3.3 Man-made Satellites
Man-made satellites are launched into space to orbit the Earth at specific altitudes to perform vital functions such as communication, weather forecasting, navigation, and Earth imaging. The motion of these satellites is governed by Newton's Universal Law of Gravitation and centripetal force principles. Understanding orbital motion also involves the concept of escape velocity, which is the minimum velocity needed for an object to overcome Earth's gravitational pull and escape completely into outer space.
Learning Objectives
- Establish and determine the linear speed of a satellite orbiting the Earth using centripetal force and Newton's Universal Law of Gravitation.
- Describe the characteristics of geostationary and non-geostationary satellites and differentiate between them.
- Define escape velocity and derive its formula using kinetic energy and gravitational potential energy.
- Solve problems involving escape velocity and understand the benefits and implications of Earth's high escape velocity.
Comparison of Geostationary vs Non-Geostationary Satellites & Escape Velocity
Illustrate orbital mechanics, difference in satellite paths relative to Earth's rotation, and the energy balance needed for escape velocity.
Geostationary vs Non-Geostationary
Escape Velocity Energy Balance
Short Explanation
Linear Speed of a Satellite
Satellites move in circular orbits around the Earth where the centripetal force is provided by the gravitational force between the satellite and Earth.
\(F = \dfrac{GMm}{r^2}\) and \(F = \dfrac{mv^2}{r}\)
Equating forces gives \(v = \sqrt{\dfrac{GM}{r}}\).
Orbital Radius
Radius of orbit, \(r = R + h\), where \(R\) is the radius of the Earth and \(h\) is the altitude of the satellite above Earth's surface.
Therefore, \(v = \sqrt{\dfrac{GM}{R + h}}\).
The linear speed depends only on the radius of its orbit, \(r\). It does not depend on the mass of the satellite, \(m\).
Geostationary Satellite
Orbits in the Geostationary Earth Orbit. Its orbit is circular and in the equatorial plane, its direction of motion is the same as Earth's rotation, and its orbital period is \(T = 24\text{ hours}\). It therefore remains above the same geographical location. Example: MEASAT. Primary use: communication satellites.
Non-Geostationary Satellite
Does not satisfy one or more conditions for a geostationary orbit. Its orbit may be lower, higher, inclined, or elliptical, and its direction need not match Earth's rotation. Its orbital period may be shorter than, longer than, or equal to 24 hours; for example, an inclined or elliptical geosynchronous satellite has the same period but appears to move in the sky. Other examples include TiungSAT, RazakSAT, Pipit, ISS, and GPS satellites. Primary uses include Earth imaging, GPS, and weather forecasting.
Escape Velocity
\(v = \sqrt{\dfrac{2GM}{R}}\)Minimum velocity needed by an object on the surface of the Earth to overcome gravitational force and escape to outer space. For Earth, \(v \approx 1.12 \times 10^4\text{ m s}^{-1} = 11.2\text{ km s}^{-1}\).
| Feature | Geostationary Satellite | Non-Geostationary Satellite |
|---|---|---|
| Orbital period | 24 hours | May be shorter than, longer than, or equal to 24 hours |
| Orbit shape and plane | Circular and in the equatorial plane | May be inclined, elliptical, or in another plane |
| Direction of motion | Same as Earth's rotation | Need not be the same as Earth's rotation |
| Position above Earth | Fixed above one geographical location | Changes over different geographical locations |
| Examples | MEASAT | TiungSAT, RazakSAT, Pipit, ISS, GPS satellites |
| Primary use | Communication | Earth imaging, GPS, weather forecasting |
Benefits and Implications of Earth's Escape Velocity
Earth's escape velocity is high (\(11\,200\text{ m s}^{-1}\)). Gas molecules in the atmosphere have an average speed of around \(500\text{ m s}^{-1}\), far below the escape velocity, so they cannot escape into outer space. This preserves the atmosphere. Commercial aircraft fly at around \(250\text{ m s}^{-1}\) and fighter jets at up to \(2\,200\text{ m s}^{-1}\), both well below escape velocity, allowing them to fly high without risk of escaping into outer space. Launching spacecraft into outer space requires massive quantities of fuel to produce enough thrust to achieve Earth's escape velocity.
