiting object. So, no matter how heavy the satellite is, the orbital velocity for a given orbital radius around Earth would be the same.

In a circular orbit, the orbital velocity is perpendicular to the gravitational force (centripetal force).

If a satellite can achieve and maintain a certain orbital velocity (dependent on the planet’s mass and radius of orbit), it can stay in orbit without additional force nor acceleration in the same direction as its motion. In other words, it is possible to undergo stable orbit with only gravitational force. Travelling at a velocity different to the orbital velocity will cause the satellite to take on a different orbit path (see below).

Orbital period is defined as the time taken for an object to complete one revolution of an orbit. The orbital period is shorter when the circumference or radius of orbit is smaller. Vice versa.

What Happens When a Satellite's Velocity Changes?

The velocity of an orbiting satellite is crucial in determining its orbit around a planet. If the satellite's velocity deviates from the required orbital velocity for its altitude, the nature of its orbit changes significantly.

Here's what happens in scenarios where the satellite's velocity is either greater or less than the needed orbital velocity:

Velocity Greater than Orbital Velocity

If a satellite's velocity exceeds the required orbital velocity for a given altitude, several outcomes are possible. If the increase is moderate, the satellite's orbit changes from circular to elliptical, with the satellite moving to a higher orbit.

If the velocity is significantly higher (equal to greater than the escape velocity), it could even reach a hyperbolic trajectory, potentially leading the satellite to leave the Earth's gravitational influence and travel out into space.

Velocity Less than Orbital Velocity

If the satellite's velocity is less than the required orbital velocity for its altitude, the satellite’s orbit will decay, bringing it closer to the Earth. This is often a controlled process in de-orbiting a satellite, where its velocity is deliberately reduced. The orbit may become more elliptical, with the satellite coming closer to the Earth at the perigee (the point in the orbit closest to Earth).

If the velocity reduction is significant, the satellite might enter the Earth's atmosphere and burn up or crash.

Calculation Example 1

Calculate the orbital velocity of a satellite in an orbit of 500 km above Earth’s surface. Use the following information from NESA HSC Physics data sheet in your calculations:

• Mass of Earth = 6.0 xx 10^{24} kg
• Radius of Earth = 6.371 xx 10^6 m

Solution:

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

$$v_{\text{orbital}} = \sqrt{\frac{(6.67 \times 10^{-11})(6.0 \times 10^{24})}{6.371 \times 10^6 + 500 \times 10^3}}$$

$$v_{\text{orbital}} = 7632 \text{ m/s}$$

Orbital Velocity of Planets

When considering Earth's orbit around the Sun, it's important to note that the orbit is elliptical, not perfectly circular.

While the formula v = \sqrt{\frac{GM}{r}} is accurate for circular orbits, it becomes an approximation for elliptical orbits, especially for orbits that are close to being circular, like Earth's. Earth's orbit has a low eccentricity, meaning it is nearly circular, but there are still some adjustments needed for more accurate calculations.

Calculate Example 2

Assume Earth orbits the Sun in a circular path with a radius of 1.50 xx 10^{11} m

Calculate the orbital velocity of the Earth around the Sun.

Use the following information in your calculations:

• Mass of Sun = 2.0 xx 10^{30} kg
• Mass of Earth = 6.0 xx 10^{24} kg

Solution:

Earth's orbit around the Sun is elliptical with a low eccentricity which means we can use the orbital velocity formula for a circular orbit to approximate its orbital velocity.

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

Here, M is mass of the Sun:

$$v_{\text{orbital}} = \sqrt{\frac{(6.67 \times 10^{-11})(2.0 \times 10^{30})}{1.50 \times 10^{11}}}$$

$$v_{\text{orbital}} = 3.0 \times 10^4 \text{ m/s}$$

Previous section: Gravitational Force and Acceleration

Next section: Kepler's Laws of Planetary Motion