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Fig. 1 - A satellite can help us in more ways than we can imagine. Here we show a satellite doing remote sensing to optimize crops and natural resources.
Orbit of planets
In the 1500s astronomer, Nicolaus Copernicus hypothesized that the Sun was stationary at the center of our universe, with the other planets, such as Earth, orbiting around it with circular revolutions. One of the major improvements to this model was originated by Johannes Kepler, who used astronomical observations to show that instead of having circular orbits, the planets actually followed an elliptical motion around the Sun.
These orbital trajectories were described by Kepler's Laws, which we will be discussing and expanding upon in this article. Even though these were originally formulated for the planets in the Solar System orbiting the Sun, they can be applied universally. They can be applied to exoplanets around other stars, as well as describing the motion of a satellite orbiting a planet. However, if we want to apply Kepler's Laws we must make sure that the center of mass of the satellite-planet system is inside the central body and that the satellite is much less massive than the planet's center of mass.
A Satellite is a celestial or artificial body orbiting the Earth or another planet.
Recall that Kepler's laws of planetary motion can be used to describe satellite orbits around the Earth.
Kepler's First Law states that planets move in elliptical orbits with the sun at its focus.
Fig. 2 - Elipses have two focus points as well as a semi-major axis and a semi-minor axis.
The shape of the orbit will be defined by its eccentricity. This quantity describes how an orbit will deviate from a perfect circle. The larger the eccentricity, the more elongated the orbit. An eccentricity of zero refers to a circular orbit.
Fig. 3 - Different types of orbital trajectories depend on eccentricity. The circular orbit is blue, the elliptical orbit is green, the parabolic orbit is red, and the hyperbolic orbit is purple.
Kepler's Second Law states that the distance vector between the planet and the Sun sweeps equal areas of the ellipse at equal times.
In a circular orbit, the satellite will move at the same speed throughout the orbit. However, according to Kepler's Second Law, in an elliptical orbit, a satellite travels faster when it is nearer the planet and moves slower when farthest away from the planet.
Fig. 4 - A visual representation of Kepler's Second Law. Notice how equal areas of space are swept across equal periods of time.
Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of the ellipse.
Kepler's Third Law explains that the period for a satellite to orbit Earth increases rapidly with the radius of its orbit. For an elliptical orbit, Kepler's Third Law can be expressed as follows,
$$T^2=\frac{4\pi^2}{GM}a^3,$$
where \(T\) is the orbit's period in seconds \(\mathrm s\), \(G\) is the gravitational constant \(6.67\times10^{-11}\;\frac{\mathrm N\;\mathrm m^2}{\mathrm{kg}^2}\), \(M\) is the planet's mass in \(\text{kg}\), and \(a\) is the semi-major axis in \(\text{m}\). The semi-major axis is equal in length to half of the longest diameter of an ellipse. In the case of a circular orbit, the semi-major axis can be replaced with the radius of the orbit.
Orbit Equation
To study satellite orbits, we consider a satellite moving towards a planet influenced by a gravitational force. To make the math simpler we reduce the two-body problem of a planet-satellite system to a one-body problem, we study the orbit in the frame of reference of the Center of Mass.
\[M=m_1+m_2.\]
We look at the orbit of the reduced mass \(\mu\) about the origin, where the total mass of the system \(M\) is at rest,
\[\begin{align*} \mu&=\frac{m_1m_2}{m_1+m_2},\\\mu&=\frac{m_1m_2}M,\end{align*}\]
where \(m_1\) is the mass of the central body in kilograms \(\text{kg}\), \(m_2\) is the mass of the orbiting body in kilograms \(\text{kg}\), \(M\) is the total mass in kilograms \(\text{kg}\), and \(\mu\) is the reduced mass in kilograms \(\text{kg}\).
Fig. 5 - Visualization of the reduced mass. A two-body problem is turned into a one-body problem by using the Center of Mass frame and the reduced mass.
The orbit equation can be expressed as,
$$r=\frac{l^2}{{\mu}Gm_1m_2}\frac1{1-e\cos\left(\theta\right)},$$
where \(G\) is the gravitational constant \(6.67\times10^{-11}\;\frac{\mathrm N\;\mathrm m^2}{\mathrm{kg}^2}\), \(r\) is the distance between the bodies in \(\mathrm m\), \(\theta\) is the angle between \(r\) and the longest axis of the orbit in degrees or radians, \(^\circ\) or \(\mathrm{rad}\), \(e\) is the eccentricity of the orbital trajectory, \(l\) is the angular momentum in \(\frac{\mathrm{kg}\;\mathrm m^2}{\mathrm s}\), \(m\) is the mass of the orbiting body in \(\mathrm{kg}\), and \(\mu\) is the reduced mass in \(\mathrm{kg}\). It's derivation is explained in the deep dive below.
