THE RESTRICTED THREE-BODY PROBLEM - CELESTIAL
MECHANICS AND ORBITAL DYNAMICS
1 NUMERICAL PROBLEMS ON THE RESTRICTED THREE-BODY PROBLEM
1. Consider a restricted three-body system with two primary bodies of masses 𝑀1=
1.98 ×1030 kg (Sun) and 𝑀2= 5.97 ×1024 kg (Earth). Calculate the mass parameter 𝜇
for this system.
Solution: The mass parameter 𝜇 is defined as:
𝜇 = 𝑀2
𝑀1+ 𝑀2
Substituting the given values:
𝜇 = 5.97 ×1024
1.98 ×1030 + 5.97 ×1024 = 3.0036 × 10−6
2. In the Earth-Moon system, the Earth’s mass is 5.97 ×1024 kg, and the Moon’s mass is
7.34 ×1022 kg. The distance between their centers is 384,400 km. Calculate the position
of the barycenter of the system relative to the Earth’s center.
Solution: The distance of the barycenter from the Earth’s center, 𝑟, is given by:
𝑟 = 𝑀𝑀𝑜𝑜𝑛
𝑀𝐸𝑎𝑟𝑡ℎ + 𝑀𝑀𝑜𝑜𝑛 ×Earth-Moon distance
𝑟 = 7.34 ×1022
5.97 ×1024 + 7.34 ×1022 ×384,400 = 4,671 km
3. For the Sun-Jupiter system, 𝜇 = 9.537 ×10−4. Calculate the positions of the L1, L2, and
L3 Lagrange points along the Sun-Jupiter line. Express your answers in terms of the
Sun-Jupiter distance 𝑅.
Solution: For small 𝜇, we can approximate the positions as:
𝐿1: 𝑥 ≈ 𝑅(1 − (𝜇/3)1/3)= 0.9326𝑅
𝐿2: 𝑥 ≈ 𝑅(1 + (𝜇/3)1/3)= 1.0686𝑅
𝐿3: 𝑥 ≈ −𝑅(1 + (5𝜇/12))= −1.0004𝑅
4. A spacecraft is located at the L4 point of the Earth-Moon system. If the Earth-Moon
distance is 384,400 km, what is the distance of the spacecraft from Earth?
Solution: The L4 point forms an equilateral triangle with the two primary bodies.
Therefore, its distance from Earth is equal to the Earth-Moon distance:
𝑑 = 384,400 km
5. In the circular restricted three-body problem, two bodies with masses 𝑀1= 2 × 1030 kg
and 𝑀2= 6 × 1024 kg orbit their common center of mass. A small satellite orbits in the
same plane. If the orbital period of the two massive bodies is 365 days, what is their
angular velocity 𝜔 in rad/s?
Solution: The angular velocity is given by:
𝜔 = 2𝜋
𝑇
where 𝑇 is the orbital period. Converting 365 days to seconds:
𝜔 = 2𝜋
365 ×24 ×3600 = 1.991 ×10−7 rad/s
6. For a system with mass parameter 𝜇 = 0.01, calculate the Jacobi constant 𝐶𝐽 for a
particle at rest at the L1 Lagrange point.
Solution: The Jacobi constant at L1 is approximately:
𝐶𝐽≈ 3 + 34/3𝜇2/3 −10𝜇/3
Substituting 𝜇 = 0.01:
𝐶𝐽≈ 3 + 34/3(0.01)2/3 −10(0.01)/3 = 3.0699
7. In the Earth-Moon system (𝜇 = 0.0123), a spacecraft is initially at rest relative to the
rotating frame at position (0.5, 0.5) in normalized coordinates. Calculate its initial
velocity in the inertial frame.
Solution: In the rotating frame, the initial velocity is (0, 0). To convert to the inertial
frame:
𝑣𝑥= −𝜔𝑦 = −(1)(0.5)= −0.5
𝑣𝑦=𝜔𝑥 =(1)(0.5)= 0.5
The initial velocity in the inertial frame is (-0.5, 0.5) in normalized units.
8. A spacecraft is located at (0.8, 0.6, 0) in the normalized rotating frame of the Sun-Earth
system (𝜇 = 3 × 10−6). Calculate the effective potential 𝑈 at this point.
Solution: The effective potential is given by:
𝑈 = − 1 − 𝜇
𝑟1−𝜇
𝑟2−1
2(𝑥2+ 𝑦2)
where 𝑟1=√(𝑥 + 𝜇)2+ 𝑦2+ 𝑧2 and 𝑟2=√(𝑥 − 1 + 𝜇)2+ 𝑦2+ 𝑧2.
Calculating 𝑟1 and 𝑟2:
𝑟1=√(0.8 + 3 × 10−6)2+ 0.62+ 02= 1.0
𝑟2=√(0.8 − 1 + 3 × 10−6)2+ 0.62+ 02= 0.28284
Substituting into the equation for 𝑈:
𝑈 = − 0.999997
1.0 −3 × 10−6
0.28284 −1
2(0.82+ 0.62)= −1.4999
9. Consider a restricted three-body system with two primary bodies of masses 𝑀1=
1.98 ×1030 kg (Sun) and 𝑀2= 5.97 ×1024 kg (Earth). Calculate the mass parameter 𝜇
for this system.
Solution: The mass parameter 𝜇 is defined as:
𝜇 = 𝑀2
𝑀1+ 𝑀2
Substituting the given values:
𝜇 = 5.97 ×1024
1.98 ×1030 + 5.97 ×1024 = 3.0036 × 10−6
10. In the Earth-Moon system, the Earth’s mass is 5.97 ×1024 kg, and the Moon’s mass is
7.34 ×1022 kg. The distance between their centers is 384,400 km. Calculate the position
of the barycenter of the system relative to the Earth’s center.
Solution: The distance of the barycenter from the Earth’s center, 𝑟, is given by:
𝑟 = 𝑀𝑀𝑜𝑜𝑛
𝑀𝐸𝑎𝑟𝑡ℎ + 𝑀𝑀𝑜𝑜𝑛 ×Earth-Moon distance
𝑟 = 7.34 ×1022
5.97 ×1024 + 7.34 ×1022 ×384,400 = 4,671 km
11. For the Sun-Jupiter system, 𝜇 = 9.537 ×10−4. Calculate the positions of the L1, L2, and
L3 Lagrange points along the Sun-Jupiter line. Express your answers in terms of the
Sun-Jupiter distance 𝑅.
Solution: For small 𝜇, we can approximate the positions as:
𝐿1: 𝑥 ≈ 𝑅(1 − (𝜇/3)1/3)= 0.9326𝑅
𝐿2: 𝑥 ≈ 𝑅(1 + (𝜇/3)1/3)= 1.0686𝑅
𝐿3: 𝑥 ≈ −𝑅(1 + (5𝜇/12))= −1.0004𝑅
12. A spacecraft is located at the L4 point of the Earth-Moon system. If the Earth-Moon
distance is 384,400 km, what is the distance of the spacecraft from Earth?
Solution: The L4 point forms an equilateral triangle with the two primary bodies.
Therefore, its distance from Earth is equal to the Earth-Moon distance:
𝑑 = 384,400 km
13. In the circular restricted three-body problem, two bodies with masses 𝑀1= 2 × 1030 kg
and 𝑀2= 6 × 1024 kg orbit their common center of mass. A small satellite orbits in the
same plane. If the orbital period of the two massive bodies is 365 days, what is their
angular velocity 𝜔 in rad/s?
Solution: The angular velocity is given by:
𝜔 = 2𝜋
𝑇
where 𝑇 is the orbital period. Converting 365 days to seconds:
𝜔 = 2𝜋
365 ×24 ×3600 = 1.991 ×10−7 rad/s
14. For a system with mass parameter 𝜇 = 0.01, calculate the Jacobi constant 𝐶𝐽 for a
particle at rest at the L1 Lagrange point.
Solution: The Jacobi constant at L1 is approximately:
𝐶𝐽≈ 3 + 34/3𝜇2/3 −10𝜇/3
Substituting 𝜇 = 0.01:
𝐶𝐽≈ 3 + 34/3(0.01)2/3 −10(0.01)/3 = 3.0699
15. In the Earth-Moon system (𝜇 = 0.0123), a spacecraft is initially at rest relative to the
rotating frame at position (0.5, 0.5) in normalized coordinates. Calculate its initial
velocity in the inertial frame.
Solution: In the rotating frame, the initial velocity is (0, 0). To convert to the inertial
frame:
𝑣𝑥= −𝜔𝑦 = −(1)(0.5)= −0.5
𝑣𝑦=𝜔𝑥 =(1)(0.5)= 0.5
The initial velocity in the inertial frame is (-0.5, 0.5) in normalized units.
16. A spacecraft is located at (0.8, 0.6, 0) in the normalized rotating frame of the Sun-Earth
system (𝜇 = 3 × 10−6). Calculate the effective potential 𝑈 at this point.
Solution: The effective potential is given by:
𝑈 = − 1 − 𝜇
𝑟1−𝜇
𝑟2−1
2(𝑥2+ 𝑦2)
where 𝑟1=√(𝑥 + 𝜇)2+ 𝑦2+ 𝑧2 and 𝑟2=√(𝑥 − 1 + 𝜇)2+ 𝑦2+ 𝑧2.
Calculating 𝑟1 and 𝑟2:
𝑟1=√(0.8 + 3 × 10−6)2+ 0.62+ 02= 1.0
𝑟2=√(0.8 − 1 + 3 × 10−6)2+ 0.62+ 02= 0.28284
Substituting into the equation for 𝑈:
𝑈 = − 0.999997
1.0 −3 × 10−6
0.28284 −1
2(0.82+ 0.62)= −1.4999
17. Consider a restricted three-body system with two primary bodies of masses 𝑀1=
1.98 ×1030 kg (Sun) and 𝑀2= 5.97 ×1024 kg (Earth). Calculate the mass parameter 𝜇
for this system.
Solution: The mass parameter 𝜇 is defined as:
𝜇 = 𝑀2
𝑀1+ 𝑀2
Substituting the given values:
𝜇 = 5.97 ×1024
1.98 ×1030 + 5.97 ×1024 = 3.0036 × 10−6
18. In the Earth-Moon system, the Earth’s mass is 5.97 ×1024 kg, and the Moon’s mass is
7.34 ×1022 kg. The distance between their centers is 384,400 km. Calculate the position
of the barycenter of the system relative to the Earth’s center.
Solution: The distance of the barycenter from the Earth’s center, 𝑟, is given by:
𝑟 = 𝑀𝑀𝑜𝑜𝑛
𝑀𝐸𝑎𝑟𝑡ℎ + 𝑀𝑀𝑜𝑜𝑛 ×Earth-Moon distance
𝑟 = 7.34 ×1022
5.97 ×1024 + 7.34 ×1022 ×384,400 = 4,671 km
19. For the Sun-Jupiter system, 𝜇 = 9.537 ×10−4. Calculate the positions of the L1, L2, and
L3 Lagrange points along the Sun-Jupiter line. Express your answers in terms of the
Sun-Jupiter distance 𝑅.
Solution: For small 𝜇, we can approximate the positions as:
𝐿1: 𝑥 ≈ 𝑅(1 − (𝜇/3)1/3)= 0.9326𝑅
𝐿2: 𝑥 ≈ 𝑅(1 + (𝜇/3)1/3)= 1.0686𝑅
𝐿3: 𝑥 ≈ −𝑅(1 + (5𝜇/12))= −1.0004𝑅
20. A spacecraft is located at the L4 point of the Earth-Moon system. If the Earth-Moon
distance is 384,400 km, what is the distance of the spacecraft from Earth?
Solution: The L4 point forms an equilateral triangle with the two primary bodies.
Therefore, its distance from Earth is equal to the Earth-Moon distance:
𝑑 = 384,400 km
21. In the circular restricted three-body problem, two bodies with masses 𝑀1= 2 × 1030 kg
and 𝑀2= 6 × 1024 kg orbit their common center of mass. A small satellite orbits in the
same plane. If the orbital period of the two massive bodies is 365 days, what is their
angular velocity 𝜔 in rad/s?
Solution: The angular velocity is given by:
𝜔 = 2𝜋
𝑇
where 𝑇 is the orbital period. Converting 365 days to seconds:
𝜔 = 2𝜋
365 ×24 ×3600 = 1.991 ×10−7 rad/s
22. For a system with mass parameter 𝜇 = 0.01, calculate the Jacobi constant 𝐶𝐽 for a
particle at rest at the L1 Lagrange point.
Solution: The Jacobi constant at L1 is approximately:
𝐶𝐽≈ 3 + 34/3𝜇2/3 −10𝜇/3
Substituting 𝜇 = 0.01:
𝐶𝐽≈ 3 + 34/3(0.01)2/3 −10(0.01)/3 = 3.0699
23. In the Earth-Moon system (𝜇 = 0.0123), a spacecraft is initially at rest relative to the
rotating frame at position (0.5, 0.5) in normalized coordinates. Calculate its initial
velocity in the inertial frame.
Solution: In the rotating frame, the initial velocity is (0, 0). To convert to the inertial
frame:
𝑣𝑥= −𝜔𝑦 = −(1)(0.5)= −0.5
𝑣𝑦=𝜔𝑥 =(1)(0.5)= 0.5
The initial velocity in the inertial frame is (-0.5, 0.5) in normalized units.
24. A spacecraft is located at (0.8, 0.6, 0) in the normalized rotating frame of the Sun-Earth
system (𝜇 = 3 × 10−6). Calculate the effective potential 𝑈 at this point.
Solution: The effective potential is given by:
𝑈 = − 1 − 𝜇
𝑟1−𝜇
𝑟2−1
2(𝑥2+ 𝑦2)
where 𝑟1=√(𝑥 + 𝜇)2+ 𝑦2+ 𝑧2 and 𝑟2=√(𝑥 − 1 + 𝜇)2+ 𝑦2+ 𝑧2.
Calculating 𝑟1 and 𝑟2:
𝑟1=√(0.8 + 3 × 10−6)2+ 0.62+ 02= 1.0
𝑟2=√(0.8 − 1 + 3 × 10−6)2+ 0.62+ 02= 0.28284
Substituting into the equation for 𝑈:
𝑈 = − 0.999997
1.0 −3 × 10−6
0.28284 −1
2(0.82+ 0.62)= −1.4999
25. Consider a restricted three-body system with two primary bodies of masses 𝑀1=
1.98 ×1030 kg (Sun) and 𝑀2= 5.97 ×1024 kg (Earth). Calculate the mass parameter 𝜇
for this system.
