Problem 12.86
•
•
•
To place a communication satellite into a
geosynchronous orbit (see problem 12.80)
at an altitude of 22,240 mi above the
surface of the earth, the satellite first is
released from a space shuttle, which in a
circular orbit at an altitude of 185mi, and
then is propelled by an upper-stage booster
to its final altitude. As the satellite passes
through A, the booster’s motor is fired to
insert the satellite into an elliptic transfer
orbit. The booster is again fired at B to insert
the satellite into a geosynchronous orbit.
Knowing that the second firing increases the
speed of the satellite by 4810 ft/s, determine
(a) the speed of the satellite as it
approaches B on the elliptic transfer orbit,
(b) the increase in speed resulting from the
first firing at A.
Problem 12.86
•
•
•
To place a communication satellite into a geosynchronous orbit (see
problem 12.80) at an altitude of 22,240 mi above the surface of the
earth, the satellite first is released from a space shuttle, which in a
circular orbit at an altitude of 185mi, and then is propelled by an upperstage booster to its final altitude. As the satellite passes through A, the
booster’s motor is fired to insert the satellite into an elliptic transfer
orbit. The booster is again fired at B to insert the satellite into a
geosynchronous orbit. Knowing that the second firing increases the
speed of the satellite by 4810 ft/s, determine
(a) the speed of the satellite as it approaches B on the elliptic transfer
orbit,
(b) the increase in speed resulting from the first firing at A.
First Notes…
R=3960 mi = 20.9088x106
rA = (3960 + 185)mi = 4145mi = 21.8856 x106 ft
rB = (3960 + 22,240)mi = 26,200mi = 138.336 x106 ft
For a circular orbit….
∑ Fn = man : F = m
v2
r
1
Problem 12.86
•
To place a communication satellite into a geosynchronous orbit (see
problem 12.80) at an altitude of 22,240 mi above the surface of the
earth, the satellite first is released from a space shuttle, which in a
circular orbit at an altitude of 185mi, and then is propelled by an upperstage booster to its final altitude. As the satellite passes through A, the
booster’s motor is fired to insert the satellite into an elliptic transfer
orbit. The booster is again fired at B to insert the satellite into a
geosynchronous orbit. Knowing that the second firing increases the
speed of the satellite by 4810 ft/s, determine
(a) the speed of the satellite as it approaches B on the elliptic transfer
orbit,
(b) the increase in speed resulting from the first firing at A.
•
•
Newton's Law of Gravitation
F =G
Mm
r2
Then
G
Mm
v2
m
=
r2
r
or
v2 =
Gm
r
Problem 12.86
•
•
•
To place a communication satellite into a geosynchronous orbit (see
problem 12.80) at an altitude of 22,240 mi above the surface of the
earth, the satellite first is released from a space shuttle, which in a
circular orbit at an altitude of 185mi, and then is propelled by an upperstage booster to its final altitude. As the satellite passes through A, the
booster’s motor is fired to insert the satellite into an elliptic transfer
orbit. The booster is again fired at B to insert the satellite into a
geosynchronous orbit. Knowing that the second firing increases the
speed of the satellite by 4810 ft/s, determine
(a) the speed of the satellite as it approaches B on the elliptic transfer
orbit,
(b) the increase in speed resulting from the first firing at A.
Newton's Law of Gravitation
F =G
Then
G
Mm
r2
Mm
v2
m
=
r2
r
or
v2 =
Gm
r
The product of the constant of gravitation G and mass M of the earth can be expressed as:
GM = gR 2 so that
v2 =
gR 2
r
for a circular orbit
2
Problem 12.86
•
•
•
To place a communication satellite into a geosynchronous orbit (see
problem 12.80) at an altitude of 22,240 mi above the surface of the
earth, the satellite first is released from a space shuttle, which in a
circular orbit at an altitude of 185mi, and then is propelled by an upperstage booster to its final altitude. As the satellite passes through A, the
booster’s motor is fired to insert the satellite into an elliptic transfer
orbit. The booster is again fired at B to insert the satellite into a
geosynchronous orbit. Knowing that the second firing increases the
speed of the satellite by 4810 ft/s, determine
(a) the speed of the satellite as it approaches B on the elliptic transfer
orbit,
(b) the increase in speed resulting from the first firing at A.
(v A ) 2circ =
Then:
32.2
ft
x(20.9088 x106 ft ) 2
s2
21.8856 x106 ft
or (v A ) circ = 25,362
2
and:
(vB ) 2circ =
32.2
ft
x(20.9088 x106 ft ) 2
s2
138.336 x106 ft
or (vB ) circ = 10,088
2
ft
s
ft
s
Problem 12.86
•
•
•
To place a communication satellite into a
geosynchronous orbit (see problem 12.80) at an
altitude of 22,240 mi above the surface of the
earth, the satellite first is released from a space
shuttle, which in a circular orbit at an altitude of
185mi, and then is propelled by an upper-stage
booster to its final altitude. As the satellite
passes through A, the booster’s motor is fired to
insert the satellite into an elliptic transfer orbit.
The booster is again fired at B to insert the
satellite into a geosynchronous orbit. Knowing
that the second firing increases the speed of the
satellite by 4810 ft/s, determine
(a) the speed of the satellite as it approaches B
on the elliptic transfer orbit,
(b) the increase in speed resulting from the first
firing at A.
(a) Have….
(vB ) circ = (vB )TR + ∆vB
(a) or…
(vB )TR = (10,088 − 4810)
ft
ft
= 5278
s
s
or… (vB )TR = 5280
ft
s
3
Problem 12.86
•
•
•
To place a communication satellite into a geosynchronous orbit (see
problem 12.80) at an altitude of 22,240 mi above the surface of the
earth, the satellite first is released from a space shuttle, which in a
circular orbit at an altitude of 185mi, and then is propelled by an upperstage booster to its final altitude. As the satellite passes through A, the
booster’s motor is fired to insert the satellite into an elliptic transfer
orbit. The booster is again fired at B to insert the satellite into a
geosynchronous orbit. Knowing that the second firing increases the
speed of the satellite by 4810 ft/s, determine
(a) the speed of the satellite as it approaches B on the elliptic transfer
orbit,
(b) the increase in speed resulting from the first firing at A.
(b) Conservation of angular momentum requires
that
rA m(v A )TR = rB m(vB )TR
or (v A )TR =
ft
ft
26,200mi
x5278 = 33,362
s
s
4145mi
Now…
(v A )TR = (v A ) circ + ∆v A
or…
∆v A = (33,362 − 25,362)
ft
s
or…
∆v A = 8000
ft
s
4