Fig. 6 - Representation of our reduced mass system. This is an elliptical orbit using the reduced mass. The maximum and minimum distances from the central body are located on the horizontal axis.
To derive the orbit equation we must first begin by describing the energy of the reduced mass system orbiting the system's center of mass. We know the kinetic energy and the gravitational potential of the system, so we can express the total energy of the system as,
$$E=\frac{1}{2}\mu v^2-\frac{Gm_1m_2}{r},$$
where \mu is the reduced mass in \(\text{kg}\) and \(v\) is the relative velocity between the two bodies.
The velocity can be defined in polar coordinates as,
\begin{align*}\vec{v}&=v_\text{r} \hat{r} +v_\theta \hat{\theta},\\v&=|\vec{v}|,\\v&=|\frac{\text{d}\vec{r}}{\text{d}t}|,\end{align*}
where \(v_\text{r}=\frac{\text{d}r}{\text{d}t}\) and \(v_\theta=r\left(\frac{\text{d}\theta}{\text{d}t}\right)\).
The energy of the system is now:
$$E=\frac{1}{2}\mu\left[\left(\frac{\text{d}r}{\text{d}t}\right)^2+\left(r\frac{\text{d}\theta}{\text{d}t}\right)^2\right]-\frac{Gm_1m_2}{r}.$$
The angular momentum of the system is given by:
$$\begin{align*}\vec{l}&=\vec{r}\times\mu\vec{v},\\\vec{l}&=r \hat{r}\times\mu\left(v_\text{r} \hat{r}+v_\theta \hat{\theta}\right),\\\vec{l}&=r\mu v_\theta \hat{k},\\\vec{l}&=\mu r^2\frac{\text{d}\theta}{\text{d}t} \hat{k},\\l&=\mu r^2\frac{\text{d}\theta}{\text{d}t},\\\frac{\text{d}\theta}{\text{d}t}&=\frac{l}{\mu r^2}.\end{align*}$$
where \(\hat{k}\) is a unit vector perpendicular to the plane of motion.
Revise your vector calculation rules if you struggled to follow the above derivation.
Now we rewrite the total energy of the system,
$$E=\frac{1}{2}\mu\left(\frac{\text{d}r}{\text{d}t}\right)^2+\frac{1}{2}\frac{l^2}{\mu r^2}-\frac{Gm_1m_2}{r}.$$
We can rearrange and write this expression in terms of \(\frac{\text{d}r}{\text{d}t}\):
$$\frac{\text{d}r}{\text{d}t}=\sqrt{\frac{2}{\mu}}\left(E-\frac{1}{2}\frac{l^2}{\mu r^2}+\frac{Gm_1m_2}{r}\right)^{\frac{1}{2}}.$$
Now we divide \(\frac{\text{d}\theta}{\text{d}t}\) by \(\frac{\text{d}r}{\text{d}t}\) to obtain \(\frac{\text{d}\theta}{\text{d}r}\), an expression that relates the distance \(r\) to the angle \(\theta\). It gives the orbit equation in differential form:
\begin{align*}\frac{\text{d}\theta}{\text{d}r}&=\frac{\frac{\text{d}\theta}{\text{d}t}}{\frac{\text{d}r}{\text{d}t}},\\\frac{\text{d}\theta}{\text{d}r}&=\frac{l}{\sqrt{2\mu}}\frac{\left(\frac{1}{r^2}\right)}{\left(E-\frac{l^2}{2\mu r^2}+\frac{Gm_1m_2}{r}\right)^{\frac{1}{2}}},\\\text{d}\theta&=\frac{l}{\sqrt{2\mu}}\frac{\left(\frac{1}{r^2}\right)}{\left(E-\frac{l^2}{2\mu r^2}+\frac{Gm_1m_2}{r}\right)^{\frac{1}{2}}}\text{d}r.\end{align*}
Before we do any integration, we need to do some substitutions. We do the substitution \(u=\frac{1}{r}\), \(\text{d}u=-\left(\frac{1}{r^2}\right)\text{d}r\,\):
$$\text{d}\theta=-\frac{l}{\sqrt{2\mu}}\frac{\text{d}u}{\left(E-\frac{l^2}{2\mu}u^2+Gm_1m_2u\right)^{\frac{1}{2}}}.$$
Next, we rearrange the \(\frac{l}{\sqrt{2\mu}}\) factor into the denominator of the right-hand side of the equation. The reciprocal of the square of this factor must be taken which equates to\(\frac{{2\mu}}{l^2}\). (The reciprocal must be taken as the factor is being moved under a denominator and the factor must be squared as the entire denominator is to the power of \(\frac 1 2\).) This factor is then multiplied by each component in the denominator,
$$\begin{align*}\text{d}\theta&=-\frac{\text{d}u}{\left(\frac{2\mu E}{l^2}-u^2+2\left(\frac{\mu Gm_1m_2}{l^2}\right)u\right)^{\frac{1}{2}}},\\\text{d}\theta&=-\frac{\text{d}u}{\left(\frac{2\mu E}{l^2}-u^2+\frac{2u}{r_0}\right)^{\frac{1}{2}}},\end{align*}$$
where we have defined \(r_0=\frac{l^2}{\mu Gm_1m_2}\).