Solution: The mass parameter 𝜇 is defined as:
𝜇 = 𝑀2
𝑀1+ 𝑀2
Substituting the given values:
𝜇 = 5.97 ×1024
1.98 ×1030 + 5.97 ×1024 = 3.0036 × 10−6
26. In the Earth-Moon system, the Earth’s mass is 5.97 ×1024 kg, and the Moon’s mass is
7.34 ×1022 kg. The distance between their centers is 384,400 km. Calculate the position
of the barycenter of the system relative to the Earth’s center.
Solution: The distance of the barycenter from the Earth’s center, 𝑟, is given by:
𝑟 = 𝑀𝑀𝑜𝑜𝑛
𝑀𝐸𝑎𝑟𝑡ℎ + 𝑀𝑀𝑜𝑜𝑛 ×Earth-Moon distance
𝑟 = 7.34 ×1022
5.97 ×1024 + 7.34 ×1022 ×384,400 = 4,671 km
27. For the Sun-Jupiter system, 𝜇 = 9.537 ×10−4. Calculate the positions of the L1, L2, and
L3 Lagrange points along the Sun-Jupiter line. Express your answers in terms of the
Sun-Jupiter distance 𝑅.
Solution: For small 𝜇, we can approximate the positions as:
𝐿1: 𝑥 ≈ 𝑅(1 − (𝜇/3)1/3)= 0.9326𝑅
𝐿2: 𝑥 ≈ 𝑅(1 + (𝜇/3)1/3)= 1.0686𝑅
𝐿3: 𝑥 ≈ −𝑅(1 + (5𝜇/12))= −1.0004𝑅
28. A spacecraft is located at the L4 point of the Earth-Moon system. If the Earth-Moon
distance is 384,400 km, what is the distance of the spacecraft from Earth?
Solution: The L4 point forms an equilateral triangle with the two primary bodies.
Therefore, its distance from Earth is equal to the Earth-Moon distance:
𝑑 = 384,400 km
29. In the circular restricted three-body problem, two bodies with masses 𝑀1= 2 × 1030 kg
and 𝑀2= 6 × 1024 kg orbit their common center of mass. A small satellite orbits in the
same plane. If the orbital period of the two massive bodies is 365 days, what is their
angular velocity 𝜔 in rad/s?
Solution: The angular velocity is given by:
𝜔 = 2𝜋
𝑇
where 𝑇 is the orbital period. Converting 365 days to seconds:
𝜔 = 2𝜋
365 ×24 ×3600 = 1.991 ×10−7 rad/s
30. For a system with mass parameter 𝜇 = 0.01, calculate the Jacobi constant 𝐶𝐽 for a
particle at rest at the L1 Lagrange point.
Solution: The Jacobi constant at L1 is approximately:
𝐶𝐽≈ 3 + 34/3𝜇2/3 −10𝜇/3
Substituting 𝜇 = 0.01:
𝐶𝐽≈ 3 + 34/3(0.01)2/3 −10(0.01)/3 = 3.0699
31. In the Earth-Moon system (𝜇 = 0.0123), a spacecraft is initially at rest relative to the
rotating frame at position (0.5, 0.5) in normalized coordinates. Calculate its initial
velocity in the inertial frame.
Solution: In the rotating frame, the initial velocity is (0, 0). To convert to the inertial
frame:
𝑣𝑥= −𝜔𝑦 = −(1)(0.5)= −0.5
𝑣𝑦=𝜔𝑥 =(1)(0.5)= 0.5
The initial velocity in the inertial frame is (-0.5, 0.5) in normalized units.
32. A spacecraft is located at (0.8, 0.6, 0) in the normalized rotating frame of the Sun-Earth
system (𝜇 = 3 × 10−6). Calculate the effective potential 𝑈 at this point.
Solution: The effective potential is given by:
𝑈 = − 1 − 𝜇
𝑟1−𝜇
𝑟2−1
2(𝑥2+ 𝑦2)
where 𝑟1=√(𝑥 + 𝜇)2+ 𝑦2+ 𝑧2 and 𝑟2=√(𝑥 − 1 + 𝜇)2+ 𝑦2+ 𝑧2.
Calculating 𝑟1 and 𝑟2:
𝑟1=√(0.8 + 3 × 10−6)2+ 0.62+ 02= 1.0
𝑟2=√(0.8 − 1 + 3 × 10−6)2+ 0.62+ 02= 0.28284
Substituting into the equation for 𝑈:
𝑈 = − 0.999997
1.0 −3 × 10−6
0.28284 −1
2(0.82+ 0.62)= −1.4999
33. Consider a restricted three-body system with two primary bodies of masses 𝑀1=
1.98 ×1030 kg (Sun) and 𝑀2= 5.97 ×1024 kg (Earth). Calculate the mass parameter 𝜇
for this system.
Solution: The mass parameter 𝜇 is defined as:
𝜇 = 𝑀2
𝑀1+ 𝑀2
Substituting the given values:
𝜇 = 5.97 ×1024
1.98 ×1030 + 5.97 ×1024 = 3.0036 × 10−6
34. In the Earth-Moon system, the Earth’s mass is 5.97 ×1024 kg, and the Moon’s mass is
7.34 ×1022 kg. The distance between their centers is 384,400 km. Calculate the position
of the barycenter of the system relative to the Earth’s center.
Solution: The distance of the barycenter from the Earth’s center, 𝑟, is given by:
𝑟 = 𝑀𝑀𝑜𝑜𝑛
𝑀𝐸𝑎𝑟𝑡ℎ + 𝑀𝑀𝑜𝑜𝑛 ×Earth-Moon distance
𝑟 = 7.34 ×1022
5.97 ×1024 + 7.34 ×1022 ×384,400 = 4,671 km
35. For the Sun-Jupiter system, 𝜇 = 9.537 ×10−4. Calculate the positions of the L1, L2, and
L3 Lagrange points along the Sun-Jupiter line. Express your answers in terms of the
Sun-Jupiter distance 𝑅.
Solution: For small 𝜇, we can approximate the positions as:
𝐿1: 𝑥 ≈ 𝑅(1 − (𝜇/3)1/3)= 0.9326𝑅
𝐿2: 𝑥 ≈ 𝑅(1 + (𝜇/3)1/3)= 1.0686𝑅
𝐿3: 𝑥 ≈ −𝑅(1 + (5𝜇/12))= −1.0004𝑅
36. A spacecraft is located at the L4 point of the Earth-Moon system. If the Earth-Moon
distance is 384,400 km, what is the distance of the spacecraft from Earth?
Solution: The L4 point forms an equilateral triangle with the two primary bodies.
Therefore, its distance from Earth is equal to the Earth-Moon distance:
𝑑 = 384,400 km
37. In the circular restricted three-body problem, two bodies with masses 𝑀1= 2 × 1030 kg
and 𝑀2= 6 × 1024 kg orbit their common center of mass. A small satellite orbits in the
same plane. If the orbital period of the two massive bodies is 365 days, what is their
angular velocity 𝜔 in rad/s?
Solution: The angular velocity is given by:
𝜔 = 2𝜋
𝑇
where 𝑇 is the orbital period. Converting 365 days to seconds:
𝜔 = 2𝜋
365 ×24 ×3600 = 1.991 ×10−7 rad/s
38. For a system with mass parameter 𝜇 = 0.01, calculate the Jacobi constant 𝐶𝐽 for a
particle at rest at the L1 Lagrange point.
Solution: The Jacobi constant at L1 is approximately:
𝐶𝐽≈ 3 + 34/3𝜇2/3 −10𝜇/3
Substituting 𝜇 = 0.01:
𝐶𝐽≈ 3 + 34/3(0.01)2/3 −10(0.01)/3 = 3.0699
39. In the Earth-Moon system (𝜇 = 0.0123), a spacecraft is initially at rest relative to the
rotating frame at position (0.5, 0.5) in normalized coordinates. Calculate its initial
velocity in the inertial frame.
Solution: In the rotating frame, the initial velocity is (0, 0). To convert to the inertial
frame:
𝑣𝑥= −𝜔𝑦 = −(1)(0.5)= −0.5
𝑣𝑦=𝜔𝑥 =(1)(0.5)= 0.5
The initial velocity in the inertial frame is (-0.5, 0.5) in normalized units.
40. A spacecraft is located at (0.8, 0.6, 0) in the normalized rotating frame of the Sun-Earth
system (𝜇 = 3 × 10−6). Calculate the effective potential 𝑈 at this point.
Solution: The effective potential is given by:
𝑈 = − 1 − 𝜇
𝑟1−𝜇
𝑟2−1
2(𝑥2+ 𝑦2)
where 𝑟1=√(𝑥 + 𝜇)2+ 𝑦2+ 𝑧2 and 𝑟2=√(𝑥 − 1 + 𝜇)2+ 𝑦2+ 𝑧2.
Calculating 𝑟1 and 𝑟2:
𝑟1=√(0.8 + 3 × 10−6)2+ 0.62+ 02= 1.0
𝑟2=√(0.8 − 1 + 3 × 10−6)2+ 0.62+ 02= 0.28284
Substituting into the equation for 𝑈:
𝑈 = − 0.999997
1.0 −3 × 10−6
0.28284 −1
2(0.82+ 0.62)= −1.4999
41. Consider a restricted three-body system with two primary bodies of masses 𝑀1=
1.98 ×1030 kg (Sun) and 𝑀2= 5.97 ×1024 kg (Earth). Calculate the mass parameter 𝜇
for this system.
Solution: The mass parameter 𝜇 is defined as:
𝜇 = 𝑀2
𝑀1+ 𝑀2
Substituting the given values:
𝜇 = 5.97 ×1024
1.98 ×1030 + 5.97 ×1024 = 3.0036 × 10−6
42. In the Earth-Moon system, the Earth’s mass is 5.97 ×1024 kg, and the Moon’s mass is
7.34 ×1022 kg. The distance between their centers is 384,400 km. Calculate the position
of the barycenter of the system relative to the Earth’s center.
Solution: The distance of the barycenter from the Earth’s center, 𝑟, is given by:
𝑟 = 𝑀𝑀𝑜𝑜𝑛
𝑀𝐸𝑎𝑟𝑡ℎ + 𝑀𝑀𝑜𝑜𝑛 ×Earth-Moon distance
𝑟 = 7.34 ×1022
5.97 ×1024 + 7.34 ×1022 ×384,400 = 4,671 km
43. For the Sun-Jupiter system, 𝜇 = 9.537 ×10−4. Calculate the positions of the L1, L2, and
L3 Lagrange points along the Sun-Jupiter line. Express your answers in terms of the
Sun-Jupiter distance 𝑅.
Solution: For small 𝜇, we can approximate the positions as:
𝐿1: 𝑥 ≈ 𝑅(1 − (𝜇/3)1/3)= 0.9326𝑅
𝐿2: 𝑥 ≈ 𝑅(1 + (𝜇/3)1/3)= 1.0686𝑅
𝐿3: 𝑥 ≈ −𝑅(1 + (5𝜇/12))= −1.0004𝑅
44. A spacecraft is located at the L4 point of the Earth-Moon system. If the Earth-Moon
distance is 384,400 km, what is the distance of the spacecraft from Earth?
Solution: The L4 point forms an equilateral triangle with the two primary bodies.
Therefore, its distance from Earth is equal to the Earth-Moon distance:
𝑑 = 384,400 km
45. In the circular restricted three-body problem, two bodies with masses 𝑀1= 2 × 1030 kg
and 𝑀2= 6 × 1024 kg orbit their common center of mass. A small satellite orbits in the
same plane. If the orbital period of the two massive bodies is 365 days, what is their
angular velocity 𝜔 in rad/s?
Solution: The angular velocity is given by:
𝜔 = 2𝜋
𝑇
where 𝑇 is the orbital period. Converting 365 days to seconds:
𝜔 = 2𝜋
365 ×24 ×3600 = 1.991 ×10−7 rad/s
46. For a system with mass parameter 𝜇 = 0.01, calculate the Jacobi constant 𝐶𝐽 for a
particle at rest at the L1 Lagrange point.
Solution: The Jacobi constant at L1 is approximately:
𝐶𝐽≈ 3 + 34/3𝜇2/3 −10𝜇/3
Substituting 𝜇 = 0.01:
𝐶𝐽≈ 3 + 34/3(0.01)2/3 −10(0.01)/3 = 3.0699
47. In the Earth-Moon system (𝜇 = 0.0123), a spacecraft is initially at rest relative to the
rotating frame at position (0.5, 0.5) in normalized coordinates. Calculate its initial
velocity in the inertial frame.
Solution: In the rotating frame, the initial velocity is (0, 0). To convert to the inertial
frame:
𝑣𝑥= −𝜔𝑦 = −(1)(0.5)= −0.5
𝑣𝑦=𝜔𝑥 =(1)(0.5)= 0.5
The initial velocity in the inertial frame is (-0.5, 0.5) in normalized units.
48. A spacecraft is located at (0.8, 0.6, 0) in the normalized rotating frame of the Sun-Earth
system (𝜇 = 3 × 10−6). Calculate the effective potential 𝑈 at this point.
Solution: The effective potential is given by:
𝑈 = − 1 − 𝜇
𝑟1−𝜇
𝑟2−1
2(𝑥2+ 𝑦2)
where 𝑟1=√(𝑥 + 𝜇)2+ 𝑦2+ 𝑧2 and 𝑟2=√(𝑥 − 1 + 𝜇)2+ 𝑦2+ 𝑧2.
Calculating 𝑟1 and 𝑟2:
𝑟1=√(0.8 + 3 × 10−6)2+ 0.62+ 02= 1.0
𝑟2=√(0.8 − 1 + 3 × 10−6)2+ 0.62+ 02= 0.28284
Substituting into the equation for 𝑈:
𝑈 = − 0.999997
1.0 −3 × 10−6
0.28284 −1
2(0.82+ 0.62)= −1.4999
49. Consider a restricted three-body system with two primary bodies of masses 𝑀1=
1.98 ×1030 kg (Sun) and 𝑀2= 5.97 ×1024 kg (Earth). Calculate the mass parameter 𝜇
for this system.
Solution: The mass parameter 𝜇 is defined as:
𝜇 = 𝑀2
𝑀1+ 𝑀2
Substituting the given values:
𝜇 = 5.97 ×1024
1.98 ×1030 + 5.97 ×1024 = 3.0036 × 10−6
50. In the Earth-Moon system, the Earth’s mass is 5.97 ×1024 kg, and the Moon’s mass is
7.34 ×1022 kg. The distance between their centers is 384,400 km. Calculate the position
of the barycenter of the system relative to the Earth’s center.
