Explanation About Earth Satellites

Explanation About 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) = \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WGhbGaamyBaiaad2eadaWgaaWcbaGaamyraaqabaaakeaadaqadaqa % aiaadkfadaWgaaWcbaGaamyraaqabaGccqGHRaWkcaWGObaacaGLOa % Gaayzkaaaaaaaa!3EC5! \frac{{Gm{M_E}}}{{\left( {{R_E} + h} \right)}}\)            (8.33)

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

F(gravitation) = \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WGhbGaamyBaiaad2eadaWgaaWcbaGaamyraaqabaaakeaadaqadaqa % aiaadkfadaWgaaWcbaGaamyraaqabaGccqGHRaWkcaWGObaacaGLOa % GaayzkaaWaaWbaaSqabeaacaaIYaaaaaaaaaa!3FAE! \frac{{Gm{M_E}}}{{{{\left( {{R_E} + h} \right)}^2}}}\)          (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

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaCa % aaleqabaGaaGOmaaaakiabg2da9maalaaabaGaam4raiaad2eadaWg % aaWcbaGaamyraaqabaaakeaadaqadaqaaiaadkfadaWgaaWcbaGaam % yraaqabaGccqGHRaWkcaWGObaacaGLOaGaayzkaaaaaaaa!40A7! {V^2} = \frac{{G{M_E}}}{{\left( {{R_E} + h} \right)}}\)                              (8.35)

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

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivaiabg2 % da9maalaaabaGaaGOmaiabec8aWnaabmaabaGaamOuamaaBaaaleaa % caWGfbaabeaakiabgUcaRiaadIgaaiaawIcacaGLPaaaaeaacaWGwb % aaaiabg2da9maalaaabaGaaGOmaiabec8aWnaabmaabaGaamOuamaa % BaaaleaacaWGfbaabeaakiabgUcaRiaadIgaaiaawIcacaGLPaaada % ahaaWcbeqaaiaaiodacaGGVaGaaGOmaaaaaOqaamaakaaabaGaam4r % aiaad2eadaWgaaWcbaGaamyraaqabaaabeaaaaaaaa!4E2D! T = \frac{{2\pi \left( {{R_E} + h} \right)}}{V} = \frac{{2\pi {{\left( {{R_E} + h} \right)}^{3/2}}}}{{\sqrt {G{M_E}} }}\)   (8.37)

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

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 R_E in Eq. (8.38). 

Hence, for such satellites, T is T_o, where 

T_o=\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOmaiabec % 8aWnaakaaabaGaamOuamaaBaaaleaacaWGfbaabeaakiaac+cacaWG % Nbaaleqaaaaa!3C00! 2\pi \sqrt {{R_E}/g} \)                   (8.39)

If we substitute the numerical values g \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeS4qISdaaa!3727! \simeq \) 9.8 m s^-2 and R = 6400 km., we get

T_o= \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0JaaG % Omaiabec8aWnaakaaabaWaaSaaaeaacaaI2aGaaiOlaiaaisdacqGH % xdaTcaaIXaGaaGimamaaCaaaleqabaGaaGOnaaaaaOqaaiaaiMdaca % GGUaGaaGioaaaaaSqabaGccaWGZbaaaa!438C! = 2\pi \sqrt {\frac{{6.4 \times {{10}^6}}}{{9.8}}} s\)

Which is approximately 85 minutes.

EXAMPLE 5

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 \(\times\) 10³ 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