Now we add and subtract \(\frac{1}{{r_0}^2}\) inside the parenthesis with the square root:
\begin{align*}\text{d}\theta&=-\frac{\text{d}u}{\left(\frac{2\mu E}{l^2}+\frac{1}{{r_0}^2}-u^2+\frac{2u}{r_0}-\frac{1}{{r_0}^2}\right)^{\frac{1}{2}}},\\\text{d}\theta&=-\frac{\text{d}u}{\left(\frac{2\mu E}{l^2}+\frac{1}{{r_0}^2}-\left(u-\frac{1}{r_0}\right)^2\right)^{\frac{1}{2}}},\\\text{d}\theta&=-\frac{r_0\text{d}u}{\left(\frac{2\mu E{r_0}^2}{l^2}+1-\left(r_0 u-1\right)^{2}\right)^{\frac{1}{2}}},\end{align*}
where we define the eccentricity as \(e=\sqrt{\frac{2\mu E{r_0}^2}{l^2}+1}\). This dimensionless quantity is responsible for the shape of the orbit. This is better discussed in the article Orbital Trajectories.
We rewrite our equation in terms of eccentricity:
$$\text{d}\theta=-\frac{r_0\text{d}u}{\left({e}^2-\left(r_0 u-1\right)^{2}\right)^{\frac{1}{2}}}.$$
Finally, we do the last substitution before solving the integral, \(r_0u-1=e\cos{\alpha}\) and \(r_0 \text{d}u=-e\sin{\alpha}\text{d}\alpha\):
\begin{align*}\text{d}\theta&=-\frac{-e\sin{\alpha}\text{d}\alpha}{\left({e}^2-{e}^2{\cos^2{\alpha}}\right)^{\frac{1}{2}}},\\\text{d}\theta&=-\frac{-\bcancel{e}\sin{\alpha}\text{d}\alpha}{\bcancel{e}\left(1-\cos^2{\alpha}\right)^{\frac{1}{2}}},\\\text{d}\theta&=\frac{\sin{\alpha}\text{d}\alpha}{\left(sin^2{\alpha}\right)^{\frac{1}{2}}},\\\text{d}\theta&=\frac{\bcancel{\sin{\alpha}}\text{d}\alpha}{\bcancel{sin{\alpha}}},\\\theta&=\int{\text{d}\alpha},\\\theta&=\alpha + \text{constant}.\end{align*}
We already know that \(r=\frac{1}{u}\). If we choose the constant to be zero, we find that:
\begin{align*}r_0u-1&=e\cos{\alpha},\\r_0u-1&=e\cos{\theta},\\u&=\frac{1-e\cos{\theta}}{r_0}.\end{align*}
Our final expression for the orbit equation is:
\begin{align*}r&=\frac{1}{u},\\r&=\frac{r_0}{1+e\cos{\theta}}.\end{align*}
Alternately, if we choose the constant to be \(\pi\) we get the same orbit equation but with the vertical axis reflected,
\begin{align*}r&=\frac{r_0}{1-e\cos{\theta}},\\r&=\frac{l^2}{{\mu}Gm_1m_2}\frac1{1-e\cos\left(\theta\right)}.\end{align*}
Orbital Speed of a Satellite
All satellites in orbit go into orbit by riding a rocket. In order to escape Earth's gravitational field, the object must reach escape velocity. The escape velocity is the initial velocity required for an object at the moment of launch, typically from the surface of a body, to escape its gravitational field.