Solution: The distance of the barycenter from the Earth’s center, 𝑟, is given by:
𝑟 = 𝑀𝑀𝑜𝑜𝑛
𝑀𝐸𝑎𝑟𝑡ℎ + 𝑀𝑀𝑜𝑜𝑛 ×Earth-Moon distance
𝑟 = 7.34 ×1022
5.97 ×1024 + 7.34 ×1022 ×384,400 = 4,671 km
51. For the Sun-Jupiter system, 𝜇 = 9.537 ×10−4. Calculate the positions of the L1, L2, and
L3 Lagrange points along the Sun-Jupiter line. Express your answers in terms of the
Sun-Jupiter distance 𝑅.
Solution: For small 𝜇, we can approximate the positions as:
𝐿1: 𝑥 ≈ 𝑅(1 − (𝜇/3)1/3)= 0.9326𝑅
𝐿2: 𝑥 ≈ 𝑅(1 + (𝜇/3)1/3)= 1.0686𝑅
𝐿3: 𝑥 ≈ −𝑅(1 + (5𝜇/12))= −1.0004𝑅
52. A spacecraft is located at the L4 point of the Earth-Moon system. If the Earth-Moon
distance is 384,400 km, what is the distance of the spacecraft from Earth?
Solution: The L4 point forms an equilateral triangle with the two primary bodies.
Therefore, its distance from Earth is equal to the Earth-Moon distance:
𝑑 = 384,400 km
53. In the circular restricted three-body problem, two bodies with masses 𝑀1= 2 × 1030 kg
and 𝑀2= 6 × 1024 kg orbit their common center of mass. A small satellite orbits in the
same plane. If the orbital period of the two massive bodies is 365 days, what is their
angular velocity 𝜔 in rad/s?
Solution: The angular velocity is given by:
𝜔 = 2𝜋
𝑇
where 𝑇 is the orbital period. Converting 365 days to seconds:
𝜔 = 2𝜋
365 ×24 ×3600 = 1.991 ×10−7 rad/s
54. For a system with mass parameter 𝜇 = 0.01, calculate the Jacobi constant 𝐶𝐽 for a
particle at rest at the L1 Lagrange point.
Solution: The Jacobi constant at L1 is approximately:
𝐶𝐽≈ 3 + 34/3𝜇2/3 −10𝜇/3
Substituting 𝜇 = 0.01:
𝐶𝐽≈ 3 + 34/3(0.01)2/3 −10(0.01)/3 = 3.0699
55. In the Earth-Moon system (𝜇 = 0.0123), a spacecraft is initially at rest relative to the
rotating frame at position (0.5, 0.5) in normalized coordinates. Calculate its initial
velocity in the inertial frame.
Solution: In the rotating frame, the initial velocity is (0, 0). To convert to the inertial
frame:
𝑣𝑥= −𝜔𝑦 = −(1)(0.5)= −0.5
𝑣𝑦=𝜔𝑥 =(1)(0.5)= 0.5
The initial velocity in the inertial frame is (-0.5, 0.5) in normalized units.
56. A spacecraft is located at (0.8, 0.6, 0) in the normalized rotating frame of the Sun-Earth
system (𝜇 = 3 × 10−6). Calculate the effective potential 𝑈 at this point.
Solution: The effective potential is given by:
𝑈 = − 1 − 𝜇
𝑟1−𝜇
𝑟2−1
2(𝑥2+ 𝑦2)
where 𝑟1=√(𝑥 + 𝜇)2+ 𝑦2+ 𝑧2 and 𝑟2=√(𝑥 − 1 + 𝜇)2+ 𝑦2+ 𝑧2.
Calculating 𝑟1 and 𝑟2:
𝑟1=√(0.8 + 3 × 10−6)2+ 0.62+ 02= 1.0
𝑟2=√(0.8 − 1 + 3 × 10−6)2+ 0.62+ 02= 0.28284
Substituting into the equation for 𝑈:
𝑈 = − 0.999997
1.0 −3 × 10−6
0.28284 −1
2(0.82+ 0.62)= −1.4999
57. Consider a restricted three-body system with two primary bodies of masses 𝑀1=
1.98 ×1030 kg (Sun) and 𝑀2= 5.97 ×1024 kg (Earth). Calculate the mass parameter 𝜇
for this system.
Solution: The mass parameter 𝜇 is defined as:
𝜇 = 𝑀2
𝑀1+ 𝑀2
Substituting the given values:
𝜇 = 5.97 ×1024
1.98 ×1030 + 5.97 ×1024 = 3.0036 × 10−6
58. In the Earth-Moon system, the Earth’s mass is 5.97 ×1024 kg, and the Moon’s mass is
7.34 ×1022 kg. The distance between their centers is 384,400 km. Calculate the position
of the barycenter of the system relative to the Earth’s center.
Solution: The distance of the barycenter from the Earth’s center, 𝑟, is given by:
𝑟 = 𝑀𝑀𝑜𝑜𝑛
𝑀𝐸𝑎𝑟𝑡ℎ + 𝑀𝑀𝑜𝑜𝑛 ×Earth-Moon distance
𝑟 = 7.34 ×1022
5.97 ×1024 + 7.34 ×1022 ×384,400 = 4,671 km
59. For the Sun-Jupiter system, 𝜇 = 9.537 ×10−4. Calculate the positions of the L1, L2, and
L3 Lagrange points along the Sun-Jupiter line. Express your answers in terms of the
Sun-Jupiter distance 𝑅.
Solution: For small 𝜇, we can approximate the positions as:
𝐿1: 𝑥 ≈ 𝑅(1 − (𝜇/3)1/3)= 0.9326𝑅
𝐿2: 𝑥 ≈ 𝑅(1 + (𝜇/3)1/3)= 1.0686𝑅
𝐿3: 𝑥 ≈ −𝑅(1 + (5𝜇/12))= −1.0004𝑅
60. A spacecraft is located at the L4 point of the Earth-Moon system. If the Earth-Moon
distance is 384,400 km, what is the distance of the spacecraft from Earth?
Solution: The L4 point forms an equilateral triangle with the two primary bodies.
Therefore, its distance from Earth is equal to the Earth-Moon distance:
𝑑 = 384,400 km
61. In the circular restricted three-body problem, two bodies with masses 𝑀1= 2 × 1030 kg
and 𝑀2= 6 × 1024 kg orbit their common center of mass. A small satellite orbits in the
same plane. If the orbital period of the two massive bodies is 365 days, what is their
angular velocity 𝜔 in rad/s?
Solution: The angular velocity is given by:
𝜔 = 2𝜋
𝑇
where 𝑇 is the orbital period. Converting 365 days to seconds:
𝜔 = 2𝜋
365 ×24 ×3600 = 1.991 ×10−7 rad/s
62. For a system with mass parameter 𝜇 = 0.01, calculate the Jacobi constant 𝐶𝐽 for a
particle at rest at the L1 Lagrange point.
Solution: The Jacobi constant at L1 is approximately:
𝐶𝐽≈ 3 + 34/3𝜇2/3 −10𝜇/3
Substituting 𝜇 = 0.01:
𝐶𝐽≈ 3 + 34/3(0.01)2/3 −10(0.01)/3 = 3.0699
63. In the Earth-Moon system (𝜇 = 0.0123), a spacecraft is initially at rest relative to the
rotating frame at position (0.5, 0.5) in normalized coordinates. Calculate its initial
velocity in the inertial frame.
Solution: In the rotating frame, the initial velocity is (0, 0). To convert to the inertial
frame:
𝑣𝑥= −𝜔𝑦 = −(1)(0.5)= −0.5
𝑣𝑦=𝜔𝑥 =(1)(0.5)= 0.5
The initial velocity in the inertial frame is (-0.5, 0.5) in normalized units.
64. A spacecraft is located at (0.8, 0.6, 0) in the normalized rotating frame of the Sun-Earth
system (𝜇 = 3 × 10−6). Calculate the effective potential 𝑈 at this point.
Solution: The effective potential is given by:
𝑈 = − 1 − 𝜇
𝑟1−𝜇
𝑟2−1
2(𝑥2+ 𝑦2)
where 𝑟1=√(𝑥 + 𝜇)2+ 𝑦2+ 𝑧2 and 𝑟2=√(𝑥 − 1 + 𝜇)2+ 𝑦2+ 𝑧2.
Calculating 𝑟1 and 𝑟2:
𝑟1=√(0.8 + 3 × 10−6)2+ 0.62+ 02= 1.0
𝑟2=√(0.8 − 1 + 3 × 10−6)2+ 0.62+ 02= 0.28284
Substituting into the equation for 𝑈:
𝑈 = − 0.999997
1.0 −3 × 10−6
0.28284 −1
2(0.82+ 0.62)= −1.4999
65. Consider a restricted three-body system with two primary bodies of masses 𝑀1=
1.98 ×1030 kg (Sun) and 𝑀2= 5.97 ×1024 kg (Earth). Calculate the mass parameter 𝜇
for this system.
Solution: The mass parameter 𝜇 is defined as:
𝜇 = 𝑀2
𝑀1+ 𝑀2
Substituting the given values:
𝜇 = 5.97 ×1024
1.98 ×1030 + 5.97 ×1024 = 3.0036 × 10−6
66. In the Earth-Moon system, the Earth’s mass is 5.97 ×1024 kg, and the Moon’s mass is
7.34 ×1022 kg. The distance between their centers is 384,400 km. Calculate the position
of the barycenter of the system relative to the Earth’s center.
Solution: The distance of the barycenter from the Earth’s center, 𝑟, is given by:
𝑟 = 𝑀𝑀𝑜𝑜𝑛
𝑀𝐸𝑎𝑟𝑡ℎ + 𝑀𝑀𝑜𝑜𝑛 ×Earth-Moon distance
𝑟 = 7.34 ×1022
5.97 ×1024 + 7.34 ×1022 ×384,400 = 4,671 km
67. For the Sun-Jupiter system, 𝜇 = 9.537 ×10−4. Calculate the positions of the L1, L2, and
L3 Lagrange points along the Sun-Jupiter line. Express your answers in terms of the
Sun-Jupiter distance 𝑅.
Solution: For small 𝜇, we can approximate the positions as:
𝐿1: 𝑥 ≈ 𝑅(1 − (𝜇/3)1/3)= 0.9326𝑅
𝐿2: 𝑥 ≈ 𝑅(1 + (𝜇/3)1/3)= 1.0686𝑅
𝐿3: 𝑥 ≈ −𝑅(1 + (5𝜇/12))= −1.0004𝑅
68. A spacecraft is located at the L4 point of the Earth-Moon system. If the Earth-Moon
distance is 384,400 km, what is the distance of the spacecraft from Earth?
Solution: The L4 point forms an equilateral triangle with the two primary bodies.
Therefore, its distance from Earth is equal to the Earth-Moon distance:
𝑑 = 384,400 km
69. In the circular restricted three-body problem, two bodies with masses 𝑀1= 2 × 1030 kg
and 𝑀2= 6 × 1024 kg orbit their common center of mass. A small satellite orbits in the
same plane. If the orbital period of the two massive bodies is 365 days, what is their
angular velocity 𝜔 in rad/s?
Solution: The angular velocity is given by:
𝜔 = 2𝜋
𝑇
where 𝑇 is the orbital period. Converting 365 days to seconds:
𝜔 = 2𝜋
365 ×24 ×3600 = 1.991 ×10−7 rad/s
70. For a system with mass parameter 𝜇 = 0.01, calculate the Jacobi constant 𝐶𝐽 for a
particle at rest at the L1 Lagrange point.
Solution: The Jacobi constant at L1 is approximately:
𝐶𝐽≈ 3 + 34/3𝜇2/3 −10𝜇/3
Substituting 𝜇 = 0.01:
𝐶𝐽≈ 3 + 34/3(0.01)2/3 −10(0.01)/3 = 3.0699
71. In the Earth-Moon system (𝜇 = 0.0123), a spacecraft is initially at rest relative to the
rotating frame at position (0.5, 0.5) in normalized coordinates. Calculate its initial
velocity in the inertial frame.
Solution: In the rotating frame, the initial velocity is (0, 0). To convert to the inertial
frame:
𝑣𝑥= −𝜔𝑦 = −(1)(0.5)= −0.5
𝑣𝑦=𝜔𝑥 =(1)(0.5)= 0.5
The initial velocity in the inertial frame is (-0.5, 0.5) in normalized units.
72. A spacecraft is located at (0.8, 0.6, 0) in the normalized rotating frame of the Sun-Earth
system (𝜇 = 3 × 10−6). Calculate the effective potential 𝑈 at this point.
Solution: The effective potential is given by:
𝑈 = − 1 − 𝜇
𝑟1−𝜇
𝑟2−1
2(𝑥2+ 𝑦2)
where 𝑟1=√(𝑥 + 𝜇)2+ 𝑦2+ 𝑧2 and 𝑟2=√(𝑥 − 1 + 𝜇)2+ 𝑦2+ 𝑧2.
Calculating 𝑟1 and 𝑟2:
𝑟1=√(0.8 + 3 × 10−6)2+ 0.62+ 02= 1.0
𝑟2=√(0.8 − 1 + 3 × 10−6)2+ 0.62+ 02= 0.28284
Substituting into the equation for 𝑈:
𝑈 = − 0.999997
1.0 −3 × 10−6
0.28284 −1
2(0.82+ 0.62)= −1.4999
73. Consider a restricted three-body system with two primary bodies of masses 𝑀1=
1.98 ×1030 kg (Sun) and 𝑀2= 5.97 ×1024 kg (Earth). Calculate the mass parameter 𝜇
for this system.
Solution: The mass parameter 𝜇 is defined as:
𝜇 = 𝑀2
𝑀1+ 𝑀2
Substituting the given values:
𝜇 = 5.97 ×1024
1.98 ×1030 + 5.97 ×1024 = 3.0036 × 10−6
74. In the Earth-Moon system, the Earth’s mass is 5.97 ×1024 kg, and the Moon’s mass is
7.34 ×1022 kg. The distance between their centers is 384,400 km. Calculate the position
of the barycenter of the system relative to the Earth’s center.
Solution: The distance of the barycenter from the Earth’s center, 𝑟, is given by:
𝑟 = 𝑀𝑀𝑜𝑜𝑛
𝑀𝐸𝑎𝑟𝑡ℎ + 𝑀𝑀𝑜𝑜𝑛 ×Earth-Moon distance
𝑟 = 7.34 ×1022
5.97 ×1024 + 7.34 ×1022 ×384,400 = 4,671 km
75. For the Sun-Jupiter system, 𝜇 = 9.537 ×10−4. Calculate the positions of the L1, L2, and
L3 Lagrange points along the Sun-Jupiter line. Express your answers in terms of the
Sun-Jupiter distance 𝑅.
