A satellite is in an elliptic orbit around the earth with aphelion of 6R and perihelion of 2R where R = 6400 km is the radius of the earth . Find the velcoity of the satellite at apogee and perigee.[ G = 6.67 xx10^(-1) SI unit and M= 6 xx 10^(24)kg ]

A satellite is in an elliptic orbit around the earth with aphelion of

1.	A satellite is in an elliptic orbit around the earth with aphelion of 6R and perihelion of 2R where R = 6400 km is the radius of the earth . Find the velcoity of the satellite at apogee and perigee.[ G = 6.67 xx10^(-1) SI unit and M= 6 xx 10^(24)kg ]
Answer» Solution :`implies` Given , `r_p` = radius of perihelion = 2R `r_a` = radius of aphelion = 6R Hence , we can write, `r_(a) = a(1+e) = 6R "" ...(i)` `r_p = a (1-e) = 2 R ""...(ii)` SOLVING equ (i) and (ii) ,we get ECCENTRICITY , `e =1/2` Angular MOMENTUM remains unchanged `:. mv_pr_p = mv_ar_a` `:. (v_a)/(v_p) = r_p/r_a = (2R)/(6R) =1/3` Energy is same at perigee and apogee, `1/2mv_p^2-(GMm)/r_p=1/2mv_a^2-(GMm)/r_a` `:. v_p^2(1-1/9)=-2GM(1/r_a-1/r_p)=2GM(1/r_p-1/r_a)` (By putting `v_a=v_p/3`) `v_p=([2GM(1/r_p-1/r_a)]^(1/2))/([1-(v_a/v_p)^2]^(1/2))=[((2GM)/R(1/2-1/6))/((1-1/9))]^(1/2)` `=((3/2GM)/(8/9R))^(1/2)=sqrt(3/4(GM)/R) ` = 6.85 km/s `v_p = 6.85km//s` For circular orbit of radius r, `v_c` = orbita velocity = `sqrt((GM)/r)` For `r = 6R ,v_c =sqrt((GM)/(6R)) = 3.23 km//s` Hence , to increase to a circular orbit at apogee, we have to increase the velocity by (3.22-2.28) = 0.95 km/s . This can be done by suitably firing rockets from the SATELLITE .