Question

Question 3 1 pts A spacecraft with mass 1,976 kg is in circular orbit around Earth as shown with the green circle in the figure, at an altitude h = 608 km. At point Pin the orbit (see figure), the spacecraft reduces its speed by 4%, causing it to be in an elliptical orbit. What is the semi-major axis of the elliptical orbit in km? Reminder the radius of the orbit is the altitude plus Re, the radius of Earth. Re - 6.371 x 103 km. For reference, the mass of Earth, Me, is 5.98 x 1024 kg. Your answer should be in km, but enter only the numerical part in the box.

Answer 1

Ans Radius o Green orbit = altitude tRE 608 + 6.3718 103 km Ro 69 79 km Spacecraft is mo im circular orbit to Newton law g motion moving according so F = mv r Gmina rx Mavz X G mi 6.67x10-11x5.98x1024 6979 X 10 3 m, mars sf earth m2 = mass og spacecraft u = $57.15 8106 V 7.55 X 103 m/s 7.55 km/s This is velocity of space craft Reducing up = to speed by 4 % at point ? v- x y 960 loo VP - 7.248 km/s.

P of the of point P is the Up is the speed of spacecraft at point elliptical or bit. forest point on semimajor axis so up is the mimimum of velocity in the the elliptical vp = Umin semimato value at r max Vmin Earth gimax rmin Imax rmax Ro from figure) Aplying conservation points rmax & ₂ horax - G Mah Lihviming -GM at two e this ไวาท 2 r max نیو N apllying conservation om dementum both angular point. h rmax Vmin = mormin & max

Page using ļ (v mox unito ) - GM { Fun in France] -3 ed from rmin r max & min Vmax from 7 G putting in Dmax - Dominumax to min) 2 GM Vmax Imaxomin mat + Honzattain) (imex Tomin ) = GM ma & max rmax Umin Imin Imax ปัทมา m 2. GM वर का Imax 246.67X 10 X 5.98x1024 -7248 6979 %109 x 7248 1.5770 XT-7248 15770 7248 8522 m/s Imax 8.52 km/s Jamax

eersing min = Vomin x rmax Vmax = 7248 X 69 79 x103 8 522 5935.67 x 103 m 5935.67 km rmin or axib = rmax trimin major a 6979 + 5 935.67 12914.67 km semi major axis 2 TAT Semi major axis - 6457.33 km