Solution: For small 𝜇, we can approximate the positions as:
𝐿1: 𝑥 ≈ 𝑅(1 − (𝜇/3)1/3)= 0.9326𝑅
𝐿2: 𝑥 ≈ 𝑅(1 + (𝜇/3)1/3)= 1.0686𝑅
𝐿3: 𝑥 ≈ −𝑅(1 + (5𝜇/12))= −1.0004𝑅
76. A spacecraft is located at the L4 point of the Earth-Moon system. If the Earth-Moon
distance is 384,400 km, what is the distance of the spacecraft from Earth?
Solution: The L4 point forms an equilateral triangle with the two primary bodies.
Therefore, its distance from Earth is equal to the Earth-Moon distance:
𝑑 = 384,400 km
77. In the circular restricted three-body problem, two bodies with masses 𝑀1= 2 × 1030 kg
and 𝑀2= 6 × 1024 kg orbit their common center of mass. A small satellite orbits in the
same plane. If the orbital period of the two massive bodies is 365 days, what is their
angular velocity 𝜔 in rad/s?
Solution: The angular velocity is given by:
𝜔 = 2𝜋
𝑇
where 𝑇 is the orbital period. Converting 365 days to seconds:
𝜔 = 2𝜋
365 ×24 ×3600 = 1.991 ×10−7 rad/s
78. For a system with mass parameter 𝜇 = 0.01, calculate the Jacobi constant 𝐶𝐽 for a
particle at rest at the L1 Lagrange point.
Solution: The Jacobi constant at L1 is approximately:
𝐶𝐽≈ 3 + 34/3𝜇2/3 −10𝜇/3
Substituting 𝜇 = 0.01:
𝐶𝐽≈ 3 + 34/3(0.01)2/3 −10(0.01)/3 = 3.0699
79. In the Earth-Moon system (𝜇 = 0.0123), a spacecraft is initially at rest relative to the
rotating frame at position (0.5, 0.5) in normalized coordinates. Calculate its initial
velocity in the inertial frame.
Solution: In the rotating frame, the initial velocity is (0, 0). To convert to the inertial
frame:
𝑣𝑥= −𝜔𝑦 = −(1)(0.5)= −0.5
𝑣𝑦=𝜔𝑥 =(1)(0.5)= 0.5
The initial velocity in the inertial frame is (-0.5, 0.5) in normalized units.
80. A spacecraft is located at (0.8, 0.6, 0) in the normalized rotating frame of the Sun-Earth
system (𝜇 = 3 × 10−6). Calculate the effective potential 𝑈 at this point.
Solution: The effective potential is given by:
𝑈 = − 1 − 𝜇
𝑟1−𝜇
𝑟2−1
2(𝑥2+ 𝑦2)
where 𝑟1=√(𝑥 + 𝜇)2+ 𝑦2+ 𝑧2 and 𝑟2=√(𝑥 − 1 + 𝜇)2+ 𝑦2+ 𝑧2.
Calculating 𝑟1 and 𝑟2:
𝑟1=√(0.8 + 3 × 10−6)2+ 0.62+ 02= 1.0
𝑟2=√(0.8 − 1 + 3 × 10−6)2+ 0.62+ 02= 0.28284
Substituting into the equation for 𝑈:
𝑈 = − 0.999997
1.0 −3 × 10−6
0.28284 −1
2(0.82+ 0.62)= −1.4999
81. Consider a restricted three-body system with two primary bodies of masses 𝑀1=
1.98 ×1030 kg (Sun) and 𝑀2= 5.97 ×1024 kg (Earth). Calculate the mass parameter 𝜇
for this system.
Solution: The mass parameter 𝜇 is defined as:
𝜇 = 𝑀2
𝑀1+ 𝑀2
Substituting the given values:
𝜇 = 5.97 ×1024
1.98 ×1030 + 5.97 ×1024 = 3.0036 × 10−6
82. In the Earth-Moon system, the Earth’s mass is 5.97 ×1024 kg, and the Moon’s mass is
7.34 ×1022 kg. The distance between their centers is 384,400 km. Calculate the position
of the barycenter of the system relative to the Earth’s center.
Solution: The distance of the barycenter from the Earth’s center, 𝑟, is given by:
𝑟 = 𝑀𝑀𝑜𝑜𝑛
𝑀𝐸𝑎𝑟𝑡ℎ + 𝑀𝑀𝑜𝑜𝑛 ×Earth-Moon distance
𝑟 = 7.34 ×1022
5.97 ×1024 + 7.34 ×1022 ×384,400 = 4,671 km
83. For the Sun-Jupiter system, 𝜇 = 9.537 ×10−4. Calculate the positions of the L1, L2, and
L3 Lagrange points along the Sun-Jupiter line. Express your answers in terms of the
Sun-Jupiter distance 𝑅.
Solution: For small 𝜇, we can approximate the positions as:
𝐿1: 𝑥 ≈ 𝑅(1 − (𝜇/3)1/3)= 0.9326𝑅
𝐿2: 𝑥 ≈ 𝑅(1 + (𝜇/3)1/3)= 1.0686𝑅
𝐿3: 𝑥 ≈ −𝑅(1 + (5𝜇/12))= −1.0004𝑅
84. A spacecraft is located at the L4 point of the Earth-Moon system. If the Earth-Moon
distance is 384,400 km, what is the distance of the spacecraft from Earth?
Solution: The L4 point forms an equilateral triangle with the two primary bodies.
Therefore, its distance from Earth is equal to the Earth-Moon distance:
𝑑 = 384,400 km
85. In the circular restricted three-body problem, two bodies with masses 𝑀1= 2 × 1030 kg
and 𝑀2= 6 × 1024 kg orbit their common center of mass. A small satellite orbits in the
same plane. If the orbital period of the two massive bodies is 365 days, what is their
angular velocity 𝜔 in rad/s?
Solution: The angular velocity is given by:
𝜔 = 2𝜋
𝑇
where 𝑇 is the orbital period. Converting 365 days to seconds:
𝜔 = 2𝜋
365 ×24 ×3600 = 1.991 ×10−7 rad/s
86. For a system with mass parameter 𝜇 = 0.01, calculate the Jacobi constant 𝐶𝐽 for a
particle at rest at the L1 Lagrange point.
Solution: The Jacobi constant at L1 is approximately:
𝐶𝐽≈ 3 + 34/3𝜇2/3 −10𝜇/3
Substituting 𝜇 = 0.01:
𝐶𝐽≈ 3 + 34/3(0.01)2/3 −10(0.01)/3 = 3.0699
87. In the Earth-Moon system (𝜇 = 0.0123), a spacecraft is initially at rest relative to the
rotating frame at position (0.5, 0.5) in normalized coordinates. Calculate its initial
velocity in the inertial frame.
Solution: In the rotating frame, the initial velocity is (0, 0). To convert to the inertial
frame:
𝑣𝑥= −𝜔𝑦 = −(1)(0.5)= −0.5
𝑣𝑦=𝜔𝑥 =(1)(0.5)= 0.5
The initial velocity in the inertial frame is (-0.5, 0.5) in normalized units.
88. A spacecraft is located at (0.8, 0.6, 0) in the normalized rotating frame of the Sun-Earth
system (𝜇 = 3 × 10−6). Calculate the effective potential 𝑈 at this point.
Solution: The effective potential is given by:
𝑈 = − 1 − 𝜇
𝑟1−𝜇
𝑟2−1
2(𝑥2+ 𝑦2)
where 𝑟1=√(𝑥 + 𝜇)2+ 𝑦2+ 𝑧2 and 𝑟2=√(𝑥 − 1 + 𝜇)2+ 𝑦2+ 𝑧2.
Calculating 𝑟1 and 𝑟2:
𝑟1=√(0.8 + 3 × 10−6)2+ 0.62+ 02= 1.0
𝑟2=√(0.8 − 1 + 3 × 10−6)2+ 0.62+ 02= 0.28284
Substituting into the equation for 𝑈:
𝑈 = − 0.999997
1.0 −3 × 10−6
0.28284 −1
2(0.82+ 0.62)= −1.4999
89. Consider a restricted three-body system with two primary bodies of masses 𝑀1=
1.98 ×1030 kg (Sun) and 𝑀2= 5.97 ×1024 kg (Earth). Calculate the mass parameter 𝜇
for this system.
Solution: The mass parameter 𝜇 is defined as:
𝜇 = 𝑀2
𝑀1+ 𝑀2
Substituting the given values:
𝜇 = 5.97 ×1024
1.98 ×1030 + 5.97 ×1024 = 3.0036 × 10−6
90. In the Earth-Moon system, the Earth’s mass is 5.97 ×1024 kg, and the Moon’s mass is
7.34 ×1022 kg. The distance between their centers is 384,400 km. Calculate the position
of the barycenter of the system relative to the Earth’s center.
Solution: The distance of the barycenter from the Earth’s center, 𝑟, is given by:
𝑟 = 𝑀𝑀𝑜𝑜𝑛
𝑀𝐸𝑎𝑟𝑡ℎ + 𝑀𝑀𝑜𝑜𝑛 ×Earth-Moon distance
𝑟 = 7.34 ×1022
5.97 ×1024 + 7.34 ×1022 ×384,400 = 4,671 km
91. For the Sun-Jupiter system, 𝜇 = 9.537 ×10−4. Calculate the positions of the L1, L2, and
L3 Lagrange points along the Sun-Jupiter line. Express your answers in terms of the
Sun-Jupiter distance 𝑅.
Solution: For small 𝜇, we can approximate the positions as:
𝐿1: 𝑥 ≈ 𝑅(1 − (𝜇/3)1/3)= 0.9326𝑅
𝐿2: 𝑥 ≈ 𝑅(1 + (𝜇/3)1/3)= 1.0686𝑅
𝐿3: 𝑥 ≈ −𝑅(1 + (5𝜇/12))= −1.0004𝑅
92. A spacecraft is located at the L4 point of the Earth-Moon system. If the Earth-Moon
distance is 384,400 km, what is the distance of the spacecraft from Earth?
Solution: The L4 point forms an equilateral triangle with the two primary bodies.
Therefore, its distance from Earth is equal to the Earth-Moon distance:
𝑑 = 384,400 km
93. In the circular restricted three-body problem, two bodies with masses 𝑀1= 2 × 1030 kg
and 𝑀2= 6 × 1024 kg orbit their common center of mass. A small satellite orbits in the
same plane. If the orbital period of the two massive bodies is 365 days, what is their
angular velocity 𝜔 in rad/s?
Solution: The angular velocity is given by:
𝜔 = 2𝜋
𝑇
where 𝑇 is the orbital period. Converting 365 days to seconds:
𝜔 = 2𝜋
365 ×24 ×3600 = 1.991 ×10−7 rad/s
94. For a system with mass parameter 𝜇 = 0.01, calculate the Jacobi constant 𝐶𝐽 for a
particle at rest at the L1 Lagrange point.
Solution: The Jacobi constant at L1 is approximately:
𝐶𝐽≈ 3 + 34/3𝜇2/3 −10𝜇/3
Substituting 𝜇 = 0.01:
𝐶𝐽≈ 3 + 34/3(0.01)2/3 −10(0.01)/3 = 3.0699
95. In the Earth-Moon system (𝜇 = 0.0123), a spacecraft is initially at rest relative to the
rotating frame at position (0.5, 0.5) in normalized coordinates. Calculate its initial
velocity in the inertial frame.
Solution: In the rotating frame, the initial velocity is (0, 0). To convert to the inertial
frame:
𝑣𝑥= −𝜔𝑦 = −(1)(0.5)= −0.5
𝑣𝑦=𝜔𝑥 =(1)(0.5)= 0.5
The initial velocity in the inertial frame is (-0.5, 0.5) in normalized units.
96. A spacecraft is located at (0.8, 0.6, 0) in the normalized rotating frame of the Sun-Earth
system (𝜇 = 3 × 10−6). Calculate the effective potential 𝑈 at this point.
Solution: The effective potential is given by:
𝑈 = − 1 − 𝜇
𝑟1−𝜇
𝑟2−1
2(𝑥2+ 𝑦2)
where 𝑟1=√(𝑥 + 𝜇)2+ 𝑦2+ 𝑧2 and 𝑟2=√(𝑥 − 1 + 𝜇)2+ 𝑦2+ 𝑧2.
Calculating 𝑟1 and 𝑟2:
𝑟1=√(0.8 + 3 × 10−6)2+ 0.62+ 02= 1.0
𝑟2=√(0.8 − 1 + 3 × 10−6)2+ 0.62+ 02= 0.28284
Substituting into the equation for 𝑈:
𝑈 = − 0.999997
1.0 −3 × 10−6
0.28284 −1
2(0.82+ 0.62)= −1.4999
97. Consider a restricted three-body system with two primary bodies of masses 𝑀1=
1.98 ×1030 kg (Sun) and 𝑀2= 5.97 ×1024 kg (Earth). Calculate the mass parameter 𝜇
for this system.
Solution: The mass parameter 𝜇 is defined as:
𝜇 = 𝑀2
𝑀1+ 𝑀2
Substituting the given values:
𝜇 = 5.97 ×1024
1.98 ×1030 + 5.97 ×1024 = 3.0036 × 10−6
98. In the Earth-Moon system, the Earth’s mass is 5.97 ×1024 kg, and the Moon’s mass is
7.34 ×1022 kg. The distance between their centers is 384,400 km. Calculate the position
of the barycenter of the system relative to the Earth’s center.
Solution: The distance of the barycenter from the Earth’s center, 𝑟, is given by:
𝑟 = 𝑀𝑀𝑜𝑜𝑛
𝑀𝐸𝑎𝑟𝑡ℎ + 𝑀𝑀𝑜𝑜𝑛 ×Earth-Moon distance
𝑟 = 7.34 ×1022
5.97 ×1024 + 7.34 ×1022 ×384,400 = 4,671 km
99. For the Sun-Jupiter system, 𝜇 = 9.537 ×10−4. Calculate the positions of the L1, L2, and
L3 Lagrange points along the Sun-Jupiter line. Express your answers in terms of the
Sun-Jupiter distance 𝑅.
Solution: For small 𝜇, we can approximate the positions as:
𝐿1: 𝑥 ≈ 𝑅(1 − (𝜇/3)1/3)= 0.9326𝑅
𝐿2: 𝑥 ≈ 𝑅(1 + (𝜇/3)1/3)= 1.0686𝑅
𝐿3: 𝑥 ≈ −𝑅(1 + (5𝜇/12))= −1.0004𝑅
100. A spacecraft is located at the L4 point of the Earth-Moon system. If the Earth-Moon
distance is 384,400 km, what is the distance of the spacecraft from Earth?
Solution: The L4 point forms an equilateral triangle with the two primary bodies.
