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. In particular, their orbits around the earth are circular or elliptic. Moon is the only natural satellite of the earth with a near circular orbit with a time period of approximately 27.3 days which is also roughly equal to the rotational period of the moon about its own axis. Since, 1957, advances in technology have enabled many countries including India to launch artificial earth satellites for practical use in fields like telecommunication, geophysics and meteorology.

We will consider a satellite in a circular orbit of a distance ($R_E + h$) from the centre of the earth, where $R_E$ = radius of the earth. If m is the mass of the satellite and V its speed, the centripetal force required for this orbit is

F(centripetal) = $\frac{mV^2}{(R_E+h)}$     (1)

directed towards the centre. This centripetal force is provided by the gravitational force, which is

F(gravitation) = $\frac{GmM_E}{(R_E+h)^2}$     (2)

where $M_E$ is the mass of the earth. 

Equating R.H.S of Eqs. (1) and (2) and cancelling out m, we get

$V^2=\frac{GM_E}{(R_E+h)}$                (3)

Thus V decreases as h increases. From equation (3),the speed V for h = 0 is

$V^2(h=0)=\frac{GM_E}{R_E}=gR_E$      (4)

where we have used the relation $g = 2GM/R_E$ . In every orbit, the satellite traverses a distance 2π($R_E + h$) with speed V. Its time period T therefore is

$T=\frac{2\pi(R_E+h)}{V}=\frac{2\pi(R_E+h)^{3/2}}{\sqrt{GM_E}}$ (5)

on substitution of value of V from Eq. (3). Squaring both sides of Eq. (5), we get

$T^2=k(R_E+h)^3$   (where $k = 4\pi^2/GM_E$)          (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 $R_E$ in Eq. (6). Hence, for such satellites, T is $T_o$, where

$T_0=2\pi\sqrt{\frac{R_E}{g}}$          (7)

If we substitute the numerical values $g \simeq 9.8 \ m.s^{-2}$ and $R_E$ = 6400 km., we get

$T_0=2\pi\sqrt{\frac{6.4 \times 10^6}{9.8}}$ s

Which is approximately 85 minutes.

Example 1

The planet Mars has two moons, phobos and delmos. (i) phobos has a period 7 hours, 39 minutes and an orbital radius of 9.4 × 10$^3$ 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 martian year in days?

Answer 

(i) We employ Eq. (7) with the sun’s mass replaced by the martian mass $M_m$

$T^2=\frac{4\pi^2}{GM_m}R^3$

$M_m=\frac{4\pi^2}{G}\frac{R^3}{T^2}$

$=\frac{4\times (3.14)^2\times (9.4)^3\times (10)^{18}}{6.67 \times 10^{-11}\times (459 \times 60)^2}$

$M_m=6.48 \times 10^{23}$ kg

(ii) Once again Kepler’s third law comes to our aid,

$\frac{T_M^2}{T_E^2}=\frac{R_{MS}^2}{R_{ES}^2}$

where $R_{MS}$ is the mars -sun distance and $R_{ES}$ is the earth-sun distance.

∴ $T_M=(1.52)^{3/2}\times 365$ = 684 days

days 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 semi-major axis for our Earth is, b/a = 0.99986.

Example 2

Weighing the Earth : You are given the following data: g = 9.81 m.s$^{–2}$, R$_E$ = 6.37 × 10$^6$ m, the distance to the moon R = 3.84 × 10$^8$ 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.

Answer 

From Eq. $g=\frac{GM_E}{R_E^2}$ we have

$M_E=\frac{gR_E^2}{G}$

$=\frac{9.81\times (6.37 \times 10^6)^2}{6.67 \times 10^{-11}}$

$=5.97 \times 10^{24}$ kg

The moon is a satellite of the Earth. From the derivation of Kepler’s third law [see Eq. (7)]

$T^2=\frac{4\pi^2R^3}{GM_E}$

$M_E=\frac{4\pi^2R^3}{GT^2}$

$=\frac{4\times (3.14)^2 \times (3.84)^3 \times 10^{24}}{6.67 \times 10^{-11}\times (27.3 \times 24 \times 60^2)^2}$

$M_E=6.02\times 10^{24}$ kg

Both methods yield almost the same answer, the difference between them being less than 1%.

Example 3 

Express the constant k of Eq. (8.38) in days and kilometres. Given $k = 10^{–13} \ s^2 m^{–3}$. The moon is at a distance of 3.84 × 10$^5$ km from the earth. Obtain its time-period of revolution in days.

Answer 

Given $k = 10^{–13} \ s^2 m^{–3}$

$=10^{-13}\left[\frac{1}{(24\times 60 \times 60)^2}d^2\right]\left[\frac{1}{(1/1000)^3km^3}\right]$

$=1.33 \times 10^{-14}d^2 \ km^{-3}$

Using Eq. (7) and the given value of k, the time period of the moon is

$T^2=(1.38 \times 10^{-14})(3.84 \times 10^5)^3$

T = 27.3 d

Note that Eq. (7) also holds for elliptical orbits if we replace ($R_E+h$) by the semi-major axis of the ellipse. The earth will then be at one of the foci of this ellipse.