two -body motion
TRANSCRIPT
8/6/2019 Two -Body Motion
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BY
Y.Jeevan kumar 08951A2173
A .Sai kiran 08951A2189
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The otion of a spacecraft in the universe is governed by an
infinite network of attractions to all celestial bodes.The motion of a spacecraft in the solar system is dominatedby one central body at a time.
above observation leads to the very useful two body assumptions
1.The motion of a spacecraft is governed by attraction to asingle body.
2.The mass of a spacecraft is negligible compared to thatof the central body.
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3.The bodies are spherically symmetric with the massesconcentrated at the centers
4.No forces act on the bodies except for gravitational forcesand centrifugal forces.
The two body assumptions are nearly very true. theinfluence of earth on the spacecraft is more than a factor of 1000
No explicit solution found for the n-body problem exceptfor N=2.
these solutions are used only for the suspecting solutionse.g:- mercury orbiter.
Or when high accuracy is required e.g:-navigation calculation.
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INSAT-2A/2B SPACECRAFT
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Newtons law of universal gravitational gas constant for two body motiongiven by
F=GRAV ITATIONAL FORCE BETWEEN TWO BODIES.
G=UNI VERSAL GRAV ITATIONAL CONSTANT
M,m=MASSES OF TWO BODIES
R= RADI AL DISTANCE BETWEEN TWO BODIES.
r
r nGmM F
3
.
!
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Circular orbits
y Above diagram shows the motion two bodies in circularorbit.
y
Centrifugal force of spacecraft isy
y
y v=velocity of two bodies
ym= mass of the spacecraft
y r=radius from the spacecraft center to the mass of thecentre body.
r
M F
v2
!
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The period of circular orbit ,derived with equal simplicity, is given by
QT
r velocity
ncecircumfere p
3
!!
From the values of the eccentricity we can determine the shape of theorbit.
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K cos1 e pr
e= 0 circle
e= 1 parabola
e< 1 ellipse
e>1 hyperbola
the ratio of its absolute distance from a given point (a focus) to its absolutedistance from a given line (a directrix) i s a positiveconstant e (the eccentricity).
e= c/a
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Conservation of Mechanical Energy. The energy constant of motion can be derived 'as follows :1 . Dot multiply equation by r
0..3
!dddd r r r r
r
Q
2.Since in general a.a=aa&v=r
0'.'.3
! r r vv
r
Q
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3.Noting that
0)()2/(2
!
r dt
d
dt
d
vQ
0)
2
(
2
!
r dt
d v Q4.
5.so from the above equation we can say that
0)
2
(
2
!
r
c
dt
d v Q
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So derivative of expression is zero so that value is constant , here we call thatconstant as specific mechanical energy
Denoted as \
Above equation is specific mechanical energy equation
potential energy considered as zero at infinity & -ve at radii less thaninfinity..
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a2 Q\
Relative energy of orbit types are
Hyperbola>parabola>ellipse>circle
for any conic shapes
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T Orbital period: when E=2 ,from the keplers law elliptical orbital period
n p /T
Q
T a p
3
!
for the elliptical orbit general radius
)cos1(
)1(
Ue
a
r e
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o Six independent quantities are required to describe about the orbitbehavior.
They are
o equatorial planeo vernal equinoxo line of nodeso perigee directiono satellite positiono orbital plane
oFIELD OF V IEW (FOV)
oField of view (fov) is defined as the view of the satellite at a single timefrom the above surface of the earth atmosphere .
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oThe max time that can be view by satellite is
1436
60
840
360
cos
5.1
1
¹¹ º ¸
©©ª¨
¼¼½
»
¬¬«
h
h
R R
R
R
t
e
e
e
e
view
o for the circular orbit the no. of revolution per day is
5.1
997.16/
¼¼½
»¬¬«
hdayrev
R Re
e