Therefore, its distance from Earth is equal to the Earth-Moon distance:
𝑑 = 384,400 km
101. In the circular restricted three-body problem, two bodies with masses 𝑀1= 2 × 1030 kg
and 𝑀2= 6 × 1024 kg orbit their common center of mass. A small satellite orbits in the
same plane. If the orbital period of the two massive bodies is 365 days, what is their
angular velocity 𝜔 in rad/s?
Solution: The angular velocity is given by:
𝜔 = 2𝜋
𝑇
where 𝑇 is the orbital period. Converting 365 days to seconds:
𝜔 = 2𝜋
365 ×24 ×3600 = 1.991 ×10−7 rad/s
102. For a system with mass parameter 𝜇 = 0.01, calculate the Jacobi constant 𝐶𝐽 for a
particle at rest at the L1 Lagrange point.
Solution: The Jacobi constant at L1 is approximately:
𝐶𝐽≈ 3 + 34/3𝜇2/3 −10𝜇/3
Substituting 𝜇 = 0.01:
𝐶𝐽≈ 3 + 34/3(0.01)2/3 −10(0.01)/3 = 3.0699
103. In the Earth-Moon system (𝜇 = 0.0123), a spacecraft is initially at rest relative to the
rotating frame at position (0.5, 0.5) in normalized coordinates. Calculate its initial
velocity in the inertial frame.
Solution: In the rotating frame, the initial velocity is (0, 0). To convert to the inertial
frame:
𝑣𝑥= −𝜔𝑦 = −(1)(0.5)= −0.5
𝑣𝑦=𝜔𝑥 =(1)(0.5)= 0.5
The initial velocity in the inertial frame is (-0.5, 0.5) in normalized units.
104. A spacecraft is located at (0.8, 0.6, 0) in the normalized rotating frame of the Sun-Earth
system (𝜇 = 3 × 10−6). Calculate the effective potential 𝑈 at this point.
Solution: The effective potential is given by:
𝑈 = − 1 − 𝜇
𝑟1−𝜇
𝑟2−1
2(𝑥2+ 𝑦2)
where 𝑟1=√(𝑥 + 𝜇)2+ 𝑦2+ 𝑧2 and 𝑟2=√(𝑥 − 1 + 𝜇)2+ 𝑦2+ 𝑧2.
Calculating 𝑟1 and 𝑟2:
𝑟1=√(0.8 + 3 × 10−6)2+ 0.62+ 02= 1.0
𝑟2=√(0.8 − 1 + 3 × 10−6)2+ 0.62+ 02= 0.28284
Substituting into the equation for 𝑈:
𝑈 = − 0.999997
1.0 −3 × 10−6
0.28284 −1
2(0.82+ 0.62)= −1.4999
105. Consider a restricted three-body system with two primary bodies of masses 𝑀1=
1.98 ×1030 kg (Sun) and 𝑀2= 5.97 ×1024 kg (Earth). Calculate the mass parameter 𝜇
for this system.
Solution: The mass parameter 𝜇 is defined as:
𝜇 = 𝑀2
𝑀1+ 𝑀2
Substituting the given values:
𝜇 = 5.97 ×1024
1.98 ×1030 + 5.97 ×1024 = 3.0036 × 10−6
106. In the Earth-Moon system, the Earth’s mass is 5.97 ×1024 kg, and the Moon’s mass is
7.34 ×1022 kg. The distance between their centers is 384,400 km. Calculate the position
of the barycenter of the system relative to the Earth’s center.
Solution: The distance of the barycenter from the Earth’s center, 𝑟, is given by:
𝑟 = 𝑀𝑀𝑜𝑜𝑛
𝑀𝐸𝑎𝑟𝑡ℎ + 𝑀𝑀𝑜𝑜𝑛 ×Earth-Moon distance
𝑟 = 7.34 ×1022
5.97 ×1024 + 7.34 ×1022 ×384,400 = 4,671 km
107. For the Sun-Jupiter system, 𝜇 = 9.537 ×10−4. Calculate the positions of the L1, L2, and
L3 Lagrange points along the Sun-Jupiter line. Express your answers in terms of the
Sun-Jupiter distance 𝑅.
Solution: For small 𝜇, we can approximate the positions as:
𝐿1: 𝑥 ≈ 𝑅(1 − (𝜇/3)1/3)= 0.9326𝑅
𝐿2: 𝑥 ≈ 𝑅(1 + (𝜇/3)1/3)= 1.0686𝑅
𝐿3: 𝑥 ≈ −𝑅(1 + (5𝜇/12))= −1.0004𝑅
108. A spacecraft is located at the L4 point of the Earth-Moon system. If the Earth-Moon
distance is 384,400 km, what is the distance of the spacecraft from Earth?
Solution: The L4 point forms an equilateral triangle with the two primary bodies.
Therefore, its distance from Earth is equal to the Earth-Moon distance:
𝑑 = 384,400 km
109. In the circular restricted three-body problem, two bodies with masses 𝑀1= 2 × 1030 kg
and 𝑀2= 6 × 1024 kg orbit their common center of mass. A small satellite orbits in the
same plane. If the orbital period of the two massive bodies is 365 days, what is their
angular velocity 𝜔 in rad/s?
Solution: The angular velocity is given by:
𝜔 = 2𝜋
𝑇
where 𝑇 is the orbital period. Converting 365 days to seconds:
𝜔 = 2𝜋
365 ×24 ×3600 = 1.991 ×10−7 rad/s
110. For a system with mass parameter 𝜇 = 0.01, calculate the Jacobi constant 𝐶𝐽 for a
particle at rest at the L1 Lagrange point.
Solution: The Jacobi constant at L1 is approximately:
𝐶𝐽≈ 3 + 34/3𝜇2/3 −10𝜇/3
Substituting 𝜇 = 0.01:
𝐶𝐽≈ 3 + 34/3(0.01)2/3 −10(0.01)/3 = 3.0699
111. In the Earth-Moon system (𝜇 = 0.0123), a spacecraft is initially at rest relative to the
rotating frame at position (0.5, 0.5) in normalized coordinates. Calculate its initial
velocity in the inertial frame.
Solution: In the rotating frame, the initial velocity is (0, 0). To convert to the inertial
frame:
𝑣𝑥= −𝜔𝑦 = −(1)(0.5)= −0.5
𝑣𝑦=𝜔𝑥 =(1)(0.5)= 0.5
The initial velocity in the inertial frame is (-0.5, 0.5) in normalized units.
112. A spacecraft is located at (0.8, 0.6, 0) in the normalized rotating frame of the Sun-Earth
system (𝜇 = 3 × 10−6). Calculate the effective potential 𝑈 at this point.
Solution: The effective potential is given by:
𝑈 = − 1 − 𝜇
𝑟1−𝜇
𝑟2−1
2(𝑥2+ 𝑦2)
where 𝑟1=√(𝑥 + 𝜇)2+ 𝑦2+ 𝑧2 and 𝑟2=√(𝑥 − 1 + 𝜇)2+ 𝑦2+ 𝑧2.
Calculating 𝑟1 and 𝑟2:
𝑟1=√(0.8 + 3 × 10−6)2+ 0.62+ 02= 1.0
𝑟2=√(0.8 − 1 + 3 × 10−6)2+ 0.62+ 02= 0.28284
Substituting into the equation for 𝑈:
𝑈 = − 0.999997
1.0 −3 × 10−6
0.28284 −1
2(0.82+ 0.62)= −1.4999
113. Consider a restricted three-body system with two primary bodies of masses 𝑀1=
1.98 ×1030 kg (Sun) and 𝑀2= 5.97 ×1024 kg (Earth). Calculate the mass parameter 𝜇
for this system.
Solution: The mass parameter 𝜇 is defined as:
𝜇 = 𝑀2
𝑀1+ 𝑀2
Substituting the given values:
𝜇 = 5.97 ×1024
1.98 ×1030 + 5.97 ×1024 = 3.0036 × 10−6
114. In the Earth-Moon system, the Earth’s mass is 5.97 ×1024 kg, and the Moon’s mass is
7.34 ×1022 kg. The distance between their centers is 384,400 km. Calculate the position
of the barycenter of the system relative to the Earth’s center.
Solution: The distance of the barycenter from the Earth’s center, 𝑟, is given by:
𝑟 = 𝑀𝑀𝑜𝑜𝑛
𝑀𝐸𝑎𝑟𝑡ℎ + 𝑀𝑀𝑜𝑜𝑛 ×Earth-Moon distance
𝑟 = 7.34 ×1022
5.97 ×1024 + 7.34 ×1022 ×384,400 = 4,671 km
115. For the Sun-Jupiter system, 𝜇 = 9.537 ×10−4. Calculate the positions of the L1, L2, and
L3 Lagrange points along the Sun-Jupiter line. Express your answers in terms of the
Sun-Jupiter distance 𝑅.
Solution: For small 𝜇, we can approximate the positions as:
𝐿1: 𝑥 ≈ 𝑅(1 − (𝜇/3)1/3)= 0.9326𝑅
𝐿2: 𝑥 ≈ 𝑅(1 + (𝜇/3)1/3)= 1.0686𝑅
𝐿3: 𝑥 ≈ −𝑅(1 + (5𝜇/12))= −1.0004𝑅
116. A spacecraft is located at the L4 point of the Earth-Moon system. If the Earth-Moon
distance is 384,400 km, what is the distance of the spacecraft from Earth?
Solution: The L4 point forms an equilateral triangle with the two primary bodies.
Therefore, its distance from Earth is equal to the Earth-Moon distance:
𝑑 = 384,400 km
117. In the circular restricted three-body problem, two bodies with masses 𝑀1= 2 × 1030 kg
and 𝑀2= 6 × 1024 kg orbit their common center of mass. A small satellite orbits in the
same plane. If the orbital period of the two massive bodies is 365 days, what is their
angular velocity 𝜔 in rad/s?
Solution: The angular velocity is given by:
𝜔 = 2𝜋
𝑇
where 𝑇 is the orbital period. Converting 365 days to seconds:
𝜔 = 2𝜋
365 ×24 ×3600 = 1.991 ×10−7 rad/s
118. For a system with mass parameter 𝜇 = 0.01, calculate the Jacobi constant 𝐶𝐽 for a
particle at rest at the L1 Lagrange point.
Solution: The Jacobi constant at L1 is approximately:
𝐶𝐽≈ 3 + 34/3𝜇2/3 −10𝜇/3
Substituting 𝜇 = 0.01:
𝐶𝐽≈ 3 + 34/3(0.01)2/3 −10(0.01)/3 = 3.0699
119. In the Earth-Moon system (𝜇 = 0.0123), a spacecraft is initially at rest relative to the
rotating frame at position (0.5, 0.5) in normalized coordinates. Calculate its initial
velocity in the inertial frame.
Solution: In the rotating frame, the initial velocity is (0, 0). To convert to the inertial
frame:
𝑣𝑥= −𝜔𝑦 = −(1)(0.5)= −0.5
𝑣𝑦=𝜔𝑥 =(1)(0.5)= 0.5
The initial velocity in the inertial frame is (-0.5, 0.5) in normalized units.
120. A spacecraft is located at (0.8, 0.6, 0) in the normalized rotating frame of the Sun-Earth
system (𝜇 = 3 × 10−6). Calculate the effective potential 𝑈 at this point.
Solution: The effective potential is given by:
𝑈 = − 1 − 𝜇
𝑟1−𝜇
𝑟2−1
2(𝑥2+ 𝑦2)
where 𝑟1=√(𝑥 + 𝜇)2+ 𝑦2+ 𝑧2 and 𝑟2=√(𝑥 − 1 + 𝜇)2+ 𝑦2+ 𝑧2.
Calculating 𝑟1 and 𝑟2:
𝑟1=√(0.8 + 3 × 10−6)2+ 0.62+ 02= 1.0
𝑟2=√(0.8 − 1 + 3 × 10−6)2+ 0.62+ 02= 0.28284
Substituting into the equation for 𝑈:
𝑈 = − 0.999997
1.0 −3 × 10−6
0.28284 −1
2(0.82+ 0.62)= −1.4999
121. Consider a restricted three-body system with two primary bodies of masses 𝑀1=
1.98 ×1030 kg (Sun) and 𝑀2= 5.97 ×1024 kg (Earth). Calculate the mass parameter 𝜇
for this system.
Solution: The mass parameter 𝜇 is defined as:
𝜇 = 𝑀2
𝑀1+ 𝑀2
Substituting the given values:
𝜇 = 5.97 ×1024
1.98 ×1030 + 5.97 ×1024 = 3.0036 × 10−6
122. In the Earth-Moon system, the Earth’s mass is 5.97 ×1024 kg, and the Moon’s mass is
7.34 ×1022 kg. The distance between their centers is 384,400 km. Calculate the position
of the barycenter of the system relative to the Earth’s center.
Solution: The distance of the barycenter from the Earth’s center, 𝑟, is given by:
𝑟 = 𝑀𝑀𝑜𝑜𝑛
𝑀𝐸𝑎𝑟𝑡ℎ + 𝑀𝑀𝑜𝑜𝑛 ×Earth-Moon distance
𝑟 = 7.34 ×1022
5.97 ×1024 + 7.34 ×1022 ×384,400 = 4,671 km
123. For the Sun-Jupiter system, 𝜇 = 9.537 ×10−4. Calculate the positions of the L1, L2, and
L3 Lagrange points along the Sun-Jupiter line. Express your answers in terms of the
Sun-Jupiter distance 𝑅.
Solution: For small 𝜇, we can approximate the positions as:
𝐿1: 𝑥 ≈ 𝑅(1 − (𝜇/3)1/3)= 0.9326𝑅
𝐿2: 𝑥 ≈ 𝑅(1 + (𝜇/3)1/3)= 1.0686𝑅
𝐿3: 𝑥 ≈ −𝑅(1 + (5𝜇/12))= −1.0004𝑅
124. A spacecraft is located at the L4 point of the Earth-Moon system. If the Earth-Moon
distance is 384,400 km, what is the distance of the spacecraft from Earth?
Solution: The L4 point forms an equilateral triangle with the two primary bodies.
Therefore, its distance from Earth is equal to the Earth-Moon distance:
𝑑 = 384,400 km
125. In the circular restricted three-body problem, two bodies with masses 𝑀1= 2 × 1030 kg
and 𝑀2= 6 × 1024 kg orbit their common center of mass. A small satellite orbits in the
same plane. If the orbital period of the two massive bodies is 365 days, what is their
angular velocity 𝜔 in rad/s?
Solution: The angular velocity is given by:
𝜔 = 2𝜋
𝑇
where 𝑇 is the orbital period. Converting 365 days to seconds:
𝜔 = 2𝜋
365 ×24 ×3600 = 1.991 ×10−7 rad/s
126. For a system with mass parameter 𝜇 = 0.01, calculate the Jacobi constant 𝐶𝐽 for a
particle at rest at the L1 Lagrange point.