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGub % WaaWbaaSqabeaacaaIYaaaaOGaeyypa0ZaaSaaaeaacaaI0aGaeqiW % da3aaWbaaSqabeaacaaIYaaaaaGcbaGaam4raiaad2eadaWgaaWcba % GaamyBaaqabaaaaOGaamOuamaaCaaaleqabaGaaG4maaaaaOqaaiaa % d2eadaWgaaWcbaGaamyBaaqabaGccqGH9aqpdaWcaaqaaiaaisdacq % aHapaCdaahaaWcbeqaaiaaikdaaaaakeaacaWGhbaaamaalaaabaGa % amOuamaaCaaaleqabaGaaG4maaaaaOqaaiaadsfadaahaaWcbeqaai % aaikdaaaaaaaGcbaGaeyypa0ZaaSaaaeaacaaI0aGaey41aq7aaeWa % aeaacaaIZaGaaiOlaiaaigdacaaI0aaacaGLOaGaayzkaaWaaWbaaS % qabeaacaaIYaaaaOGaey41aq7aaeWaaeaacaaI5aGaaiOlaiaaisda % aiaawIcacaGLPaaadaahaaWcbeqaaiaaiodaaaGccqGHxdaTcaaIXa % GaaGimamaaCaaaleqabaGaaGymaiaaiIdaaaaakeaacaaI2aGaaiOl % aiaaiAdacaaI3aGaey41aqRaaGymaiaaicdadaahaaWcbeqaaiabgk % HiTiaaigdacaaIXaaaaOGaey41aq7aaeWaaeaacaaI0aGaaGynaiaa % iAdacqGHxdaTcaaI2aGaaGimaaGaayjkaiaawMcaamaaCaaaleqaba % GaaGOmaaaaaaaakeaacaWGnbGaaCjaVpaaBaaaleaacaWGTbaabeaa % kiabg2da9maalaaabaGaaGinaiabgEna0oaabmaabaGaaG4maiaac6 % cacaaIXaGaaGinaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaa % kiabgEna0oaabmaabaGaaGyoaiaac6cacaaI0aaacaGLOaGaayzkaa % WaaWbaaSqabeaacaaIZaaaaOGaey41aqRaaGymaiaaicdadaahaaWc % beqaaiaaigdacaaI4aaaaaGcbaGaaGOnaiaac6cacaaI2aGaaG4nai % abgEna0kaaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaIXaGaaGym % aaaakiabgEna0oaabmaabaGaaGinaiaaiwdacaaI2aGaey41aqRaaG % OnaiaaicdaaiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGccqGH % xdaTcaaIXaGaaGimamaaCaaaleqabaGaeyOeI0IaaGynaaaaaaaake % aacqGH9aqpcaaI2aGaaiOlaiaaisdacaaI4aGaey41aqRaaGymaiaa % icdadaahaaWcbeqaaiaaikdacaaIZaaaaOGaam4AaiaadEgacaGGUa % aaaaa!B225! \begin{array}{ccccc} {T^2} = \frac{{4{\pi ^2}}}{{G{M_m}}}{R^3}\\ {M_m} = \frac{{4{\pi ^2}}}{G}\frac{{{R^3}}}{{{T^2}}}\\ = \frac{{4 \times {{\left( {3.14} \right)}^2} \times {{\left( {9.4} \right)}^3} \times {{10}^{18}}}}{{6.67 \times {{10}^{ - 11}} \times {{\left( {456 \times 60} \right)}^2}}}\\ M{ _m} = \frac{{4 \times {{\left( {3.14} \right)}^2} \times {{\left( {9.4} \right)}^3} \times {{10}^{18}}}}{{6.67 \times {{10}^{ - 11}} \times {{\left( {456 \times 60} \right)}^2} \times {{10}^{ - 5}}}}\\ = 6.48 \times {10^{23}}kg. \end{array}\)

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

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WGubWaa0baaSqaaiaad2eaaeaacaaIYaaaaaGcbaGaamivamaaDaaa % leaacaWGfbaabaGaaGOmaaaaaaGccqGH9aqpdaWcaaqaaiaadkfada % qhaaWcbaGaamytaiaadofaaeaacaaIZaaaaaGcbaGaamOuamaaDaaa % leaacaWGfbGaam4uaaqaaiaaiodaaaaaaaaa!4328! \frac{{T_M^2}}{{T_E^2}} = \frac{{R_{MS}^3}}{{R_{ES}^3}}\)

where R_MS is the mars -sun distance and R_ES is the earth-sun distance.

