Problem 3 (19 polnts): A satellite Is in a circular low earth orblt (LEO) that i

Question

Problem 3 (19 polnts): A satellite Is in a circular low earth orblt (LEO) that is concentric with Earth and 2000 km above the surface of Earth. A Hohmann transfer Is used to put the satellite Into a geostationary orbit, which is a circular orblt concentric with Earth with a perlod of 24 hours. (For your interest, you may want to look into why one would want to put a satellte into a geostationary orblt.) a) (2 points) Calculate the speed of the satellite when it is in LEO. b) (3 polnts) Calculate the radlus of the geostationary orblt and the speed of the satellite when in geostationary orblt. c) (4 points) Show that the latus rectum and eccentricity of the transfer orbit are respectively given by TGs + TLEO and TLEO where rLEo and ras represent the radius of the LEO and geostationary orbit, respectively. Calculate the numerical value of Er. What does this tell you about the shape of the transfer orbit? (3 points) Show that the perigee speed of the transfer orbit is given by d) 2GMGS where Mg represents the mass of Earth and G represents the gravitational constant. e) (3 points) Use the result of part d) to calculate the change in speed required to "kick" the satellite out of LEO into the transfer orbit. Also calculate the speed required to "kick the satellite out of its transfer orbit into geostationary orbit f (4 points) Suppose a rocket with an exhaust velocity of 40 km/s is used for the apogee and perigee kicks. How much fuel is required for every kilogram of payload transferred from the LEO to the geostationary orbit? Hint: assume the length of the perigee and apogee burns are negligible compared to the circumference of the orbits

Explanation / Answer

for the satellite in LEO, h = 2000,000 m

radius of earth, R = 6371,000 m

a. in LEO

speed of satellite = v

in circular orbit

mv^2/(R + h) = GMm/(R + h)^2

v^2 = GM/(R + h)

mass of earth M = 5.972*10^24 kg

G = 6.67*10^-11

hence

v = 6898.1740107678 m/s

b. radius of geostationery orbit = R'

speed of satellite in geostationary orbit = v'

then  

v'^2 = GM/(R')

also

T = 2*pi*R'/v'

hence

R' = Tv'/2*pi

v'^3 = 2*piGM/T

T = 24 hourts = 24*60*60 s

hence

v' = 3071.170444 m/s

R' = 42231625.1058899 m

R = 8371000 m

c. latus rectum of ellipse = 2b^2/a

where a is semi major axis and b is semi minor axis

now, for the transfer orbit

a = (R + R')/2

now apogee = R'

perigee = R

hence

R' = a(1 + e)

R = a(1 - e)

hence

R'/R = (1 + e)/(1 - e)

where e is eccentrity

hence

(R' - R)/(R' + R) = e

hence eccentrity = e

R' = Rg

R = Rl

hence

e = (Rg - Rl)/(Rg + Rl)

and hence

b = a*sqrt( 1 - e^2)

hence

L = 2b^2/a = 2a^2*(1 - e^2)/a = 2a(1 - e^2) = 2a(1 - (Rg - Rl)^2/(Rg + Rl)^2)

L = 2a((Rg + Rl)^2 - (Rg - Rl)^2)/(Rg + Rl)^2

2a = (Rg + Rl)

hence

L = (Rg + Rl + Rg - Rl)(Rg + Rl - Rg + Rl)/(Rg + Rl)

L = 2Rg*Rl/(Rg + Rl)

d. perigee speed of transfer orbit= vp

now, vp = sqrt(GM(2/Rl - 2/(Rl + Rg))) = sqrt(GM(2/R - 2/(R + R'))

vp = sqrt(2GM(Rl + Rg - Rl)/Rl*(Rg + Rl) = sqrt(2GM*Rg/(Rl*(Rl + Rg)))

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