In real life, it's not so easy to reach escape velocity, as we need to consider atmospheric drag and heating at high velocities. Traveling at high velocities through the Earth's atmosphere would cause most objects to burn up, destroying the rocket and everything inside it. As a consequence, objects must slowly accelerate until they reach escape velocity at a higher altitude.
If the kinetic energy \(K\) of an object launched from Earth is equal to its gravitational potential energy \(U\), then it could escape from the Earth,
\begin{align*}K&=U,\\\frac12mv^2&=\frac{GMm}r,\\v_e&=\sqrt{\frac{2GM}r}.\end{align*}
As we can see, the satellite's mass \(m\) does not affect the escape velocity. The minimum velocity to escape the Earth's gravitational pull from the surface is \(11.2\;{\textstyle\frac{\mathrm{km}}{\mathrm s}}\). At this speed, the satellite would fly off into space. But with satellites, we really do not want to escape Earth's gravity but to balance it. If the satellite goes too fast, it will drift away with a constant velocity until a force acts upon it, as described by the law of inertia or Newton's First Law. If the satellite goes too slow, it will return to the Earth.
The speed needed to balance Earth's gravity and the satellite's inertia is called the orbital speed. To find the orbital speed of a satellite in a circular orbit, we can equate the centripetal force of the satellite to the gravitational force between the satellite and the planet. Once again, we see that the satellite's mass does not affect its orbit,
\begin{align*}m\frac{v^2}r&=\frac{GMm}{r^2},\\v_o&=\sqrt{\frac{GM}r}.\end{align*}
We know that a satellite travels \(2\pi r\) to complete one orbit and that \(T\) is the time period it takes to complete one orbit, so we can divide the distance by the time period to derive another useful expression for the orbital velocity,
$$v_o=\frac{2\pi r}T.$$
Weight of Satellite in Orbit
As we learned in the previous section, a satellite's mass or weight does not affect the speed necessary to orbit a planet. Still, the satellite has a weight in orbit. Its weight will depend on the satellite's distance from the center of the planet. We can determine a satellite's weight in orbit by calculating the force of gravity acting on the satellite while it is in its orbit:
$$F=\frac{GMm}{r^2},$$
where \(G\) is the gravitational constant \(6.67\times10^{-11}\;\frac{\mathrm N\;\mathrm m^2}{\mathrm{kg}^2}\), \(M\) is the planet's mass in kilograms \(\left(\text{kg}\right)\), \(m\) is the satellite's mass in kilograms \(\left(\text{kg}\right)\), and \(r\) is the distance between the satellite and the center of the planet in meters \(\left(\text{m}\right)\).
A satellite orbiting Earth is at a distance that is two times the radius of the Earth. We know that the satellite's weight on the Earth's surface is \(3500\,\text{N}\). What is the satellite's weight in orbit?
Because we know the satellite's weight on the surface of the Earth, we can determine its mass. We assume that the acceleration due to gravity on Earth's surface is \(g=10\,\frac{m}{s^2}\):
$$\begin{align*}w_s&=mg,\\m&=\frac{w_s}{g},\\m&=\frac{3500\,\text{N}}{10\,\frac{m}{s^2}},\\m&=350\,\text{kg}.\end{align*}$$
We know Earth's mass is \(M=5.94\times10^24\,\text{kg}\) and its radius is \(R=6.371\times10^6\,\text{m}\).Now we can determine the satellite's weight in orbit:
\begin{align*}F&=\frac{GMm}{r^2},\\F&=\frac{GMm}{\left(2R\right)^2},\\F&=\frac{\left(6.67\times10^{-11}\;\frac{\mathrm N\;\mathrm m^2}{\mathrm{kg}^2}\right)\left(5.94\times10^{24}\,\text{kg}\right)\left(350\,\text{kg}\right)}{\left(2\times6.371\times10^6\,\text{m}\right)^2},\\F&=854\,\text{N}.\end{align*}
The weight of the satellite in orbit is \(854\,\text{N}\).
Satellites Orbiting the Earth
As of 2021, there are more than 6,000 satellites orbiting the Earth and about 40% of them are operational. There are various types of satellite orbits, and the orbit for each satellite depends on what the satellite was designed to achieve.
Fig. 7 - Example orbits for Geostationary, Low Earth, and Medium Earth Orbits.
