PHYS-1500 PHYSICAL MODELING FALL 2006
Class 16: Transfer Orbit NAME ______ SOLUTION ______________
1. If conservation of mechanical energy and angular momentum are applied at points P and Q , the following equations result.
1
2 mv
P
2 
GMm
R
A

1
2 mv
Q
2 
GMm and
R
B mv
P
R
A
 mv
Q
R
B
Then,
2
1 v
P
2 
GM
R
A
1
2 v
P
2 
GM
R
A

1
2 v
2
Q

GM
R
B and v
Q

R
A
R
B v
P
, which leads to

1
2


R
A
R
B v
P


2

GM or 1
2
R
B v
P
2

 1

R
A
2
R
B
2



GM


1
R
A

1
R
B


The last expression can be re-written in a form that allows some cancellation. v
2
P


R
2
B

R
R
B
2
2
A


2 GM


R
B

R
R R
B
A


or v
2
P



R
B

R
A

R
R R
B
B

R
A



2 GM



R
B

R
R R
B
A



With some cancellation, this becomes, v
P
2


R
B

R
A
R
B


2 GM
R
A
or v
2
P

1

R
A
R
B


2 GM
R
A
and v
2
P

2 GM
R
A


 1

1

R
A
/ R
B




Then, v
P

2 GM
R
A


 1

1

R
A
/ R
B




2. For R
A
= 7.00

10
6
m and R
B
= 4.22 ×10 v
P

2

6 .
67

10

11
N
 m
7 .
00

2
/
7
m, kg
2

5 .
98
10
6 m

10
24 kg



1

1

7 .
00

10
6 m / 4 .
22

10
7 m
 

 v
P
= 9887 m/s = 9.89 km/s
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
3. To find the speed in a circular orbit, just apply F = ma ,
GMm
 m v
2
, then
GM
 v
2
and v

GM r 2 r r r
For r = R
A
= 7.00

10
6
m, this becomes, v
A

GM
R
A


6 .
67

10

11
N
 m
7 .
00
2
/
 kg
10
2
6

5 .
98 m

10
24 kg


7549 m / s

7 .
55 m / s
Then,
 v
P
= v
P
– v
A
= 9887 m/s – 7549 m/s

v
P
= 2338 m/s = 2.34 km/s
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

4. For R
A
= 7.00

10
6
m and R
B
= 4.22 ×10
7
m, v
Q

R
A
R
B v
P

7 .
00

10
6 m
4 .
22

10
7 m
( 9887 m / s) v
Q

1640 m / s

1 .
64 km / s
5. In the geosynchronous orbit, r = R
B
= 4.22 ×10
7
m, so the speed is, v
B

GM
R
B


6 .
67

10

11
N
 m
4 .
22
2

/ kg
10
2
7

5 .
98 m

10
24 kg


3070 m / s

3 .
07 km / s
Then,
 v
Q
= v
B
– v
Q
= 3070 m/s – 1640 m/s
 v
Q
= 1430 m/s = 1.43 km/s
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
6. Let T
B
be the period of the satellite in the orbit of radius R
B
, and T
T
be the period of the shuttle in the transfer orbit. Then, in going from point P to point Q , the shuttle goes through one half of its orbit, so the time it takes is ½
T
T
. In that time, the satellite must go from point S to point
Q
, an angular displacement of 180° – 
. Since the satellite covers 360° in one period, its angular speed is 360°/
T
B
. If the shuttle and the satellite are to arrive at point Q at the same time, we must have,
360

T
B
T
T
2

180
 
, or
   
T
T
T
B
  

T
T
T
B


. Therefore, we need T
B
and T
T
. These can be obtained from the computer model, or from Kepler’s Law of
Periods. That law is T
2 
( 2

)
GM
2 a
3
, or T

2

GM a , where a is the semi-major axis of the orbit. For the satellite, a is just the radius of the orbit, R
B
. Then, T
B

2

GM
R
B
3/2
. For the transfer orbit, the semi-major axis is given by a

R
A

R
B
2
, so T
T

2

GM


R
A

R
B
2


.
T
T
Then,
T
B



R
A

R
B
2 R
B

, and

180




1



R
A
2

R
B
R
B


3 / 2




180




1



7 .
00

10
6 m

4 .
22

10
7 m
2 ( 4 .
22

10
7 m)


3 / 2




= 99.9°
T
T
= 38386 s