Solution: The Jacobi constant at L1 is approximately:
𝐶𝐽≈ 3 + 34/3𝜇2/3 −10𝜇/3
Substituting 𝜇 = 0.01:
𝐶𝐽≈ 3 + 34/3(0.01)2/3 −10(0.01)/3 = 3.0699
127. In the Earth-Moon system (𝜇 = 0.0123), a spacecraft is initially at rest relative to the
rotating frame at position (0.5, 0.5) in normalized coordinates. Calculate its initial
velocity in the inertial frame.
Solution: In the rotating frame, the initial velocity is (0, 0). To convert to the inertial
frame:
𝑣𝑥= −𝜔𝑦 = −(1)(0.5)= −0.5
𝑣𝑦=𝜔𝑥 =(1)(0.5)= 0.5
The initial velocity in the inertial frame is (-0.5, 0.5) in normalized units.
128. A spacecraft is located at (0.8, 0.6, 0) in the normalized rotating frame of the Sun-Earth
system (𝜇 = 3 × 10−6). Calculate the effective potential 𝑈 at this point.
Solution: The effective potential is given by:
𝑈 = − 1 − 𝜇
𝑟1−𝜇
𝑟2−1
2(𝑥2+ 𝑦2)
where 𝑟1=√(𝑥 + 𝜇)2+ 𝑦2+ 𝑧2 and 𝑟2=√(𝑥 − 1 + 𝜇)2+ 𝑦2+ 𝑧2.
Calculating 𝑟1 and 𝑟2:
𝑟1=√(0.8 + 3 × 10−6)2+ 0.62+ 02= 1.0
𝑟2=√(0.8 − 1 + 3 × 10−6)2+ 0.62+ 02= 0.28284
Substituting into the equation for 𝑈:
𝑈 = − 0.999997
1.0 −3 × 10−6
0.28284 −1
2(0.82+ 0.62)= −1.4999
129. Consider a restricted three-body system with two primary bodies of masses 𝑀1=
1.98 ×1030 kg (Sun) and 𝑀2= 5.97 ×1024 kg (Earth). Calculate the mass parameter 𝜇
for this system.
Solution: The mass parameter 𝜇 is defined as:
𝜇 = 𝑀2
𝑀1+ 𝑀2
Substituting the given values:
𝜇 = 5.97 ×1024
1.98 ×1030 + 5.97 ×1024 = 3.0036 × 10−6
130. In the Earth-Moon system, the Earth’s mass is 5.97 ×1024 kg, and the Moon’s mass is
7.34 ×1022 kg. The distance between their centers is 384,400 km. Calculate the position
of the barycenter of the system relative to the Earth’s center.
Solution: The distance of the barycenter from the Earth’s center, 𝑟, is given by:
𝑟 = 𝑀𝑀𝑜𝑜𝑛
𝑀𝐸𝑎𝑟𝑡ℎ + 𝑀𝑀𝑜𝑜𝑛 ×Earth-Moon distance
𝑟 = 7.34 ×1022
5.97 ×1024 + 7.34 ×1022 ×384,400 = 4,671 km
131. For the Sun-Jupiter system, 𝜇 = 9.537 ×10−4. Calculate the positions of the L1, L2, and
L3 Lagrange points along the Sun-Jupiter line. Express your answers in terms of the
Sun-Jupiter distance 𝑅.
Solution: For small 𝜇, we can approximate the positions as:
𝐿1: 𝑥 ≈ 𝑅(1 − (𝜇/3)1/3)= 0.9326𝑅
𝐿2: 𝑥 ≈ 𝑅(1 + (𝜇/3)1/3)= 1.0686𝑅
𝐿3: 𝑥 ≈ −𝑅(1 + (5𝜇/12))= −1.0004𝑅
132. A spacecraft is located at the L4 point of the Earth-Moon system. If the Earth-Moon
distance is 384,400 km, what is the distance of the spacecraft from Earth?
Solution: The L4 point forms an equilateral triangle with the two primary bodies.
Therefore, its distance from Earth is equal to the Earth-Moon distance:
𝑑 = 384,400 km
133. In the circular restricted three-body problem, two bodies with masses 𝑀1= 2 × 1030 kg
and 𝑀2= 6 × 1024 kg orbit their common center of mass. A small satellite orbits in the
same plane. If the orbital period of the two massive bodies is 365 days, what is their
angular velocity 𝜔 in rad/s?
Solution: The angular velocity is given by:
𝜔 = 2𝜋
𝑇
where 𝑇 is the orbital period. Converting 365 days to seconds:
𝜔 = 2𝜋
365 ×24 ×3600 = 1.991 ×10−7 rad/s
134. For a system with mass parameter 𝜇 = 0.01, calculate the Jacobi constant 𝐶𝐽 for a
particle at rest at the L1 Lagrange point.
Solution: The Jacobi constant at L1 is approximately:
𝐶𝐽≈ 3 + 34/3𝜇2/3 −10𝜇/3
Substituting 𝜇 = 0.01:
𝐶𝐽≈ 3 + 34/3(0.01)2/3 −10(0.01)/3 = 3.0699
135. In the Earth-Moon system (𝜇 = 0.0123), a spacecraft is initially at rest relative to the
rotating frame at position (0.5, 0.5) in normalized coordinates. Calculate its initial
velocity in the inertial frame.
Solution: In the rotating frame, the initial velocity is (0, 0). To convert to the inertial
frame:
𝑣𝑥= −𝜔𝑦 = −(1)(0.5)= −0.5
𝑣𝑦=𝜔𝑥 =(1)(0.5)= 0.5
The initial velocity in the inertial frame is (-0.5, 0.5) in normalized units.
136. A spacecraft is located at (0.8, 0.6, 0) in the normalized rotating frame of the Sun-Earth
system (𝜇 = 3 × 10−6). Calculate the effective potential 𝑈 at this point.
Solution: The effective potential is given by:
𝑈 = − 1 − 𝜇
𝑟1−𝜇
𝑟2−1
2(𝑥2+ 𝑦2)
where 𝑟1=√(𝑥 + 𝜇)2+ 𝑦2+ 𝑧2 and 𝑟2=√(𝑥 − 1 + 𝜇)2+ 𝑦2+ 𝑧2.
Calculating 𝑟1 and 𝑟2:
𝑟1=√(0.8 + 3 × 10−6)2+ 0.62+ 02= 1.0
𝑟2=√(0.8 − 1 + 3 × 10−6)2+ 0.62+ 02= 0.28284
Substituting into the equation for 𝑈:
𝑈 = − 0.999997
1.0 −3 × 10−6
0.28284 −1
2(0.82+ 0.62)= −1.4999
137. Consider a restricted three-body system with two primary bodies of masses 𝑀1=
1.98 ×1030 kg (Sun) and 𝑀2= 5.97 ×1024 kg (Earth). Calculate the mass parameter 𝜇
for this system.
Solution: The mass parameter 𝜇 is defined as:
𝜇 = 𝑀2
𝑀1+ 𝑀2
Substituting the given values:
𝜇 = 5.97 ×1024
1.98 ×1030 + 5.97 ×1024 = 3.0036 × 10−6
138. In the Earth-Moon system, the Earth’s mass is 5.97 ×1024 kg, and the Moon’s mass is
7.34 ×1022 kg. The distance between their centers is 384,400 km. Calculate the position
of the barycenter of the system relative to the Earth’s center.
Solution: The distance of the barycenter from the Earth’s center, 𝑟, is given by:
𝑟 = 𝑀𝑀𝑜𝑜𝑛
𝑀𝐸𝑎𝑟𝑡ℎ + 𝑀𝑀𝑜𝑜𝑛 ×Earth-Moon distance
𝑟 = 7.34 ×1022
5.97 ×1024 + 7.34 ×1022 ×384,400 = 4,671 km
139. For the Sun-Jupiter system, 𝜇 = 9.537 ×10−4. Calculate the positions of the L1, L2, and
L3 Lagrange points along the Sun-Jupiter line. Express your answers in terms of the
Sun-Jupiter distance 𝑅.
Solution: For small 𝜇, we can approximate the positions as:
𝐿1: 𝑥 ≈ 𝑅(1 − (𝜇/3)1/3)= 0.9326𝑅
𝐿2: 𝑥 ≈ 𝑅(1 + (𝜇/3)1/3)= 1.0686𝑅
𝐿3: 𝑥 ≈ −𝑅(1 + (5𝜇/12))= −1.0004𝑅
140. A spacecraft is located at the L4 point of the Earth-Moon system. If the Earth-Moon
distance is 384,400 km, what is the distance of the spacecraft from Earth?
Solution: The L4 point forms an equilateral triangle with the two primary bodies.
Therefore, its distance from Earth is equal to the Earth-Moon distance:
𝑑 = 384,400 km
141. In the circular restricted three-body problem, two bodies with masses 𝑀1= 2 × 1030 kg
and 𝑀2= 6 × 1024 kg orbit their common center of mass. A small satellite orbits in the
same plane. If the orbital period of the two massive bodies is 365 days, what is their
angular velocity 𝜔 in rad/s?
Solution: The angular velocity is given by:
𝜔 = 2𝜋
𝑇
where 𝑇 is the orbital period. Converting 365 days to seconds:
𝜔 = 2𝜋
365 ×24 ×3600 = 1.991 ×10−7 rad/s
142. For a system with mass parameter 𝜇 = 0.01, calculate the Jacobi constant 𝐶𝐽 for a
particle at rest at the L1 Lagrange point.
Solution: The Jacobi constant at L1 is approximately:
𝐶𝐽≈ 3 + 34/3𝜇2/3 −10𝜇/3
Substituting 𝜇 = 0.01:
𝐶𝐽≈ 3 + 34/3(0.01)2/3 −10(0.01)/3 = 3.0699
143. In the Earth-Moon system (𝜇 = 0.0123), a spacecraft is initially at rest relative to the
rotating frame at position (0.5, 0.5) in normalized coordinates. Calculate its initial
velocity in the inertial frame.
Solution: In the rotating frame, the initial velocity is (0, 0). To convert to the inertial
frame:
𝑣𝑥= −𝜔𝑦 = −(1)(0.5)= −0.5
𝑣𝑦=𝜔𝑥 =(1)(0.5)= 0.5
The initial velocity in the inertial frame is (-0.5, 0.5) in normalized units.
144. A spacecraft is located at (0.8, 0.6, 0) in the normalized rotating frame of the Sun-Earth
system (𝜇 = 3 × 10−6). Calculate the effective potential 𝑈 at this point.
Solution: The effective potential is given by:
𝑈 = − 1 − 𝜇
𝑟1−𝜇
𝑟2−1
2(𝑥2+ 𝑦2)
where 𝑟1=√(𝑥 + 𝜇)2+ 𝑦2+ 𝑧2 and 𝑟2=√(𝑥 − 1 + 𝜇)2+ 𝑦2+ 𝑧2.
Calculating 𝑟1 and 𝑟2:
𝑟1=√(0.8 + 3 × 10−6)2+ 0.62+ 02= 1.0
𝑟2=√(0.8 − 1 + 3 × 10−6)2+ 0.62+ 02= 0.28284
Substituting into the equation for 𝑈:
𝑈 = − 0.999997
1.0 −3 × 10−6
0.28284 −1
2(0.82+ 0.62)= −1.4999
145. Consider a restricted three-body system with two primary bodies of masses 𝑀1=
1.98 ×1030 kg (Sun) and 𝑀2= 5.97 ×1024 kg (Earth). Calculate the mass parameter 𝜇
for this system.
Solution: The mass parameter 𝜇 is defined as:
𝜇 = 𝑀2
𝑀1+ 𝑀2
Substituting the given values:
𝜇 = 5.97 ×1024
1.98 ×1030 + 5.97 ×1024 = 3.0036 × 10−6
146. In the Earth-Moon system, the Earth’s mass is 5.97 ×1024 kg, and the Moon’s mass is
7.34 ×1022 kg. The distance between their centers is 384,400 km. Calculate the position
of the barycenter of the system relative to the Earth’s center.
Solution: The distance of the barycenter from the Earth’s center, 𝑟, is given by:
𝑟 = 𝑀𝑀𝑜𝑜𝑛
𝑀𝐸𝑎𝑟𝑡ℎ + 𝑀𝑀𝑜𝑜𝑛 ×Earth-Moon distance
𝑟 = 7.34 ×1022
5.97 ×1024 + 7.34 ×1022 ×384,400 = 4,671 km
147. For the Sun-Jupiter system, 𝜇 = 9.537 ×10−4. Calculate the positions of the L1, L2, and
L3 Lagrange points along the Sun-Jupiter line. Express your answers in terms of the
Sun-Jupiter distance 𝑅.
Solution: For small 𝜇, we can approximate the positions as:
𝐿1: 𝑥 ≈ 𝑅(1 − (𝜇/3)1/3)= 0.9326𝑅
𝐿2: 𝑥 ≈ 𝑅(1 + (𝜇/3)1/3)= 1.0686𝑅
𝐿3: 𝑥 ≈ −𝑅(1 + (5𝜇/12))= −1.0004𝑅
148. A spacecraft is located at the L4 point of the Earth-Moon system. If the Earth-Moon
distance is 384,400 km, what is the distance of the spacecraft from Earth?
Solution: The L4 point forms an equilateral triangle with the two primary bodies.
Therefore, its distance from Earth is equal to the Earth-Moon distance:
𝑑 = 384,400 km
149. In the circular restricted three-body problem, two bodies with masses 𝑀1= 2 × 1030 kg
and 𝑀2= 6 × 1024 kg orbit their common center of mass. A small satellite orbits in the
same plane. If the orbital period of the two massive bodies is 365 days, what is their
angular velocity 𝜔 in rad/s?
Solution: The angular velocity is given by:
𝜔 = 2𝜋
𝑇
where 𝑇 is the orbital period. Converting 365 days to seconds:
𝜔 = 2𝜋
365 ×24 ×3600 = 1.991 ×10−7 rad/s
150. For a system with mass parameter 𝜇 = 0.01, calculate the Jacobi constant 𝐶𝐽 for a
particle at rest at the L1 Lagrange point.
Solution: The Jacobi constant at L1 is approximately:
𝐶𝐽≈ 3 + 34/3𝜇2/3 −10𝜇/3
Substituting 𝜇 = 0.01:
𝐶𝐽≈ 3 + 34/3(0.01)2/3 −10(0.01)/3 = 3.0699
151. In the Earth-Moon system (𝜇 = 0.0123), a spacecraft is initially at rest relative to the
rotating frame at position (0.5, 0.5) in normalized coordinates. Calculate its initial
velocity in the inertial frame.
