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Space Dynamics
(GATE Aerospace) by Mr Dinesh Kumar (IIT Madras Fellow)
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SPACE DYNAMICS
Topics to study –
Gravitation field Kepler’s law Mechanics of orbital trajectory Orbits
Ref: -
Introduction to flight by Anderson NPTEL- Space Technology https://nptel.ac.in/courses/101/106/101106046/
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Some constants –
o Universal gravitation constant, G = 6.67× ퟏퟎ ퟏퟏ 푵 풎ퟐ
풌품ퟐ
o 흁 = GM = 3.98× ퟏퟎퟏퟒ 풎ퟑ
풔ퟐ
o Earth radius, 푹풆 = 6371 km o Gravity acceleration, 품풐 = 9.81 풎
풔ퟐ
o Mass of the Earth, 푴풆 = 5.98× ퟏퟎퟐퟒ kg
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1. GRAVITATION FIELD
1.1 Newton’s law of Gravitation
Force between two objects of masses 푚 and 푚 is inversely proportional to the squire of the distance between two centers.
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F ∝
→It is also proportional to product of masses 푚 and 푚
F ∝ 푚 푚
F ∝
F = 푮풎ퟏ풎ퟐ 풓ퟐ
G = universal gravitational const.
G = 6.67× 10
If one of the mass is earth mass (M)
F = 푮푴풎풓ퟐ
Where, M = 5.98× 10 kg
Radius of the earth, 푅 = 6371 km ≅ 6400 km
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휇 = GM = 3.98× 10
1.2 Gravitational potential energy (U)
The potential energy (GPE) is zero when distance between two masses is infinite (∞).
Below the distance of∞, GPE is negative.
U(r) = -
r = separation between particles
푚 푎푛푑푚 = masses of two objects
→If a particle is brought from a position to new position under gravitational force, change in P.E. is given by
푈 - 푈 = - ∫ 퐹푑푟
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1.3 Gravitational potential (V)
It is the gravitational potential energy per mass.
푉 - 푉 =
1.4 Gravitational potential field (푬)
It is the intensity of gravitational force per unit mass at a point.
E =
The intensity of gravitational force per unit mass at earth surface is known as gravity acceleration (g).
E = = = = g
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1.5 The variation of g from earth surface
Above earth at an altitude h
F = ( )
∵ g =
F = mg
g =
( )
If h = 0, then g = 푔 = 9.81 m/푠
g =
( ) = 푔 1 + ≅ 푔 (1-2h/R)
Here g decreases with height.
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Below earth surface
Gravitational potential force, F = ( )
(R-h) (h→ depth)
g = =
( )(1-h/R)
Here g decreases with depth up to zero.
1.6 Effect on gravity due to rotation of earth
Centrifugal force = m휔 푟 sin휃
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g’ =
F = mg - m휔 푟 sin 휃 =
= g - 휔 푟 sin휃
g’ = g - 휔 푟 sin휃
So g is maximum at poles as centrifugal force is zero and minimum at equator as centrifugal force is maximum.
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2. Kepler’s law
First law
A satellite or object describes an elliptical path around its centre of attraction.
Its path depends on velocity.
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Second law
The area swept out by radius vector of object in equal time remains same.
dA = . 푟.푑ℎ
dh = rd휃
dA = . 푟 . d휃
= . 푟 .
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= . 푟 .휔 = . 푟 . 휃 =
→ Angular momentum is constant under conservative forces.
m푟 휃 = const.
푟 휃 = const. = h = 푟 × 푣
So, = const.
Third law
The square of time period is proportional to cube of semi-major axis.
푇 ∝ 푎
=
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3. Mechanics of orbital trajectory
Lagrange equation of motion
Kinetic energy of the body
T = T (푞 , 푞 , 푞 ,푞′ ,푞′ ,푞′ )
Potential energy of the body
∅ = ∅ (푞 , 푞 , 푞 )
푞 , 푞 , 푞 → Coordinates
Lagrange’s function
휷 = T - ∅
- = 0
- = 0
- = 0
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Orbit equation
Kinetic energy, q = 푚푣
푣 = 푟 + 푟휃
T = 푚(푟 + 푟휃 ) = 푚(푟 + r 휃 )
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Potential energy
∅ = -
→ 훽 = T - ∅ = 푚(푟 + r 휃 ) + ... (1)
For 휃 co-ordinate
- = 0 .... (2)
By equation (1),
= (mr θ)
Angular momentum = mr θ = const.