Low Earth Orbit (LEO)
Low Earth orbits have an orbital velocity of about \(7.8\,\frac{\text{km}}{\text{s}}\) and an altitude range from \(160\;\mathrm{km}\) to \(1,000\;\mathrm{km}\). These are the lowest altitude orbits, but they are still very far from the Earth. For reference, commercial airplanes fly about \(14\;\mathrm{km}\) above the surface, so low Earth orbit is in itself very high. These orbits offer advantages, as they do not have to follow along the equator and their plane can be tilted. This allows for more traffic of satellites. This type of orbit is used for the International Space Station for cheaper and more-efficient transfer of cargo and people between Earth and the ISS.
Satellites in low Earth orbits are mainly used for imaging, as being closer to the surface allows for higher-resolution pictures. These remote-sensing satellites are being used to discover natural resources, maximize agricultural production, and even monitor climate change. Some are placed in a type of low Earth orbit called polar orbit, as the satellite passes over the Earth's poles. Most of these orbits are also sun-synchronous as they are set to be in a fixed position with respect to the Sun. This allows for consistent illumination to take data for a long period of time.
Geostationary Orbit (GEO)
Satellites in geostationary orbits go around the equator of the Earth following its rotation, from west to east. A satellite in this type of orbit travels at the same rate as the Earth rotates. This means that for an observer on Earth, the satellite would appear to be stationary in the sky. Because of this, an antenna on Earth can stay fixed in one place and always be pointing at the satellite. Geostationary orbits are mainly used for telecommunications and weather satellites. The orbital velocity for geostationary orbits is \(3\,{\textstyle\frac{\mathrm{km}}{\mathrm s}}\) at an altitude of almost \(35,786\,\mathrm{km}\).
Q: What is the altitude and speed that a satellite must have so its orbit is considered to be a geostationary orbit?
A: The two expressions for the orbital speed discussed earlier are equal to each other. This is a form of Kepler's Third Law:
\begin{align*}\sqrt{\frac{GM}R}&=\frac{2\pi R}{T},\\\frac{GM}R&=\frac{4\pi^2R^2}{T^2}.\end{align*}
We solve for R,
\begin{align*}R^3=\frac{GMT^2}{4\pi^2},\\R=\sqrt[3]{\frac{GMT^2}{4\pi^2}}.\end{align*}
Because the satellite is in a geostationary orbit, its period will be equal to the period of the Earth's rotation, \(T=24\,\mathrm{hr}\) or \(T=86,400\,\mathrm{s}\). The mass of Earth is \(5.98x10^{24}\,\mathrm{kg}\).
\begin{align*}R&=\sqrt[3]{\frac{(6.67\times10^{-11}\,\frac{\mathrm m^3}{\mathrm{kg}\,\mathrm s^2})(5.98\times10^{24}\,\mathrm{kg}){(86,400\,\mathrm s)}^2}{4\pi^2}},\\R&=4.23x10^7\,\mathrm m.\end{align*}
This orbit radius includes the radius of the Earth, so to know the satellite's altitude above Earth's surface, we must solve for \(h\).
\begin{align*}h&=R-r_E,\\h&=(4.23\times10^7\,\mathrm m)-(6.37\times10^6),\\h&=3.59\times10^7\,m.\end{align*}
Now we use the equation for the orbital velocity.
\begin{align*}v_o&=\sqrt{\frac{GM}R},\\v_o&=\sqrt{\frac{(6.67x10^{-11}\frac{m^3}{kg\;s^2})(5.98x10^{24}\;\mathrm{kg})}{4.23x10^7\;\mathrm m}},\\v_o&=3,070\,\frac ms,\\v_o&=3.07\times10^3\,{\textstyle\frac{\mathrm m}{\mathrm s}}.\end{align*}
Medium Earth Orbit (MEO) and High Earth Orbit (HEO)
Medium Earth orbits have an altitude above the low Earth orbits and below high Earth orbits. These orbits are mainly used for navigation, like the Global Positioning System (GPS). Medium Earth orbits have an altitude range of between \(2000\,\text{km}\) and \(35786\,\text{km}\). Geostationary orbits are considered medium Earth orbits.
High Earth orbits have an altitude greater than geostationary orbits. Satellites in this type of orbit have an orbital period longer than the duration of a day on Earth. Consequently, for people watching from the surface of the Earth, the satellite looks like it is undergoing retrograde motion.
Lagrange Points
Lagrange points allow orbits that are very far away from the Earth and that do not orbit the Earth. These are spots in space where the gravity of the Earth and Sun combine in a way that they can be orbited by satellites by maintaining a stable position that does not lead them to Earth or to the Sun.