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGH0i % cxcaWGubWaaSbaaSqaaiaad2eaaeqaaOGaeyypa0ZaaeWaaeaacaaI % XaGaaiOlaiaaiwdacaaIYaaacaGLOaGaayzkaaWaaWbaaSqabeaaca % aIZaGaai4laiaaikdaaaGccqGHxdaTcaaIZaGaaGOnaiaaiwdaaeaa % cqGH9aqpcaaI2aGaaGioaiaaisdacaaMc8UaaGPaVlaaykW7caWGKb % GaamyyaiaadMhacaWGZbaaaaa!50F5! \begin{array}{l} \therefore {T_M} = {\left( {1.52} \right)^{3/2}} \times 365\\ = 684\,\,\,days \end{array}\)

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 6

Weighing the Earth : You are given the following data: g = 9.81 ms^–2, R_E = 6.37\(\times\)10⁶ m, the distance to the moon R = 3.84\(\times\)10⁸ 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 \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamytamaaBa % aaleaacaWGfbaabeaakiabg2da9maalaaabaGaam4zaiaadkfadaqh % aaWcbaGaamyraaqaaiaaikdaaaaakeaacaWGhbaaaaaa!3D2A! {M_E} = \frac{{gR_E^2}}{G}\)

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGH9a % qpdaWcaaqaaiaaiMdacaGGUaGaaGioaiaaigdacqGHxdaTdaqadaqa % aiaaiAdacaGGUaGaaG4maiaaiEdacqGHxdaTcaaIXaGaaGimamaaCa % aaleqabaGaaGOnaaaaaOGaayjkaiaawMcaamaaCaaaleqabaGaaGOm % aaaaaOqaaiaaiAdacaGGUaGaaGOnaiaaiEdacqGHxdaTcaaIXaGaaG % imamaaCaaaleqabaGaeyOeI0IaaGymaiaaigdaaaaaaaGcbaGaeyyp % a0JaaGynaiaac6cacaaI5aGaaG4naiabgEna0kaaigdacaaIWaWaaW % baaSqabeaacqGHsislcaaIYaGaaGinaaaakiaadUgacaWGNbGaaiOl % aaaaaa!5BD7! \begin{array}{l} = \frac{{9.81 \times {{\left( {6.37 \times {{10}^6}} \right)}^2}}}{{6.67 \times {{10}^{ - 11}}}}\\ = 5.97 \times {10^{ - 24}}kg. \end{array}\)

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

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGub % WaaWbaaSqabeaacaaIYaaaaOWaaSaaaeaacaaI0aGaeqiWda3aaWba % aSqabeaacaaIYaaaaOGaamOuamaaCaaaleqabaGaaG4maaaaaOqaai % aadEeacaWGnbWaaSbaaSqaaiaadweaaeqaaaaaaOqaaiaad2eadaWg % aaWcbaGaamyraaqabaGccqGH9aqpdaWcaaqaaiaaisdacqaHapaCda % ahaaWcbeqaaiaaikdaaaGccaWGsbWaaWbaaSqabeaacaaIZaaaaaGc % baGaam4raiaadsfadaahaaWcbeqaaiaaikdaaaaaaaGcbaGaeyypa0 % ZaaSaaaeaacaaI0aGaey41aqRaaG4maiaac6cacaaIXaGaaGinaiab % gEna0kaaiodacaGGUaGaaGymaiaaisdacqGHxdaTdaqadaqaaiaaio % dacaGGUaGaaGioaiaaisdaaiaawIcacaGLPaaadaahaaWcbeqaaiaa % iodaaaGccqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaaGOmaiaais % daaaaakeaacaaI2aGaaiOlaiaaiAdacaaI3aGaey41aqRaaGymaiaa % icdadaahaaWcbeqaaiabgkHiTiaaigdacaaIXaaaaOGaey41aq7aae % WaaeaacaaIYaGaaG4naiaac6cacaaIZaGaey41aqRaaGOmaiaaisda % cqGHxdaTcaaI2aGaaGimaiabgEna0kaaiAdacaaIWaaacaGLOaGaay % zkaaWaaWbaaSqabeaacaaIYaaaaaaaaOqaaiabg2da9iaaiAdacaGG % UaGaaGimaiaaikdacqGHxdaTcaaIXaGaaGimamaaCaaaleqabaGaaG % OmaiaaisdaaaGccaWGRbGaam4zaaaaaa!894F! \begin{array}{l} {T^2}\frac{{4{\pi ^2}{R^3}}}{{G{M_E}}}\\ {M_E} = \frac{{4{\pi ^2}{R^3}}}{{G{T^2}}}\\ = \frac{{4 \times 3.14 \times 3.14 \times {{\left( {3.84} \right)}^3} \times {{10}^{24}}}}{{6.67 \times {{10}^{ - 11}} \times {{\left( {27.3 \times 24 \times 60 \times 60} \right)}^2}}}\\ = 6.02 \times {10^{24}}kg \end{array}\)

