Mathematics EARTH SATELLITES, ENERGY OF AN ORBITING SATELLITE, GEOSTATIONARY, POLAR SATELLITES AND WEIGHTLESSNESS

### Topic covered

star EARTH SATELLITES
star ENERGY OF AN ORBITING SATELLITE
star GEOSTATIONARY AND POLAR SATELLITES
star WEIGHTLESSNESS

### EARTH SATELLITES

color{green} ✍️ 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.

● 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

color{blue}{F("centripetal") = (mV^2)/((R_E+h)).......................(8.33)}

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

color{blue}{F(gravitation)=(GmM_E)/((R_E+h)^2)......................(8.34)} where M_E is the mass of the earth.

Equating R.H.S of Eqs. (8.33) and (8.34) and cancelling out m, we get

color{blue}{V^2=(GM_E)/((R_E +h))....................(8.35)}

● Thus V decreases as h increases. From equation (8.35),the speed V for h = 0 is

color{blue}{V^2(h=0)= GM//R_E=gR_E........................(8.36)}

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

color{blue}{T=(2 pi(R_E+h))/V = (2 pi (R_E +h)^(3//2))/(sqrt(G M_E))............................(8.37)}

on substitution of value of V from Eq. (8.35). Squaring both sides of Eq. (8.37), we get

color{blue}{T^2=k(R_E+h)^3 " "("where "k = 4 pi^2 // GM_E).................... (8.38)}

=> 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 RE in Eq. (8.38). Hence, for such satellites, T is T_o, where

color{blue}{T_0= 2 pi(sqrt(R_E//g)).....................(8.39)}

● If we substitute the numerical values g approx 9.8 m s^(-2) and R_E = 6400 km., we get

color{blue}{T_0 = 2 pi sqrt((6.4 xx 10^6)/(9.8)) s}

Which is approximately 85 minutes.
Q 3260312215

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 xx10^3 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 ?

Solution:

(i) We employ Eq. (8.38) with the sun’s mass replaced by the martian mass M_m

color{green} {T^2=(4 pi^2)/(GM_m) R^3}

color{purple}{M_m =(4 pi^2 )/G R^3/T^2}

=(4 xx (3.14)^3 xx (9.4)^3 xx 10^18)/(6.67 xx 10^(-11) xx (459 xx 60)^2)

M_m =(4 xx (3.14)^2 xx (9.4)^3 xx 10^18)/(6.67 xx (4.59 xx 6)^2 xx 10^(-5))

=6.48 xx 10^23 kg

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

color{orange} {(T_M^2)/(T_E^2) =(R_(MS)^3)/(R_(ES)^3)}

where R_(MS) is the mars -sun distance and R_(ES) is the earth-sun distance

:. T_M=(1.52)^(3//2) xx 365

=684 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.
Q 3210412310

Weighing the Earth : You are given the following data: g = 9.81 ms^(–2), R_E = 6.37xx10^6 m, the distance to the moon
R = 3.84xx10^8 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.

Solution:

From Eq. (8.12) we have

color{purple} {M_E=(g R_E^2)/G}

=(9.81 xx (6.37 xx 10^6)^2)/(6.67 xx 10^(-11))

=5.97 xx 10^(24) kg

The moon is a satellite of the Earth. From the derivation of Kepler’s third law [see Eq. (8.38)]

color{green} {T^2=(4 pi^2 R^3)/(G M_E)}

color{orange} {M_E=(4 pi^2 R^3)/(GT^2)}

=(4 xx 3.14 xx 3.14 xx (3.84)^3 xx 10^24)/(6.67 xx 10^(-11) xx (27.3 xx 24 xx 60 xx 60)^2)

= 6.02 xx 10^24  kg

Both methods yield almost the same answer, the difference between them being less than 1%.
Q 3210512410

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 xx 10^5 km from the earth. Obtain its time-period of revolution in days.

Solution:

Given
color{orange} {k=10^(-13) s^2 m^(-3)}

=10^(-13) [1/((24 xx 60 xx 60)^2) d^2][1/((1//1000)^3 km^3)]

=1.33 xx 10^(-14) d^2 km^(-3)

Using Eq. (8.38) and the given value of k, the time period of the moon is

T^2 = (1.33 xx 10^(-14))(3.84 xx 10^5)^3

T = 27.3 d

color{blue} {"Note :"} Eq. (8.38) 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.

### ENERGY OF AN ORBITING SATELLITE

● Using Eq. (8.35), the kinetic energy of the satellite in a circular orbit with speed v is

K* E =1/2 mv^2

color{blue}{=(GmM_E)/(2(R_E +h))...........................(8.40)}

● Considering gravitational potential energy at infinity to be zero, the potential energy at distance (R+h) from the centre of the earth is

color{blue}{P*E =-(GmM_E)/((R_E +h))......................(8.41)}

● The K.E is positive whereas the P.E is negative. However, in magnitude the K.E. is half the P.E, so that the total E is

color{blue}{E=K*E+P*E=-(GmM_E)/(2(R_E+h))......................(8.42)}

● The total energy of an circularly orbiting satellite is thus negative, with the potential energy being negative but twice is magnitude of the positive kinetic energy. When the orbit of a satellite becomes elliptic, both the K.E. and P.E. vary from point to point.