r θ = = h (angular momentum per mass is constant)
So, = (mr θ) = 0
By equation (1),
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= 0
For 푟 co-ordinate
- = 0 .... (3)
By equation (1),
= mr
= mr휃 -
From equation (3),
(mr) - mr휃 + = 0
m푟 + - mr휃 = 0
푟 - + = 0
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푟 - + = 0 ..... (4) (h =푟 휃 = specific angular momentum)
This is differential equation of 2nd order.
Sol for above equation is given by
r = ( )
= . . ( )
= . ( )
(p = )
Here A and C (phase angle) are constant depends on initial condition.
And 퐴. = 푒 = eccentricity
So, r = . ( )
.... (5)
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The type of orbit is decided by e
(1) If e = 0 → path is circular
(2) If e < 1 → path is elliptical
(3) If e = 1 → path is parabolic
(4) If e > 1 → path is hyperbolic
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Orbital energy
It is the summation of K.E. and P.E.
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In General
Eccentricity can be written in terms of orbit energy per unit mass
e = ퟏ + ퟐ풉ퟐ
흁ퟐ
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Escape velocity
Orbit Energy
퐸 = 푣 -
At escape velocity, the kinetic energy is converted into potential energy.
퐸 = 푣 - = 0
푉 = = 2푔 푟
Example1. What will be the escape velocity, if R is doubled and mass remains same?
Sol.
(I) if mass remains same
푉 = 2푔 푅
g = ( )
= ( )
= ( )
= (h = R)
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푉 , = 2. . 2푅 = ,
√
(II) If density remains same
M = 휌푉
푀 = 휌 × 휋푅
푀 = 휌 × 휋(2푅) = 8푀 (R→ 2푅)
푔 = ( )
= 2푔
푉 , = 2.2푔 . 2푅 = 2푉 ,
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4. Orbits
4.1 For circular orbit
e = 0, for circular orbit
퐸 = -
퐸 = 푣 -
- = 푣 -
For Circular orbit r =
- = 푣 -
v = =
So velocity for circular orbit, 푉 = =
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Escape velocity from circular orbit
∆푉 = -
Time period
T =
T =
T = √
푅 /
푇 = 푅
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4.2 For parabolic orbit
e = 1, for parabolic orbit
e = 1 + = 1
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퐸 = 0
So, 퐸 = 푣 - = 0
v = (escape velocity)
For parabolic trajectory
r =
At 휃 = 0
푟 = (perigee)
푉 = 푉 =
= .
ℎ2 = 2휇
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4.3 For an elliptical orbit
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e< 1, for elliptical orbit
a = semi – major axis
b = semi – minor axis
푟 + 푟 = 2a
푟 = apogee = 푟
푟 = perigee = 푟
푟 - 푟 = 2ae
푟 = a(1+ e)
푟 = a(1- e)
= ( )( )
e =
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e = 1 − from geometry
For Elliptical trajectory
r = .
푟 =
푟 =
→ = 푟 (1+e)
= a(1-e)(1+e)
= a(1-푒 )
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We know
e = 1 +
Then orbit energy
퐸 = - ( )
=− { = a(1-푒 )}
퐸 = 푣 -
푣 = - ( )
= −
v = − 1
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Velocity at perigee r → 푟
, 푉 = ( )( )
= ( )
Velocity at apogee r → 푟
, 푉 = ( )( )
= ( )
= ( )( )
Escape velocity from elliptical orbit
∆푉 = - − 1
Time period
T = = /
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T = √
( ) =
√푎 /
T2 = 푎
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4.4 For hyperbolic orbit
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e > 1, for hyperbolic orbit
P.E. < K.E.
e = 1 +
퐸 = - ( )
For hyperbolic trajectory
r = .
e = 1 + (for hyperbolic orbit)
푟 . = 푟 = = a(e-1)
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a =
Orbital energy = - ( ) =
퐸 = 푣 -
휇2푎
= 12푣 −
휇푟
v = + 1
Velocity at Perigee
푉 = ( )( )
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GATE QUESTIONS
Gate 2020
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