Solution: In the rotating frame, the initial velocity is (0, 0). To convert to the inertial
frame:
𝑣𝑥= −𝜔𝑦 = −(1)(0.5)= −0.5
𝑣𝑦=𝜔𝑥 =(1)(0.5)= 0.5
The initial velocity in the inertial frame is (-0.5, 0.5) in normalized units.
152. A spacecraft is located at (0.8, 0.6, 0) in the normalized rotating frame of the Sun-Earth
system (𝜇 = 3 × 10−6). Calculate the effective potential 𝑈 at this point.
Solution: The effective potential is given by:
𝑈 = − 1 − 𝜇
𝑟1−𝜇
𝑟2−1
2(𝑥2+ 𝑦2)
where 𝑟1=√(𝑥 + 𝜇)2+ 𝑦2+ 𝑧2 and 𝑟2=√(𝑥 − 1 + 𝜇)2+ 𝑦2+ 𝑧2.
Calculating 𝑟1 and 𝑟2:
𝑟1=√(0.8 + 3 × 10−6)2+ 0.62+ 02= 1.0
𝑟2=√(0.8 − 1 + 3 × 10−6)2+ 0.62+ 02= 0.28284
Substituting into the equation for 𝑈:
𝑈 = − 0.999997
1.0 −3 × 10−6
0.28284 −1
2(0.82+ 0.62)= −1.4999
153. Consider a restricted three-body system with two primary bodies of masses 𝑀1=
1.98 ×1030 kg (Sun) and 𝑀2= 5.97 ×1024 kg (Earth). Calculate the mass parameter 𝜇
for this system.
Solution: The mass parameter 𝜇 is defined as:
𝜇 = 𝑀2
𝑀1+ 𝑀2
Substituting the given values:
𝜇 = 5.97 ×1024
1.98 ×1030 + 5.97 ×1024 = 3.0036 × 10−6
154. In the Earth-Moon system, the Earth’s mass is 5.97 ×1024 kg, and the Moon’s mass is
7.34 ×1022 kg. The distance between their centers is 384,400 km. Calculate the position
of the barycenter of the system relative to the Earth’s center.
Solution: The distance of the barycenter from the Earth’s center, 𝑟, is given by:
𝑟 = 𝑀𝑀𝑜𝑜𝑛
𝑀𝐸𝑎𝑟𝑡ℎ + 𝑀𝑀𝑜𝑜𝑛 ×Earth-Moon distance
𝑟 = 7.34 ×1022
5.97 ×1024 + 7.34 ×1022 ×384,400 = 4,671 km
155. For the Sun-Jupiter system, 𝜇 = 9.537 ×10−4. Calculate the positions of the L1, L2, and
L3 Lagrange points along the Sun-Jupiter line. Express your answers in terms of the
Sun-Jupiter distance 𝑅.
Solution: For small 𝜇, we can approximate the positions as:
𝐿1: 𝑥 ≈ 𝑅(1 − (𝜇/3)1/3)= 0.9326𝑅
𝐿2: 𝑥 ≈ 𝑅(1 + (𝜇/3)1/3)= 1.0686𝑅
𝐿3: 𝑥 ≈ −𝑅(1 + (5𝜇/12))= −1.0004𝑅
156. A spacecraft is located at the L4 point of the Earth-Moon system. If the Earth-Moon
distance is 384,400 km, what is the distance of the spacecraft from Earth?
Solution: The L4 point forms an equilateral triangle with the two primary bodies.
Therefore, its distance from Earth is equal to the Earth-Moon distance:
𝑑 = 384,400 km
157. In the circular restricted three-body problem, two bodies with masses 𝑀1= 2 × 1030 kg
and 𝑀2= 6 × 1024 kg orbit their common center of mass. A small satellite orbits in the
same plane. If the orbital period of the two massive bodies is 365 days, what is their
angular velocity 𝜔 in rad/s?
Solution: The angular velocity is given by:
𝜔 = 2𝜋
𝑇
where 𝑇 is the orbital period. Converting 365 days to seconds:
𝜔 = 2𝜋
365 ×24 ×3600 = 1.991 ×10−7 rad/s
158. For a system with mass parameter 𝜇 = 0.01, calculate the Jacobi constant 𝐶𝐽 for a
particle at rest at the L1 Lagrange point.
Solution: The Jacobi constant at L1 is approximately:
𝐶𝐽≈ 3 + 34/3𝜇2/3 −10𝜇/3
Substituting 𝜇 = 0.01:
𝐶𝐽≈ 3 + 34/3(0.01)2/3 −10(0.01)/3 = 3.0699
159. In the Earth-Moon system (𝜇 = 0.0123), a spacecraft is initially at rest relative to the
rotating frame at position (0.5, 0.5) in normalized coordinates. Calculate its initial
velocity in the inertial frame.
Solution: In the rotating frame, the initial velocity is (0, 0). To convert to the inertial
frame:
𝑣𝑥= −𝜔𝑦 = −(1)(0.5)= −0.5
𝑣𝑦=𝜔𝑥 =(1)(0.5)= 0.5
The initial velocity in the inertial frame is (-0.5, 0.5) in normalized units.
160. A spacecraft is located at (0.8, 0.6, 0) in the normalized rotating frame of the Sun-Earth
system (𝜇 = 3 × 10−6). Calculate the effective potential 𝑈 at this point.
Solution: The effective potential is given by:
𝑈 = − 1 − 𝜇
𝑟1−𝜇
𝑟2−1
2(𝑥2+ 𝑦2)
where 𝑟1=√(𝑥 + 𝜇)2+ 𝑦2+ 𝑧2 and 𝑟2=√(𝑥 − 1 + 𝜇)2+ 𝑦2+ 𝑧2.
Calculating 𝑟1 and 𝑟2:
𝑟1=√(0.8 + 3 × 10−6)2+ 0.62+ 02= 1.0
𝑟2=√(0.8 − 1 + 3 × 10−6)2+ 0.62+ 02= 0.28284
Substituting into the equation for 𝑈:
𝑈 = − 0.999997
1.0 −3 × 10−6
0.28284 −1
2(0.82+ 0.62)= −1.4999
161. Consider a restricted three-body system with two primary bodies of masses 𝑀1=
1.98 ×1030 kg (Sun) and 𝑀2= 5.97 ×1024 kg (Earth). Calculate the mass parameter 𝜇
for this system.
Solution: The mass parameter 𝜇 is defined as:
𝜇 = 𝑀2
𝑀1+ 𝑀2
Substituting the given values:
𝜇 = 5.97 ×1024
1.98 ×1030 + 5.97 ×1024 = 3.0036 × 10−6
162. In the Earth-Moon system, the Earth’s mass is 5.97 ×1024 kg, and the Moon’s mass is
7.34 ×1022 kg. The distance between their centers is 384,400 km. Calculate the position
of the barycenter of the system relative to the Earth’s center.
Solution: The distance of the barycenter from the Earth’s center, 𝑟, is given by:
𝑟 = 𝑀𝑀𝑜𝑜𝑛
𝑀𝐸𝑎𝑟𝑡ℎ + 𝑀𝑀𝑜𝑜𝑛 ×Earth-Moon distance
𝑟 = 7.34 ×1022
5.97 ×1024 + 7.34 ×1022 ×384,400 = 4,671 km
163. For the Sun-Jupiter system, 𝜇 = 9.537 ×10−4. Calculate the positions of the L1, L2, and
L3 Lagrange points along the Sun-Jupiter line. Express your answers in terms of the
Sun-Jupiter distance 𝑅.
Solution: For small 𝜇, we can approximate the positions as:
𝐿1: 𝑥 ≈ 𝑅(1 − (𝜇/3)1/3)= 0.9326𝑅
𝐿2: 𝑥 ≈ 𝑅(1 + (𝜇/3)1/3)= 1.0686𝑅
𝐿3: 𝑥 ≈ −𝑅(1 + (5𝜇/12))= −1.0004𝑅
164. A spacecraft is located at the L4 point of the Earth-Moon system. If the Earth-Moon
distance is 384,400 km, what is the distance of the spacecraft from Earth?
Solution: The L4 point forms an equilateral triangle with the two primary bodies.
Therefore, its distance from Earth is equal to the Earth-Moon distance:
𝑑 = 384,400 km
165. In the circular restricted three-body problem, two bodies with masses 𝑀1= 2 × 1030 kg
and 𝑀2= 6 × 1024 kg orbit their common center of mass. A small satellite orbits in the
same plane. If the orbital period of the two massive bodies is 365 days, what is their
angular velocity 𝜔 in rad/s?
Solution: The angular velocity is given by:
𝜔 = 2𝜋
𝑇
where 𝑇 is the orbital period. Converting 365 days to seconds:
𝜔 = 2𝜋
365 ×24 ×3600 = 1.991 ×10−7 rad/s
166. For a system with mass parameter 𝜇 = 0.01, calculate the Jacobi constant 𝐶𝐽 for a
particle at rest at the L1 Lagrange point.
Solution: The Jacobi constant at L1 is approximately:
𝐶𝐽≈ 3 + 34/3𝜇2/3 −10𝜇/3
Substituting 𝜇 = 0.01:
𝐶𝐽≈ 3 + 34/3(0.01)2/3 −10(0.01)/3 = 3.0699
167. In the Earth-Moon system (𝜇 = 0.0123), a spacecraft is initially at rest relative to the
rotating frame at position (0.5, 0.5) in normalized coordinates. Calculate its initial
velocity in the inertial frame.
Solution: In the rotating frame, the initial velocity is (0, 0). To convert to the inertial
frame:
𝑣𝑥= −𝜔𝑦 = −(1)(0.5)= −0.5
𝑣𝑦=𝜔𝑥 =(1)(0.5)= 0.5
The initial velocity in the inertial frame is (-0.5, 0.5) in normalized units.
168. A spacecraft is located at (0.8, 0.6, 0) in the normalized rotating frame of the Sun-Earth
system (𝜇 = 3 × 10−6). Calculate the effective potential 𝑈 at this point.
Solution: The effective potential is given by:
𝑈 = − 1 − 𝜇
𝑟1−𝜇
𝑟2−1
2(𝑥2+ 𝑦2)
where 𝑟1=√(𝑥 + 𝜇)2+ 𝑦2+ 𝑧2 and 𝑟2=√(𝑥 − 1 + 𝜇)2+ 𝑦2+ 𝑧2.
Calculating 𝑟1 and 𝑟2:
𝑟1=√(0.8 + 3 × 10−6)2+ 0.62+ 02= 1.0
𝑟2=√(0.8 − 1 + 3 × 10−6)2+ 0.62+ 02= 0.28284
Substituting into the equation for 𝑈:
𝑈 = − 0.999997
1.0 −3 × 10−6
0.28284 −1
2(0.82+ 0.62)= −1.4999
169. Consider a restricted three-body system with two primary bodies of masses 𝑀1=
1.98 ×1030 kg (Sun) and 𝑀2= 5.97 ×1024 kg (Earth). Calculate the mass parameter 𝜇
for this system.
Solution: The mass parameter 𝜇 is defined as:
𝜇 = 𝑀2
𝑀1+ 𝑀2
Substituting the given values:
𝜇 = 5.97 ×1024
1.98 ×1030 + 5.97 ×1024 = 3.0036 × 10−6
170. In the Earth-Moon system, the Earth’s mass is 5.97 ×1024 kg, and the Moon’s mass is
7.34 ×1022 kg. The distance between their centers is 384,400 km. Calculate the position
of the barycenter of the system relative to the Earth’s center.
Solution: The distance of the barycenter from the Earth’s center, 𝑟, is given by:
𝑟 = 𝑀𝑀𝑜𝑜𝑛
𝑀𝐸𝑎𝑟𝑡ℎ + 𝑀𝑀𝑜𝑜𝑛 ×Earth-Moon distance
𝑟 = 7.34 ×1022
5.97 ×1024 + 7.34 ×1022 ×384,400 = 4,671 km
171. For the Sun-Jupiter system, 𝜇 = 9.537 ×10−4. Calculate the positions of the L1, L2, and
L3 Lagrange points along the Sun-Jupiter line. Express your answers in terms of the
Sun-Jupiter distance 𝑅.
Solution: For small 𝜇, we can approximate the positions as:
𝐿1: 𝑥 ≈ 𝑅(1 − (𝜇/3)1/3)= 0.9326𝑅
𝐿2: 𝑥 ≈ 𝑅(1 + (𝜇/3)1/3)= 1.0686𝑅
𝐿3: 𝑥 ≈ −𝑅(1 + (5𝜇/12))= −1.0004𝑅
172. A spacecraft is located at the L4 point of the Earth-Moon system. If the Earth-Moon
distance is 384,400 km, what is the distance of the spacecraft from Earth?
Solution: The L4 point forms an equilateral triangle with the two primary bodies.
Therefore, its distance from Earth is equal to the Earth-Moon distance:
𝑑 = 384,400 km
173. In the circular restricted three-body problem, two bodies with masses 𝑀1= 2 × 1030 kg
and 𝑀2= 6 × 1024 kg orbit their common center of mass. A small satellite orbits in the
same plane. If the orbital period of the two massive bodies is 365 days, what is their
angular velocity 𝜔 in rad/s?
Solution: The angular velocity is given by:
𝜔 = 2𝜋
𝑇
where 𝑇 is the orbital period. Converting 365 days to seconds:
𝜔 = 2𝜋
365 ×24 ×3600 = 1.991 ×10−7 rad/s
174. For a system with mass parameter 𝜇 = 0.01, calculate the Jacobi constant 𝐶𝐽 for a
particle at rest at the L1 Lagrange point.
Solution: The Jacobi constant at L1 is approximately:
𝐶𝐽≈ 3 + 34/3𝜇2/3 −10𝜇/3
Substituting 𝜇 = 0.01:
𝐶𝐽≈ 3 + 34/3(0.01)2/3 −10(0.01)/3 = 3.0699
175. In the Earth-Moon system (𝜇 = 0.0123), a spacecraft is initially at rest relative to the
rotating frame at position (0.5, 0.5) in normalized coordinates. Calculate its initial
velocity in the inertial frame.
Solution: In the rotating frame, the initial velocity is (0, 0). To convert to the inertial
frame:
𝑣𝑥= −𝜔𝑦 = −(1)(0.5)= −0.5
𝑣𝑦=𝜔𝑥 =(1)(0.5)= 0.5
The initial velocity in the inertial frame is (-0.5, 0.5) in normalized units.
176. A spacecraft is located at (0.8, 0.6, 0) in the normalized rotating frame of the Sun-Earth
system (𝜇 = 3 × 10−6). Calculate the effective potential 𝑈 at this point.
