8.10 Earth satellites

Earth satellites are objects which revolve around the earth. Their motion is very similar to the motion of planets around the Sun and hence Kepler’s laws of planetary motion are equally applicable to them.

We will consider a satellite in a circular orbit of a distance \(\left(R+h\right)\) from the centre of the earth, where \(R=\) radius of the earth. If \(m\) is the mass of the satellite and \(V\) its speed, the centripetal force required for this orbit is
\(
\mathrm{F}(\text { centripetal })=\frac{m V^2}{\left(R_E+h\right)} \dots(1.1)
\)
directed towards the centre. This centripetal force is provided by the gravitational force, which is
\(
\text { F(gravitation })=\frac{G m M}{\left(R+h\right)^2} \dots(1.2)
\)
where \(M\) is the mass of the earth.
Equating R.H.S of Eqs. (1.1) and (1.2) and canceling out \(m\), we get
\(
V^2=\frac{G M}{\left(R+h\right)} \dots(1.3)
\)
Thus \(V\) decreases as \(h\) increases. From equation (1.3),the speed \(V\) for \(h=0\) is
\(
V^2 \quad(h=0)=G M / R=g R \dots(1.4)
\)
where we have used the relation \(\mathrm{g}=G M / R^2\). In every orbit, the satellite traverses a distance \(2 \pi\left(R+h\right)\) with speed \(V\). Its time period \(T\) therefore is
\(
T=\frac{2 \pi\left(R+h\right)}{V}=\frac{2 \pi\left(R+h\right)^{3 / 2}}{\sqrt{G M}} \dots(1.5)
\)
on substitution of value of \(V\) from Eq. (1.3). Squaring both sides of Eq. (1.5), we get
\(T^2=k\left(R+h\right)^3\left(\right.\) where \(\left.k=4 \pi^2 / G M\right) \dots(1.6)\)

which is Kepler’s law of periods, as applied to motion of satellites around the earth. For a satellite very close to the surface of earth \(h\) can be neglected in comparison to \(R\) in Eq. (1.6). Hence, for such satellites, \(T\) is \(T_{\mathrm{o}}\), where
\(
\begin{gathered}
T_0=2 \pi \sqrt{R / g} \dots(1.7) \\
\text { If we substitute the numerical values } \\
g \simeq 9.8 \mathrm{~m} \mathrm{~s}^{-2} \text { and } R=6400 \mathrm{~km} ., \text { we get } \\
T_0=2 \pi \sqrt{\frac{6.4 \times 10^6}{9.8}} \mathrm{~s}
\end{gathered}
\)
Which is approximately 85 minutes.

Example 1: The planet Mars has two moons, phobos and delmos. (1) phobos has a perlod 7 hours, 39 minutes and an orbital radius of \(9.4 \times 10^3 \mathrm{~km}\). Calculate the mass of mars. (ii) Assume that earth and mars move in circular orbits around the sun, with the martian orbit being \(1.52\) times the orbital radius of the earth. What is the length of the martlan year in days?

Solution:

(i) We employ Eq. (1.6) with the sun’s mass replaced by the martian mass \(M_m\)
\(T^2=\frac{4 \pi^2}{G M_{\text {pl }}} R^3\) \(\mathrm{M}_m=\frac{4 \pi^2}{G} \frac{R^3}{T^2}\) \(=\frac{4 \times(3.14)^2 \times(9.4)^3 \times 10^{18}}{6.67 \times 10^{-11} \times(459 \times 60)^2}\) \(\mathrm{M}_m=\frac{4 \times(3.14)^2 \times(9.4)^3 \times 10^{18}}{6.67 \times(4.59 \times 6)^2 \times 10^{-5}}\) \(=6.48 \times 10^{23} \mathrm{~kg}\).
(ii) Once again Kepler’s third law comes to our ald,
\(
\frac{T_M^2}{T_E^2}=\frac{R_{M S}^3}{R_{E S}^3}
\)
where \(R_{M S}\) is the mars -sun distance and \(R_{E S}\) is the earth-sun distance.
\(
\begin{aligned}
\therefore T_M &=(1.52)^{3 / 2} \times 365 \\
&=684 \text { days }
\end{aligned}
\)
We note that the orbits of all planets except Mercury, Mars, and Pluto are very close to being circular. For example, the ratio of the semi-minor to the semi-major axis for our Earth 1s, \(b / a=0.99986\).

Example 2: Weighing the Earth: You are given the following data: \(g=9.81 \mathrm{~ms}^{-2}\), \(R_E=6.37 \times 10^6 \mathrm{~m}\), the distance to the moon \(R=3.84 \times 10^8 \mathrm{~m}\) and the time period of the moon’s revolution is \(27.3\) days. Obtain the mass of the Earth \(M_E\) in two different ways.

Solution:

The acceleration experienced by the mass \(\mathrm{m}\), which is usually denoted by the symbol \(g\) is related to \(\mathrm{F}\) by Newton’s \(2^{\text {nd }}\) law by relation \(F=m g\). Thus

\(
g=\frac{F}{m}=\frac{G M_E}{R_E^2}
\)
\(
M_E=\frac{g R_E^2}{G}
\)
\(
\begin{aligned}
&=\frac{9.81 \times\left(6.37 \times 10^6\right)^2}{6.67 \times 10^{-11}} \\
&=5.97 \times 10^{24} \mathrm{~kg} .
\end{aligned}
\)
The moon is a satellite of the Earth. From the derivation of Kepler’s third law [see Eq (1.6)]
\(
\begin{aligned}
&T^2=\frac{4 \pi^2 R^3}{G M_E} \\
&M_E=\frac{4 \pi^2 R^3}{G T^2} \\
&=\frac{4 \times 3.14 \times 3.14 \times(3.84)^3 \times 10^{24}}{6.67 \times 10^{-11} \times(27.3 \times 24 \times 60 \times 60)^2} \\
&=6.02 \times 10^{24} \mathrm{~kg}
\end{aligned}
\)
Both methods yield almost the same answer, the difference between them being less than \(1 \%\).

Example 3: Express the constant \(\mathrm{k}\) of Eq. (1.6) in days and kilometers. Given \(\mathrm{k}=10^{-13} \mathrm{~s}^2 \mathrm{~m}^{-3}\). The moon \(1 \mathrm{~s}\) at a distance of \(3.84 \times 10^5 \mathrm{~km}\) from the earth. Obtain its time-period of revolution in days.

Solution: Given
\(
\begin{aligned}
& k=10^{-13} \mathrm{~s}^2 \mathrm{~m}^{-3} \\
=& 10^{-13}\left[\frac{1}{(24 \times 60 \times 60)^2} \mathrm{~d}^2\right]\left[\frac{1}{(1 / 1000)^3 \mathrm{~km}^3}\right] \\
=& 1.33 \times 10^{-14} \mathrm{~d}^2 \mathrm{~km}^{-3}
\end{aligned}
\)
Using Eq. (1.6) and the given value of \(\mathrm{k}\), the time period of the moon is
\(
\begin{aligned}
&T^2=\left(1.33 \times 10^{-14}\right)\left(3.84 \times 10^5\right)^3 \\
&T=27.3 \mathrm{~d}
\end{aligned}
\)
Note that Eq. (1.6) also holds for elliptical orbits if we replace \(\left(R_E+h\right)\) by the semi-major axis of the ellipse. The earth will then be at one of the foci of this ellipse.