Try to Answer First
Answer in your mind, then press “Check Answer”.
Does a heavier satellite require a higher linear speed than a lighter satellite to maintain the same orbit? Explain.
Check Answer
State two main differences between a geostationary satellite and a non-geostationary satellite.
Check Answer
Why is Earth able to retain its atmosphere while gases on Mars escape more easily?
Check Answer
Calculate the escape velocity from Earth's surface using \(G=6.67\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}\), \(M=5.97\times10^{24}\ \mathrm{kg}\), and \(R=6.37\times10^6\ \mathrm{m}\).
Check Answer
Compare the escape velocities of the Moon and Mars using \(M_{Moon}=7.35\times10^{22}\ \mathrm{kg}\), \(R_{Moon}=1.74\times10^6\ \mathrm{m}\), \(M_{Mars}=6.42\times10^{23}\ \mathrm{kg}\), and \(R_{Mars}=3.39\times10^6\ \mathrm{m}\).
Check Answer
Common Mistakes
- Using the altitude \(h\) directly instead of the orbital radius \(r = R + h\) in calculations for orbital linear speed.
- Assuming that heavier satellites require greater orbital speeds to remain in orbit.
- Assuming every non-geostationary satellite must have a period different from 24 hours; an inclined or elliptical geosynchronous orbit can have a 24-hour period without remaining above one location.
- Omitting the factor of 2 in the escape velocity formula \(v = \sqrt{\dfrac{2GM}{R}}\) and confusing it with orbital speed \(v = \sqrt{\dfrac{GM}{r}}\).
- Forgetting to convert altitude/radius from kilometres (\(\text{km}\)) to metres (\(\text{m}\)) before substituting into formulas with \(G\).
Concept Misunderstandings
A rocket launched into orbit must achieve escape velocity.
A satellite placed into orbit only needs to reach orbital linear speed (\(v = \sqrt{\dfrac{GM}{r}}\)), which is lower than escape velocity (\(v = \sqrt{\dfrac{2GM}{r}}\)). Escape velocity is only required when a spacecraft needs to leave Earth's gravitational field completely.
Escape velocity depends on the mass of the spacecraft being launched.
Escape velocity depends strictly on the mass of the celestial body (\(M\)) and its radius (\(R\)). Both light and heavy spacecraft require the exact same escape velocity from Earth.
Summary
- Orbital speed is determined by equating centripetal force and gravitational force: \(v = \sqrt{\dfrac{GM}{r}} = \sqrt{\dfrac{GM}{R + h}}\). It depends solely on orbital radius, not the satellite's mass.
- Geostationary satellites have a circular equatorial orbit, a 24-hour period, move with Earth's rotation, and remain above the same location (e.g., MEASAT for communication).
- Non-geostationary satellites fail at least one geostationary condition and change position relative to Earth's surface. Their periods may be shorter than, longer than, or equal to 24 hours.
- Escape velocity is \(v = \sqrt{\dfrac{2GM}{R}}\), the minimum speed to escape gravitational pull. For Earth, \(v \approx 11.2\text{ km s}^{-1}\).
- Earth's high escape velocity prevents atmosphere loss and ensures aeroplanes cannot accidentally exit into outer space.
Short Activity
Objective: Identify orbital speed, geostationary and non-geostationary satellites, and escape velocity.
A. Objective Quiz
1 What happens to a satellite in orbit if its linear speed becomes less than the required orbital speed?
2 Which factor influences the escape velocity of an object from the surface of a planet?
B. Fill in the Blanks
3 A geostationary satellite has an orbital period of hours, which is identical to the period of rotation of the Earth.
4 The value of escape velocity from the surface of the Earth is approximately km s⁻¹.
C. Matching / Drag and Drop
Drag each term to the correct meaning. If using a phone, tap the answer first, then tap the matching box.
