SMAD ASSIGNMENT NO 01 B.docx

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INSTITUTE OF SPACE TECHNOLOGY, ISLAMABAD

DEPARTMENT OF AERONAUTICS AND ASTRONAUTICS

SPACE MISSION ANALYSIS ANDDESIGN

CLASS ASSIGNMENT : 01 b

Prepared By: -Shakeel AhmeRe! N"# 1$0$1%001MS &AE'0()Da*e: %$ Ma+h, %01-

Submitted To: -D+ Sal.m/. ah.+P"P Ae+"23aeE!.ee+.!



Q. No 1 (a):To transfer a satellite on an elliptic orbit, the most energy ecieimpulse would be colinear with the !elocity and applied at the instant when tsatellite is at the elliptical orbit"s perigee, since at that point, the !elocityma#imum. the energy of th etransfer orbit is greater than the energy of the innorbit (a $ r 1 ), and smaller than the energy of the outer orbit (a $ r % ). &onsidecommunication satellite to be carried by a space !ehicle into low earth orbit (')an altitude of *%% +m and is to be transferred to a geostationary orbit ()*-,/0 +m using a ohmann transfer. 2irst calculate the !elocity of the inner aouter circular orbits by stating all assumptions and appropriate calculations.

v π

2= μ (2

r1

− 2

r1+r

2

)

vα 2= μ ( 2r

2

− 2

r1+r

2 )

∆ vπ =v π −vc 1=

√ μ

r1 (√

2 r2

r1+r

2−1

)

∆ vα =vc 2−vα =√ μ

r2(1−√

2 r1

r1+r

2)

3olution:

To fnd velocity o inner and outer circular orbit, we can assume that they are separ

ciruclar orbits with no relation to each other and the tangential velocity can be applied to b

o them seperatley. The tangential velocity is given by the ormula: -

v=√G .M

R

here, G=6.673×10−11

N m

2

Kg∧massof earth, M =6×10

24 Kg

Rinner= Rearth+ Altitude of inner orbit =6.4×106+322000=6.722×10

6meters

Putting the values in above e!uation gives the velocity o inner orbit: -

"

∆

v2

r1+r

vα

∆ vα r2



vinner=√

6.67×10−11

×6×1024

6.722×106

vinner=7715.947 m

s

¿>v inner=7.716 m/s Ans!er

Similarly

Route r= R earth+ Altitude of outer orbit =6.4×106+35860000=4.226×10

7meters

Putting values or the outer orbit, we get : -

vouter=√6.67×10

−11×6×10

24

4.226×107

vouter=3077.328m

s

vouter=3.077 m/s Ans!er

b). The !elocities of the transfer orbit at perigee and apogee.

S"l/*."

#elocity o the transer orbit at perigee is given by the ormula: -

v π

2= μ (2

r1

− 2

r1+r

2

)

here, μ=G × M =6.67×10

−11

×6×10

24

=4.002×10

14

N m

2

/ Kg

and r" and r$ are the initial and fnal radius, ie radius o the inner and outer circular orb

respectivley. so putting the values in above e!uation, we get: -

v π

2=4.002×1014(

2

6.722×106−

2

6.722×106+4.226×10

7 )

$







4.226×107

¿(6.67×10

6)+(¿¿)

2×6.67×106

¿1−√ ¿

∆ vα =

√ 4.002×10

14

4.226×107 ¿

∆ vα =1470.5196m

s =1.47

m

s Ans!er

Ee*+..*5 o the transer orbit is given by the relation: -

e=1−2× r

1

r1+r2

Putting the values we get: -

e=1− 2×6.722×10

6

6.722×106+4.226×10

7

e=0.7255 Ans!er

T.me "7 8.!h* is given by the relation: -

# =π √atrans

3

μ

here atrans=rinit ial+r final

2=

6.722×106+4.226×10

7

2=2.4491×10

7m

and μ=G × M =6.67×10−11

×6×1024=4.002×10

14 N m

2/ Kg

Putting values in above e!uation gives us: -

# =3.14×√(2.4491×10

7)3

(4.002×1014)

# =19033.60744 sec ¿5hrs17mins Ans!er





Q No %: &onsidering 7ars as the target planet and arth as the laun

planet, each planet is in a circular orbit, and by using ohmann transfer orbit to

from arth to 7ars. &alculate 89 and time of 6ight with some a!ailable e5uatio

and numbers gi!en:

mearth / 0 "1$+ g rearth / .+ 0 "1 m ac / v$2' %c / mv$2'

T / $3'2v / "2T 4 / ')

2T$

4 orbits earth / ".1" & ") m)

2s$

4 orbits sun / ).) & "5 m)2s$ %g / 6m72'$ 6 / .8 & -"" (.m$2g$

g / 672'$ v / 9672'"2$ r" / rearth / rp / " ;< r$ / rmars / ra

".$+ ;<

vc 1=√ μ/r1∧vc 2=√ μ /r

2

S"l/*.":

Total ∆V is given by the equation:-

∆ $ t =∆ $ π +∆$ α

here ∆ vπ =v π −vc 1=√ μ

r1(√

2r2

r1+r

2

−1)in this case, r" / distance between center o Sun and &arth / " ;< / ".+= 0 "15 m / ".+=

"1"" m

and r$ / >istance between center o Sun and 7ars / ".$+ ;< / $.$8=58 0 "15 m / $.$8

0 "1"" m

and μ=G × M sun=6.67×10−11

×1.989×1030=1.326663×10

20 N m

2/g

Putting the values in above e!uation: -

∆ vπ =√1.326663×10

20

1.496×1011 (√ 2×2.27987 %10

11

(1.496 %1011)+(2.27987 % 10

11)−1)

∆ vπ =2945.425m /s=2.945 m. /s

Similarly ∆ v α =vc 2−vα =√ μ

r2(1−√

2 r1

r1+r

2)

Putting values: -

8



∆ vα =√1.326663×10

20

2.27987 %1011 (1−√

2×1.496×1011

(1.496 %1011)+(2.27987 %10

11))∆ vα =2649.4256m /s=2.649 m/s

So the total 89 becomes:

∆ $ t =2945.425+2649.4256

∆ $ t =5594.85m /s=5.595 m/s Ans!er

The T.me "7 8.!h* is given by the relation:-

# =π √atrans

3

μ

here in this case μ=G × M sun=6.67×10−11

×1.989×1030=1.326663×10

20 N m

2/g

atrans=rinitial+rfinal

2

in this case rinitial / >istance between center o Sun and &arth / " ;< / ".+= 0 "1 5 m / ".

0 "1"" m

and rfnal / >istance between center o Sun and 7ars / ".$+ ;< / $.$8=58 0 "15 m / $.$8

0 "1"" m

atrans=(1.496×10

11)+(2.27987×1011)

2=1.887935×10

11meters

(ow putting the values in above e!uation or time o ?ight: -

# =3.14×√(1.887935×10

11)3

(1.326663×1020)

# =22374224.21 seconds

# =6215.092 hrs

# =258.9622da&s

5



# =258da&s23hrs

# =8.6 Months Ans!er

T N

=

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SMAD ASSIGNMENT NO 01 B.docx