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

EXAMPLE 7

Express the constant k of Eq. (8.38) in days and kilometres. Given k = 10^–13 s² m^–3. The moon is at a distance of 3.84\(\times\)10 ⁵ km from the earth. Obtain its time-period of revolution in days.

SOLUTION

Given

k = 10^–13 s² m^–3

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGH9a % qpcaaIXaGaaGimamaaCaaaleqabaGaaGymaiaaiodaaaGcdaWadaqa % amaalaaabaGaaGymaaqaamaabmaabaGaaGOmaiaaisdacqGHxdaTca % aI2aGaaGimaiabgEna0kaaiAdacaaIWaaacaGLOaGaayzkaaWaaWba % aSqabeaacaaIYaaaaaaakiaadsgadaahaaWcbeqaaiaaikdaaaaaki % aawUfacaGLDbaadaWadaqaamaalaaabaGaaGymaaqaamaabmaabaGa % aGymaiaac+cacaaIXaGaaGimaiaaicdacaaIWaaacaGLOaGaayzkaa % WaaWbaaSqabeaacaaIZaaaaOGaam4Aaiaad2gadaahaaWcbeqaaiaa % iodaaaaaaaGccaGLBbGaayzxaaaabaGaeyypa0JaaGymaiaac6caca % aIZaGaaG4maiabgEna0kaaigdacaaIWaWaaWbaaSqabeaacqGHsisl % caaIXaGaaGinaaaakiaadsgadaahaaWcbeqaaiaaikdaaaGccaWGRb % GaamyBamaaCaaaleqabaGaeyOeI0IaaG4maaaaaaaa!65EA! \begin{array}{l} = {10^{13}}\left[ {\frac{1}{{{{\left( {24 \times 60 \times 60} \right)}^2}}}{d^2}} \right]\left[ {\frac{1}{{{{\left( {1/1000} \right)}^3}k{m^3}}}} \right]\\ = 1.33 \times {10^{ - 14}}{d^2}k{m^{ - 3}} \end{array}\)

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

T ² = (1.33 \(\times\) 10^-14)(3.84 \(\times\) 10⁵)³

T  = 27.3 d

Note that 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 Satellites

Energy of an Orbiting Satellites

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

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGlb % GaaiOlaiaadweacqGH9aqpdaWcaaqaaiaaigdaaeaacaaIYaaaaiaa % d2gacaWG2bWaaWbaaSqabeaacaaIYaaaaaGcbaGaeyypa0ZaaSaaae % aacaWGhbGaamyBaiaad2eadaWgaaWcbaGaamyraaqabaaakeaacaaI % YaWaaeWaaeaacaWGsbWaaSbaaSqaaiaadweaaeqaaOGaey4kaSIaam % iAaaGaayjkaiaawMcaaaaacaGGSaGaaGPaVlaaykW7caaMc8UaaGPa % VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8 % UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 % caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl % aaykW7daqadaqaaiaaiIdacaGGUaGaaGinaiaaicdaaiaawIcacaGL % Paaaaaaa!7A2B! \begin{array}{l} K.E = \frac{1}{2}m{v^2}\\ = \frac{{Gm{M_E}}}{{2\left( {{R_E} + h} \right)}},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {8.40} \right) \end{array}\)

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

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuaiaac6 % cacaWGfbGaeyypa0JaeyOeI0YaaSaaaeaacaWGhbGaamyBaiaad2ea % daWgaaWcbaGaamyraaqabaaakeaadaqadaqaaiaadkfadaWgaaWcba % GaamyraaqabaGccqGHRaWkcaWGObaacaGLOaGaayzkaaaaaaaa!4309! P.E = - \frac{{Gm{M_E}}}{{\left( {{R_E} + h} \right)}}\)                           (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

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyraiabg2 % da9iaadUeacaGGUaGaamyraiabgUcaRiaadcfacaGGUaGaamyraiab % g2da9iabgkHiTmaalaaabaGaam4raiaad2gacaWGnbWaaSbaaSqaai % aadweaaeqaaaGcbaGaaGOmamaabmaabaGaamOuamaaBaaaleaacaWG % fbaabeaakiabgUcaRiaadIgaaiaawIcacaGLPaaaaaaaaa!48C3! E = K.E + P.E = - \frac{{Gm{M_E}}}{{2\left( {{R_E} + h} \right)}}\)       (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.