One example of a satellite in a Lagrange Point is the James Webb Space Telescope (JWST), whose mission is to photograph deep space in a way that it can capture infrared radiation from the faintest galaxies. If this orbit were closer to Earth, the telescope would not work as well, because of the light pollution coming from the Earth. Orbiting a Lagrange point also allows the JWST to improve its resolution and minimize the energy required to maintain its orbit. So you have an idea of how far this orbit is, its distance is \(1.5\;\text{million km}\) or 4 times the distance between the Earth and the Moon.
Fig. 8 - Lagrange Points L1, L2, L3, L4, and L5 are stable points that satellites can orbit.
Satellite Orbits - Key takeaways
- A satellite is a celestial or artificial body orbiting the Earth or another planet.
- Kepler's laws of planetary motion can be used to describe satellite orbits around the Earth. For a circular orbit, Kepler's Third Law can be expressed as follows, \(T^2=\frac{4\pi^2}{GM}R^3\).
- An eccentricity of zero refers to a circular orbit. The larger the eccentricity, the more elongated is the orbit. In a circular orbit, the satellite will move at the same speed throughout the orbit. In an elliptical orbit, the satellite moves faster when it is near the planet and moves slower when farthest away from the planet.
- The escape velocity is the minimum velocity an object needs to escape Earth's gravity. The object would continue moving at constant velocity away from the Earth.
- The velocity needed to balance Earth's gravity and the satellite's inertia is called the orbital velocity. At this speed, the satellite orbits the Earth.
- A satellite's mass does not affect its orbit.
- Satellites in Geostationary Orbits go around the equator of the Earth following its rotation, from west to east. A satellite in this type of orbit travels at the same rate as the Earth rotates.
- Lagrange points allow orbits that are very far away from the Earth and that do not orbit the Earth. These are spots in space where the gravity of the Earth and Sun combine in a way that they can be orbited by satellites by maintaining a stable position.
References
- Fig. 1 - A satellite can help us in more ways than we can imagine. Here we show a satellite doing remote sensing to optimize crops and natural resources (https://openclipart.org/detail/244734/satellite-imaging-remote-sensing), by Juhel (https://openclipart.org/artist/Juhele), licensed by CC0 1.0 Universal (https://creativecommons.org/publicdomain/zero/1.0/)
- Fig. 2 - Elipses have two focus points as well as a semi-major axis and a semi-minor axis, StudySmarter Originals
- Fig. 3 - Different types of orbital trajectories depend on the eccentricity: in blue we have a circular orbit, in green we have an elliptical orbit, in red we have a parabolic orbit, and in purple we have a hyperbolic orbit, StudySmarter Originals
- Fig. 4 - A visual representation of Kepler’s Second Law, StudySmarter Originals
- Fig. 5 - Visualization of the reduced mass. A two-body problem is turned into a one-body problem by using the Center of Mass frame and the reduced mass, StudySmarter Originals
- Fig. 6 - Representation of our reduced mass system. This is an elliptical orbit using the reduced mass. The maximum and minimum distances from the central body are located on the horizontal axis, StudySmarter Originals
- Fig. 7 - Example orbits for Geostationary, Low Earth, and Medium Earth Orbits (https://commons.wikimedia.org/wiki/File:Medium_Earth_Orbit.png), by DanielT29 (https://commons.wikimedia.org/w/index.php?title=User:DanielT29&action=edit&redlink=1), licensed by CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/)
- Fig. 8 - Lagrange Points L1, L2, L3, L4, and L5 are stable points that satellites can orbit (https://commons.wikimedia.org/wiki/File:Lagrange_points2.svg), by Xander89 (https://commons.wikimedia.org/wiki/User:Xander89), licensed by CC BY-SA 3.0 (https://creativecommons.org/licenses/by/3.0/deed.en)
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Frequently Asked Questions about Satellite Orbits
How many satellites orbit earth?
As of 2021, there are more than 6,000 satellites orbiting the Earth and about 40% of them are operational.
How does the satellite stay in orbit?
If the satellite goes too fast it will drift away with constant velocity until a force acts upon it, as described by the law of inertia or Newton's First Law. If the satellite goes too slow it will return to the Earth. The velocity needed to balance Earth's gravity and the satellite's inertia is called the orbital velocity.
How fast do satellites travel in orbit?
It depends on the type of orbit. For example, Geostationary Orbits have an orbital velocity of 3 km/s and Low Earth Orbits have an orbital velocity of 7.8 km/s.
How fast does a low earth orbit satellite travel?
Low Earth Orbits have an orbital velocity of 7.8 km/s.
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