# 2.2.4.4: Satellite Motion

- Page ID
- 84586

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In this subsection special attention will be paid to the perfectly circular trajectory of the projectile discussed in the preceding. Note that the value of acceleration due to gravity at the launching platform 1000 km above the Earth surface, as obtained from Eq. 2.16, is \( g = 7.3362 \; m/s^2 \). It has been found that the projectile flies along a perfectly circular orbit when it's launched with a speed of 7353.6 m/s. The radius of the orbit is 6371 km + 1000 km = 7371,000 m. Let's then ask what is the centripetal acceleration \(a_{cp}\) in circular motion which such parameters? From the Eq. 2.28 we get:

\[ a_{cp} = \dfrac{(7353.6 \; m/s )^2}{7371,000 \; m} = 7.3362 \; {m/s}^2 \]

Note: Exactly the same as the \( g \) value 7371 km from the Earth center.

So, we can conclude: in a flight along a circular path of radius 7371 km with a speed of 7353.6 m/s an object experiences a centripetal acceleration of \(a_{cp} = 7.3362 \; m/s^2\) -or, one can say it in a different but equivalent way: if the projectile's mass is \(m\), such a motion has to be maintained by a \textbf{centripetal force} of

\[ F_{\rm cp} =m\times a_{\rm cp} = m\times7.3362{\rm ~m/s}^2 \]

And this is exactly the force supplied by gravity on this orbit!

And if we want to have a satellite on circular orbit at a different altitude? The rule is the same for all circular orbits: the centripetal acceleration must be equal to acceleration due to gravity at the orbit's altitude, from which we can find the linear speed needed for maintaining a satellite at any altitude \(H\):

\[ \dfrac{v^2}{R_{\rm e} + H} = \dfrac{GM_{\rm e}}{(R_{\rm e} + H)^2}\;\;\;\Rightarrow \;\;\; v(H) = \sqrt{\dfrac{GM_{\rm e}}{(R_{\rm e} + H)}}\]

If the Earth's surface were perfectly smooth, and there were no atmosphere, a satellite could fly at a nearly-zero altitude. The speed of such satellite should be:

\[ v(H=0) = \sqrt{\dfrac{GM_{\rm e}}{R_{\rm e}}} = \sqrt{g(H=0)\times R_{\rm e}} =7910{\rm ~m/s}\]

This velocity is commonly referred to as “the first cosmic velocity''.

As follows from the discussions in the preceding section and this section, any kind of motion in the gravitational field of Earth, as well as of another planet, moon, or the Sun can be fully described by taking into consideration \textbf{just one} force -the force of gravity -and

solving the equations of motion based on the Second Law of Dynamics, \(\vec{F} = m\vec{a}\). There is no need to introduce any other factors. It should be noted that this is exactly the method professional astronomers, including NASA specialists, use for analyzing/predicting the motion of all celestial bodies in the Solar System. The only complication in such calculations is that the total force acting on one object is usually a vector sum of several forces, each of which which corresponds to attractive interaction with another celestial body. For instance, for calculating the motion of Moon -the Earth's natural satellite -relative to Earth, one should take into consideration the attractive forces exerted on Moon both by Earth and Sun. Because its enormous mass, the Sun can be assumed as a “stationary body'' -this is the so-called “heliocentric model'' -but Earth has to be treated as a body moving under the influence of **two** forces, the attraction by Sun **and** the attraction by Moon. The equations-of-motion for this three-body system are:

\[ \begin{matrix}

m_{m}\vec{a}_{m} & = & \vec{F}_{m \rightarrow e}+\vec{F}_{m \rightarrow s} \\

m_{e}\vec{a}_{e} & = & \vec{F}_{e \rightarrow m}+\vec{F}_{e \rightarrow s} \\

\end{matrix} \]

where \(\vec{F}_{m \rightarrow e}\) means: “force attracting Moon towards Earth'', and similarly for other symbols. Only such a model is able to reproduce correctly the motion of Moon as seen from Earth.^{1}

In the case of small artificial satellites, due to their negligibly small masses compared with that of Earth, a simpler math, such as the Eqs. 1.32, is already sufficient for analyzing their orbits. However, even such math may look scary for students who are at early stages of their education. Therefore, instructors and textbooks often choose to present an explanation of satellite motion that is surely much clearer conceptually, and needs only some pretty simple math.

Namely, it says -the motion along a circular orbit produces an outwards-pointing \textbf{centrifugal} force, which exactly compensates the inwards-pointing force of gravity.

Because of such compensation, the explanation goes on, nothing is pulling the satellite out of its orbit. This reasoning is backed by math: the centrifugal force, \(mv^2/r\), is equated with \(GmM/r^2\), which leads to the correct result: \(v = \sqrt{GM/r^2}\),essentially the same result as has been obtained in the Eq. 2.33.

However, the fact that a theory yields a correct results does not mean automatically that it is itself correct. There are some elements of doubtful validity in the reasoning -the Newton's First Law of Dynamics states that if all forces acting on a body are compensated, then the body may move, yes, but along a straight line. So how comes that moves along a circular route? Another weakness is that the explanation is based on the concept of centrifugal force. We all are very well familiar with this force, we can experience it in a car taking a turn, on a carousel, or swinging a stone tied to a string overhead. Yet, it is not a real force, it is the force of inertia that emerges in a rotating frame as a response to the centripetal force delivered by the car seat or the carousel seat to our body, or conveyed from our hand to the stone through the string. And, notice, in an empty space where the satellites move there no such thing as a "frame" of any kind.

That simple explanation can be fixed by adding some “extra'' elements -which is usually not done by instructors, though. The extra element is abstract, it is an imaginary rotating frame -sort of an invisible string tied one end to the satellite, the other end to the center of the circular orbit -and the tension force in the string equal to the force of gravity acting on the stone. The imaginary rotating frame does not exist, of course -but if it did, the motion of the satellite would be the same as it is without it. So, the model works.

Yet, such a “fixed'' model has one major disadvantage: it does work, right, but only in the case of a perfectly circular orbit. But try to use the same reasoning for a satellite on an elliptic orbit. For any point at a continuous curve, there exists an osculating circle -i.e., a circle that touches the curve at this point and has the same tangent and curvature at that point. The radius of such circle is the radius of curvature of the curve at that point. So, for a satellite moving along an elliptic path, if its velocity at a given point is \(v\), and the radius of curvature at this point is \(r\), one can find the “centrifugal force" at this very point equal to \(mv^2/r\). The problem is, however, that the force is directed towards the center of the osculating circle -which, in general, is not the same direction as of the force of gravity vector at the given point. So, the two forces can not compensate one another! So, what happens next to the satellite?

Again, the model can be fixed and used for calculating the parameters of the elliptical orbit -but now the “extras'' needed for it are far more complicated than in the case of a circular orbit.

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1. Even though the equations, as written in this text, may look pretty simple -solving them is definitely not. It took astronomers nearly 200 years to develop sophisticated mathematical tools that enabled them to obtain, using “paper and pencil'', satisfactory predictions of the Moon's position for the next tens of years. It should be stressed that the motions of Moon are pretty complicated -one manifestation of that are the highly irregular patterns of the solar and lunar eclipses. But the good news is that today high quality solutions of the equations discussed here can be obtained in minutes using computational methods.}