EXAMPLE 8

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,

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyramaaBa % aaleaacaWGPbaabeaakiabg2da9iabgkHiTmaalaaabaGaam4raiaa % d2eadaWgaaWcbaGaamyraaqabaGccaWGTbaabaGaaGinaiaadkfada % WgaaWcbaGaamyraaqabaaaaaaa!4002! {E_i} = - \frac{{G{M_E}m}}{{4{R_E}}}\)

While finally

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyramaaBa % aaleaacaWGMbaabeaakiabg2da9iabgkHiTmaalaaabaGaam4raiaa % d2eadaWgaaWcbaGaamyraaqabaGccaWGTbaabaGaaGioaiaadkfada % WgaaWcbaGaamyraaqabaaaaaaa!4003! {E_f} = - \frac{{G{M_E}m}}{{8{R_E}}}\)

The change in the total energy is 

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGHuo % arcaWGfbGaeyypa0JaamyramaaBaaaleaacaWGMbaabeaakiabgkHi % TiaadweadaWgaaWcbaGaamyAaaqabaaakeaacqGH9aqpdaWcaaqaai % aadEeacaWGnbWaaSbaaSqaaiaadweaaeqaaOGaamyBaaqaaiaaiIda % caWGsbWaaSbaaSqaaiaadweaaeqaaaaakiabg2da9maabmaabaWaaS % aaaeaacaWGhbGaamytamaaBaaaleaacaWGfbaabeaaaOqaaiaadkfa % daqhaaWcbaGaamyraaqaaiaaikdaaaaaaaGccaGLOaGaayzkaaWaaS % aaaeaacaWGTbGaamOuamaaBaaaleaacaWGfbaabeaaaOqaaiaaiIda % aaaabaGaeyiLdqKaamyraiabg2da9maalaaabaGaam4zaiaad2gaca % WGsbWaaSbaaSqaaiaadweaaeqaaaGcbaGaaGioaaaacqGH9aqpdaWc % aaqaaiaaiMdacaGGUaGaaGioaiaaigdacqGHxdaTcaaI0aGaaGimai % aaicdacqGHxdaTcaaI2aGaaiOlaiaaiodacaaI3aGaey41aqRaaGym % aiaaicdadaahaaWcbeqaaiaaiAdaaaaakeaacaaI4aaaaiabg2da9i % aaiodacaGGUaGaaGymaiaaiodacqGHxdaTcaaIXaGaaGimamaaCaaa % leqabaGaaGyoaaaakiaadQeaaaaa!7443! \begin{array}{l} \Delta E = {E_f} - {E_i}\\ = \frac{{G{M_E}m}}{{8{R_E}}} = \left( {\frac{{G{M_E}}}{{R_E^2}}} \right)\frac{{m{R_E}}}{8}\\ \Delta E = \frac{{gm{R_E}}}{8} = \frac{{9.81 \times 400 \times 6.37 \times {{10}^6}}}{8} = 3.13 \times {10^9}J \end{array}\)

The kinetic energy is reduced and it mimics \(\Delta\)E, namely, \(\Delta\)K = K_f – K_i= – 3.13\(\times\) 10⁹ J.

