Question

Estimate the followings (note: it's enough to show how to calculate) : a) velocity of an Earth satellite in a low-altitude orbit. b) the escape velocity from the Earth. c) the altitude of a geostationary satellite above ground level.

Answer 1

(A) velocity of Earth satellite Earth Satellite in low altitude orbit Vo = orbital velocity satellite ms = mass vo fms . Fcentripetal = Fgravitation GMEoms msvo? = - 2 DIRE G ME mass of Earth in orbit of earth satellite i. velocity Vo = rame centre of Earth or distance from = htRE . Vo = htRE . By putting value of h we can calculate to Corbital velocity)

from the Earth (B) Escape velocity ... Earth's Radius (RE) & mass (ME) & A mass im is projected from point the surface of the planet earth. i minimum velocity required from the body to escape the earth be → Ve a i. KE = 2 m bé & ME EARTH Me & m. At point & gravitation force btw GMM GMM F = - 0 • To take body from p to a , wook done du = For = GmEm oda -2 to gravitational force earth's surface to • work-done against take the body from infinity to da W = i ) GmEom se IFRE

wa amempadre.umn FT. - GME -?,da hmEm "/W = am GME.M. RE i KE = W Imve² = GME.M - RE -> ve JR Game [ve = fag RE - do we know g= go8 m/s² RE(Radius of Earth) = m 6400km = 6400 x 103 m ve = J 2x g. 8 x 6 4X106 ve = 1102 x 103 misec of Ive = 1102 km 1 sec |

. Altitude of a geostationary satellite (H=?) my² . Fcentripetal - iReth) V mass DOD=m GME.M - & Fgravitational = (REth) ? - Fc = Fg GME.m RE - y = TREthi? Reth ME → = -O Reth 2T CREth) ... Time period (T) = T= 27 (Reth) & CREth) 12 Tume 27 (Reth) * -ID O or TOME 49? (Reth) T² = GME 47² (REth)3 = GME 72 Reth= (hme 72 ) - h= lame + " - RE - 47?

"h= (GME 72) - RE 472 i G = 6.67 x 16-11 m3 kgalg-2 RE (Radius of Earth) = 6.38x106m ME (mass of earth) = 5.972 X 1024 kg " T = 24 has coo geostationary = 24 x 60 x 60 = 86400 sec satellites 8064X1ou see - 16067x16-11 x 5.972 x 1024 x 18.64) 2x 1087 - 6. 38x106 4 x(3•14)2 h= 42.239 8 X 106 - 60388106 → h = 35.8598 X106 m or h= 35859.8 x 103 m i h = 35859.8 km [ approximately 36000 km]