● The total energy which remains constant is negative as in the circular orbit case. This is what we expect, since as we have discussed before if the total energy is positive or zero, the object escapes to infinity. Satellites are always at finite distance from the earth and hence their energies cannot be positive or zero.
Q 3250512414

A 400 kg satellite is in a circular orbit of radius 2R_E about the Earth. How much energy is required to transfer it to a circular orbit of radius 4R_E ? What are the changes in the kinetic and potential energies ?

Solution:

Initially,

color{green} {E_i=-(GM_E m)/(4 R_E)}

While finally

E_f=-(GM_E m)/(8 R_E)

The change in the total energy is

color{orange}{Delta E= E_f-E_i}

color{purple} {=(GM_Em)/(8R_E)=((GM_E)/(R_E^2))(mR_E)/8}

Delta E=(G m R_E)/8 =(9.81 xx 400 xx 6.37 xx 10^6)/8 =3.13 xx 10^9 J

The kinetic energy is reduced and it mimics

Delta E, namely, Delta K = K_f – K_i = – 3.13 xx 10^9 J.

The change in potential energy is twice the change in the total energy, namely

color{purple} {Delta V = V_f – V_i = – 6.25 xx 10^9 J}

### GEOSTATIONARY AND POLAR SATELLITES

color{green} ✍️  An interesting phenomenon arises if in we arrange the value of (R_E + h) such that T in Eq. (8.37) becomes equal to 24 hours.
● If the circular orbit is in the equatorial plane of the earth, such a satellite, having the same period as the period of rotation of the earth about its own axis would appear stationery viewed from a point on earth. The (R _E + h) for this purpose works out to be large as compared to R_E :

color{blue}{R_E +h = ((T^2 GM_E)/(4 pi^2))^(1//3)...........................(8.43)}

and for T = 24 hours, h works out to be 35800 km.
=> which is much larger than R_E. Satellites in a circular orbits around the earth in the equatorial plane with T = 24 hours are called Geostationery Satellites.
● Clearly, since the earth rotates with the same period, the satellite would appear fixed from any point on earth. It takes very powerful rockets to throw up a satellite to such large heights above the earth but this has been done in view of the several benefits of many practical applications.

● It is known that electromagnetic waves above a certain frequency are not reflected from ionosphere. Radio waves used for radio broadcast which are in the frequency range 2 MHz to 10 MHz, are below the critical frequency.
● They are therefore reflected by the ionosphere. Thus radio waves broadcast from an antenna can be received at points far away where the direct wave fail to reach on account of the curvature of the earth.
● Waves used in television broadcast or other forms of communication have much higher frequencies and thus cannot be received beyond the line of sight.
● A Geostationery satellite, appearing fixed above the broadcasting station can however receive these signals and broadcast them back to a wide area on earth. The INSAT group of satellites sent up by India are one such group of Geostationary satellites widely used for telecommunications in India.

● Another class of satellites are called the Polar satellites (Fig. 8.11). These are low altitude (h approx 500 " to " 800 "km") satellites, but they go around the poles of the earth in a north-south direction whereas the earth rotates around its axis in an east-west direction.
● Since its time period is around 100 minutes it crosses any altitude many times a day. However, since its height h above the earth is about 500-800 km, a camera fixed on it can view only small strips of the earth in one orbit. Adjacent strips are viewed in the next orbit, so that in effect the whole earth can be viewed strip by strip during the entire day.
● These satellites can view polar and equatorial regions at close distances with good resolution. Information gathered from such satellites is extremely useful for remote sensing, meterology as well as for environmental studies of the earth.

### WEIGHTLESSNESS

color{green} ✍️ Weight of an object is the force with which the earth attracts it. We are conscious of our own weight when we stand on a surface, since the surface exerts a force opposite to our weight to keep us at rest.
● The same principle holds good when we measure the weight of an object by a spring balance hung from a fixed point e.g. the ceiling. The object would fall down unless it is subject to a force opposite to gravity.
● This is exactly what the spring exerts on the object. This is because the spring is pulled down a little by the gravitational pull of the object and in turn the spring exerts a force on the object vertically upwards.

color{green} ✍️ Now, imagine that the top end of the balance is no longer held fixed to the top ceiling of the room. Both ends of the spring as well as the object move with identical acceleration g.
● The spring is not stretched and does not exert any upward force on the object which is moving down with acceleration g due to gravity. The reading recorded in the spring balance is zero since the spring is not stretched at all. If the object were a human being, he or she will not feel his weight since there is no upward force on him.
● Thus, when an object is in free fall, it is weightless and this phenomenon is usually called the phenomenon of weightlessness.

color{green} ✍️ In a satellite around the earth, every part and parcel of the satellite has an acceleration towards the centre of the earth which is exactly the value of earth’s acceleration due to gravity at that position.
● Thus in the satellite everything inside it is in a state of free fall. This is just as if we were falling towards the earth from a height. Thus, in a manned satellite, people inside experience no gravity.
● Gravity for us defines the vertical direction and thus for them there are no horizontal or vertical directions, all directions are the same. Pictures of astronauts floating in a satellite show this fact.