The change in potential energy is twice the change in the total energy, namely \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiLdqKaam % Ovaiabg2da9iaadAfadaWgaaWcbaGaamOzaaqabaGccqGHsislcaWG % wbWaaSbaaSqaaiaadMgaaeqaaOGaeyypa0JaeyOeI0IaaGOnaiaac6 % cacaaIYaGaaGynaiabgEna0kaaigdacaaIWaWaaWbaaSqabeaacaaI % 5aaaaOGaamOsaaaa!485B! \Delta V = {V_f} - {V_i} = - 6.25 \times {10^9}J\)

Geostationary and Polar Satellites

Geostationary and Polar Satellites

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 :

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOuamaaBa % aaleaacaWGfbaabeaakiabgUcaRiaadIgacqGH9aqpdaqadaqaamaa % laaabaGaamivamaaCaaaleqabaGaaGOmaaaakiaadEeacaWGnbWaaS % baaSqaaiaadweaaeqaaaGcbaGaaGinaiabec8aWnaaCaaaleqabaGa % aGOmaaaaaaaakiaawIcacaGLPaaadaahaaWcbeqaaiaaigdacaGGVa % GaaGOmaaaaaaa!466A! {R_E} + h = {\left( {\frac{{{T^2}G{M_E}}}{{4{\pi ^2}}}} \right)^{1/2}}\)                 (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.

Fig. 8.11 A Polar satellite. A strip on earth’s surface (shown shaded) is visible from the satellite during one cycle. For the next revolution of the satellite, the earth has rotated a little on its axis so that an adjacent strip becomes visible.

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 \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyisISlaaa!37A7! \approx \) 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 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.

Geostationary and Polar Satellites

Geostationary and Polar Satellites

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 :

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOuamaaBa % aaleaacaWGfbaabeaakiabgUcaRiaadIgacqGH9aqpdaqadaqaamaa % laaabaGaamivamaaCaaaleqabaGaaGOmaaaakiaadEeacaWGnbWaaS % baaSqaaiaadweaaeqaaaGcbaGaaGinaiabec8aWnaaCaaaleqabaGa % aGOmaaaaaaaakiaawIcacaGLPaaadaahaaWcbeqaaiaaigdacaGGVa % GaaGOmaaaaaaa!466A! {R_E} + h = {\left( {\frac{{{T^2}G{M_E}}}{{4{\pi ^2}}}} \right)^{1/2}}\)                 (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.

Fig. 8.11 A Polar satellite. A strip on earth’s surface (shown shaded) is visible from the satellite during one cycle. For the next revolution of the satellite, the earth has rotated a little on its axis so that an adjacent strip becomes visible.

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 \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyisISlaaa!37A7! \approx \) 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 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.

India's Leap into Space

India's Leap into Space

India started its space programme in 1962 when Indian National Committee for Space Research was set up by the Government of India which was superseded by the Indian Space Research Organisation (ISRO) in 1969. ISRO identified the role and importance of space technology in nation’s development and bringing space to the service of the common man. India launched its first low orbit satellite Aryabhata in 1975, for which the launch vehicle was provided by the erstwhile Soviet Union. ISRO started employing its indigenous launching vehicle in 1979 by sending Rohini series of satellites into space from its main launch site at Satish Dhawan Space Center, Sriharikota, Andhra Pradesh. The tremendous progress in India’s space programme has made ISRO one of the six largest space agencies in the world. ISRO develops and delivers application specific satellite products and tools for broadcasts, communication, weather forecasts, disaster management tools, Geographic Information System, cartography, navigation, telemedicine, dedicated distance education satellite etc. In order to achieve complete self-reliance in these applications, cost effective and reliable Polar Satellite Launch Vehicle (PSLV) was developed in early 1990s. PSLV has thus become a favoured carrier for satellites of various countries, promoting unprecedented international collaboration. In 2001, the Geosynchronous Satellite Launch Vehicle (GSLV) was developed for launching heavier and more demanding Geosynchronous communication satellites. Various research centers and autonomous institutions for remote sensing, astronomy and astrophysics, atmospheric sciences and space research are functioning under the aegis of the Department of Space, Government of India. Success of lunar (Chandrayaan) and inter planetary (Mangalyaan) missions along with other scientific projects has been landmark achievements of ISRO. Future endeavors of ISRO in- clude human space flight projects, the development of heavy lift launchers, reusable launch vehicles, semi-cryogenic engines, single and two stage to orbit (SSTO and TSTO) vehicles, development and use of composite materials for space application etc. In 1984 Rakesh Sharma became the first Indian to go into

outer space aboard in a USSR spaceship.