Solution: The effective potential is given by:
𝑈 = − 1 − 𝜇
𝑟1−𝜇
𝑟2−1
2(𝑥2+ 𝑦2)
where 𝑟1=√(𝑥 + 𝜇)2+ 𝑦2+ 𝑧2 and 𝑟2=√(𝑥 − 1 + 𝜇)2+ 𝑦2+ 𝑧2.
Calculating 𝑟1 and 𝑟2:
𝑟1=√(0.8 + 3 × 10−6)2+ 0.62+ 02= 1.0
𝑟2=√(0.8 − 1 + 3 × 10−6)2+ 0.62+ 02= 0.28284
Substituting into the equation for 𝑈:
𝑈 = − 0.999997
1.0 −3 × 10−6
0.28284 −1
2(0.82+ 0.62)= −1.4999
177. Consider a restricted three-body system with two primary bodies of masses 𝑀1=
1.98 ×1030 kg (Sun) and 𝑀2= 5.97 ×1024 kg (Earth). Calculate the mass parameter 𝜇
for this system.
Solution: The mass parameter 𝜇 is defined as:
𝜇 = 𝑀2
𝑀1+ 𝑀2
Substituting the given values:
𝜇 = 5.97 ×1024
1.98 ×1030 + 5.97 ×1024 = 3.0036 × 10−6
178. In the Earth-Moon system, the Earth’s mass is 5.97 ×1024 kg, and the Moon’s mass is
7.34 ×1022 kg. The distance between their centers is 384,400 km. Calculate the position
of the barycenter of the system relative to the Earth’s center.
Solution: The distance of the barycenter from the Earth’s center, 𝑟, is given by:
𝑟 = 𝑀𝑀𝑜𝑜𝑛
𝑀𝐸𝑎𝑟𝑡ℎ + 𝑀𝑀𝑜𝑜𝑛 ×Earth-Moon distance
𝑟 = 7.34 ×1022
5.97 ×1024 + 7.34 ×1022 ×384,400 = 4,671 km
179. For the Sun-Jupiter system, 𝜇 = 9.537 ×10−4. Calculate the positions of the L1, L2, and
L3 Lagrange points along the Sun-Jupiter line. Express your answers in terms of the
Sun-Jupiter distance 𝑅.
Solution: For small 𝜇, we can approximate the positions as:
𝐿1: 𝑥 ≈ 𝑅(1 − (𝜇/3)1/3)= 0.9326𝑅
𝐿2: 𝑥 ≈ 𝑅(1 + (𝜇/3)1/3)= 1.0686𝑅
𝐿3: 𝑥 ≈ −𝑅(1 + (5𝜇/12))= −1.0004𝑅
180. A spacecraft is located at the L4 point of the Earth-Moon system. If the Earth-Moon
distance is 384,400 km, what is the distance of the spacecraft from Earth?
Solution: The L4 point forms an equilateral triangle with the two primary bodies.
Therefore, its distance from Earth is equal to the Earth-Moon distance:
𝑑 = 384,400 km
181. In the circular restricted three-body problem, two bodies with masses 𝑀1= 2 × 1030 kg
and 𝑀2= 6 × 1024 kg orbit their common center of mass. A small satellite orbits in the
same plane. If the orbital period of the two massive bodies is 365 days, what is their
angular velocity 𝜔 in rad/s?
Solution: The angular velocity is given by:
𝜔 = 2𝜋
𝑇
where 𝑇 is the orbital period. Converting 365 days to seconds:
𝜔 = 2𝜋
365 ×24 ×3600 = 1.991 ×10−7 rad/s
182. For a system with mass parameter 𝜇 = 0.01, calculate the Jacobi constant 𝐶𝐽 for a
particle at rest at the L1 Lagrange point.
Solution: The Jacobi constant at L1 is approximately:
𝐶𝐽≈ 3 + 34/3𝜇2/3 −10𝜇/3
Substituting 𝜇 = 0.01:
𝐶𝐽≈ 3 + 34/3(0.01)2/3 −10(0.01)/3 = 3.0699
183. In the Earth-Moon system (𝜇 = 0.0123), a spacecraft is initially at rest relative to the
rotating frame at position (0.5, 0.5) in normalized coordinates. Calculate its initial
velocity in the inertial frame.
Solution: In the rotating frame, the initial velocity is (0, 0). To convert to the inertial
frame:
𝑣𝑥= −𝜔𝑦 = −(1)(0.5)= −0.5
𝑣𝑦=𝜔𝑥 =(1)(0.5)= 0.5
The initial velocity in the inertial frame is (-0.5, 0.5) in normalized units.
184. A spacecraft is located at (0.8, 0.6, 0) in the normalized rotating frame of the Sun-Earth
system (𝜇 = 3 × 10−6). Calculate the effective potential 𝑈 at this point.
Solution: The effective potential is given by:
𝑈 = − 1 − 𝜇
𝑟1−𝜇
𝑟2−1
2(𝑥2+ 𝑦2)
where 𝑟1=√(𝑥 + 𝜇)2+ 𝑦2+ 𝑧2 and 𝑟2=√(𝑥 − 1 + 𝜇)2+ 𝑦2+ 𝑧2.
Calculating 𝑟1 and 𝑟2:
𝑟1=√(0.8 + 3 × 10−6)2+ 0.62+ 02= 1.0
𝑟2=√(0.8 − 1 + 3 × 10−6)2+ 0.62+ 02= 0.28284
Substituting into the equation for 𝑈:
𝑈 = − 0.999997
1.0 −3 × 10−6
0.28284 −1
2(0.82+ 0.62)= −1.4999
185. Consider a restricted three-body system with two primary bodies of masses 𝑀1=
1.98 ×1030 kg (Sun) and 𝑀2= 5.97 ×1024 kg (Earth). Calculate the mass parameter 𝜇
for this system.
Solution: The mass parameter 𝜇 is defined as:
𝜇 = 𝑀2
𝑀1+ 𝑀2
Substituting the given values:
𝜇 = 5.97 ×1024
1.98 ×1030 + 5.97 ×1024 = 3.0036 × 10−6
186. In the Earth-Moon system, the Earth’s mass is 5.97 ×1024 kg, and the Moon’s mass is
7.34 ×1022 kg. The distance between their centers is 384,400 km. Calculate the position
of the barycenter of the system relative to the Earth’s center.
Solution: The distance of the barycenter from the Earth’s center, 𝑟, is given by:
𝑟 = 𝑀𝑀𝑜𝑜𝑛
𝑀𝐸𝑎𝑟𝑡ℎ + 𝑀𝑀𝑜𝑜𝑛 ×Earth-Moon distance
𝑟 = 7.34 ×1022
5.97 ×1024 + 7.34 ×1022 ×384,400 = 4,671 km
187. For the Sun-Jupiter system, 𝜇 = 9.537 ×10−4. Calculate the positions of the L1, L2, and
L3 Lagrange points along the Sun-Jupiter line. Express your answers in terms of the
Sun-Jupiter distance 𝑅.
Solution: For small 𝜇, we can approximate the positions as:
𝐿1: 𝑥 ≈ 𝑅(1 − (𝜇/3)1/3)= 0.9326𝑅
𝐿2: 𝑥 ≈ 𝑅(1 + (𝜇/3)1/3)= 1.0686𝑅
𝐿3: 𝑥 ≈ −𝑅(1 + (5𝜇/12))= −1.0004𝑅
188. A spacecraft is located at the L4 point of the Earth-Moon system. If the Earth-Moon
distance is 384,400 km, what is the distance of the spacecraft from Earth?
Solution: The L4 point forms an equilateral triangle with the two primary bodies.
Therefore, its distance from Earth is equal to the Earth-Moon distance:
𝑑 = 384,400 km
189. In the circular restricted three-body problem, two bodies with masses 𝑀1= 2 × 1030 kg
and 𝑀2= 6 × 1024 kg orbit their common center of mass. A small satellite orbits in the
same plane. If the orbital period of the two massive bodies is 365 days, what is their
angular velocity 𝜔 in rad/s?
Solution: The angular velocity is given by:
𝜔 = 2𝜋
𝑇
where 𝑇 is the orbital period. Converting 365 days to seconds:
𝜔 = 2𝜋
365 ×24 ×3600 = 1.991 ×10−7 rad/s
190. For a system with mass parameter 𝜇 = 0.01, calculate the Jacobi constant 𝐶𝐽 for a
particle at rest at the L1 Lagrange point.
Solution: The Jacobi constant at L1 is approximately:
𝐶𝐽≈ 3 + 34/3𝜇2/3 −10𝜇/3
Substituting 𝜇 = 0.01:
𝐶𝐽≈ 3 + 34/3(0.01)2/3 −10(0.01)/3 = 3.0699
191. In the Earth-Moon system (𝜇 = 0.0123), a spacecraft is initially at rest relative to the
rotating frame at position (0.5, 0.5) in normalized coordinates. Calculate its initial
velocity in the inertial frame.
Solution: In the rotating frame, the initial velocity is (0, 0). To convert to the inertial
frame:
𝑣𝑥= −𝜔𝑦 = −(1)(0.5)= −0.5
𝑣𝑦=𝜔𝑥 =(1)(0.5)= 0.5
The initial velocity in the inertial frame is (-0.5, 0.5) in normalized units.
192. A spacecraft is located at (0.8, 0.6, 0) in the normalized rotating frame of the Sun-Earth
system (𝜇 = 3 × 10−6). Calculate the effective potential 𝑈 at this point.
Solution: The effective potential is given by:
𝑈 = − 1 − 𝜇
𝑟1−𝜇
𝑟2−1
2(𝑥2+ 𝑦2)
where 𝑟1=√(𝑥 + 𝜇)2+ 𝑦2+ 𝑧2 and 𝑟2=√(𝑥 − 1 + 𝜇)2+ 𝑦2+ 𝑧2.
Calculating 𝑟1 and 𝑟2:
𝑟1=√(0.8 + 3 × 10−6)2+ 0.62+ 02= 1.0
𝑟2=√(0.8 − 1 + 3 × 10−6)2+ 0.62+ 02= 0.28284
Substituting into the equation for 𝑈:
𝑈 = − 0.999997
1.0 −3 × 10−6
0.28284 −1
2(0.82+ 0.62)= −1.4999
193. Consider a restricted three-body system with two primary bodies of masses 𝑀1=
1.98 ×1030 kg (Sun) and 𝑀2= 5.97 ×1024 kg (Earth). Calculate the mass parameter 𝜇
for this system.
Solution: The mass parameter 𝜇 is defined as:
𝜇 = 𝑀2
𝑀1+ 𝑀2
Substituting the given values:
𝜇 = 5.97 ×1024
1.98 ×1030 + 5.97 ×1024 = 3.0036 × 10−6
194. In the Earth-Moon system, the Earth’s mass is 5.97 ×1024 kg, and the Moon’s mass is
7.34 ×1022 kg. The distance between their centers is 384,400 km. Calculate the position
of the barycenter of the system relative to the Earth’s center.
Solution: The distance of the barycenter from the Earth’s center, 𝑟, is given by:
𝑟 = 𝑀𝑀𝑜𝑜𝑛
𝑀𝐸𝑎𝑟𝑡ℎ + 𝑀𝑀𝑜𝑜𝑛 ×Earth-Moon distance
𝑟 = 7.34 ×1022
5.97 ×1024 + 7.34 ×1022 ×384,400 = 4,671 km
195. For the Sun-Jupiter system, 𝜇 = 9.537 ×10−4. Calculate the positions of the L1, L2, and
L3 Lagrange points along the Sun-Jupiter line. Express your answers in terms of the
Sun-Jupiter distance 𝑅.
Solution: For small 𝜇, we can approximate the positions as:
𝐿1: 𝑥 ≈ 𝑅(1 − (𝜇/3)1/3)= 0.9326𝑅
𝐿2: 𝑥 ≈ 𝑅(1 + (𝜇/3)1/3)= 1.0686𝑅
𝐿3: 𝑥 ≈ −𝑅(1 + (5𝜇/12))= −1.0004𝑅
196. A spacecraft is located at the L4 point of the Earth-Moon system. If the Earth-Moon
distance is 384,400 km, what is the distance of the spacecraft from Earth?
Solution: The L4 point forms an equilateral triangle with the two primary bodies.
Therefore, its distance from Earth is equal to the Earth-Moon distance:
𝑑 = 384,400 km
197. In the circular restricted three-body problem, two bodies with masses 𝑀1= 2 × 1030 kg
and 𝑀2= 6 × 1024 kg orbit their common center of mass. A small satellite orbits in the
same plane. If the orbital period of the two massive bodies is 365 days, what is their
angular velocity 𝜔 in rad/s?
Solution: The angular velocity is given by:
𝜔 = 2𝜋
𝑇
where 𝑇 is the orbital period. Converting 365 days to seconds:
𝜔 = 2𝜋
365 ×24 ×3600 = 1.991 ×10−7 rad/s
198. For a system with mass parameter 𝜇 = 0.01, calculate the Jacobi constant 𝐶𝐽 for a
particle at rest at the L1 Lagrange point.
Solution: The Jacobi constant at L1 is approximately:
𝐶𝐽≈ 3 + 34/3𝜇2/3 −10𝜇/3
Substituting 𝜇 = 0.01:
𝐶𝐽≈ 3 + 34/3(0.01)2/3 −10(0.01)/3 = 3.0699
199. In the Earth-Moon system (𝜇 = 0.0123), a spacecraft is initially at rest relative to the
rotating frame at position (0.5, 0.5) in normalized coordinates. Calculate its initial
velocity in the inertial frame.
Solution: In the rotating frame, the initial velocity is (0, 0). To convert to the inertial
frame:
𝑣𝑥= −𝜔𝑦 = −(1)(0.5)= −0.5
𝑣𝑦=𝜔𝑥 =(1)(0.5)= 0.5
The initial velocity in the inertial frame is (-0.5, 0.5) in normalized units.
200. A spacecraft is located at (0.8, 0.6, 0) in the normalized rotating frame of the Sun-Earth
system (𝜇 = 3 × 10−6). Calculate the effective potential 𝑈 at this point.
Solution: The effective potential is given by:
𝑈 = − 1 − 𝜇
𝑟1−𝜇
𝑟2−1
2(𝑥2+ 𝑦2)
where 𝑟1=√(𝑥 + 𝜇)2+ 𝑦2+ 𝑧2 and 𝑟2=√(𝑥 − 1 + 𝜇)2+ 𝑦2+ 𝑧2.
Calculating 𝑟1 and 𝑟2:
𝑟1=√(0.8 + 3 × 10−6)2+ 0.62+ 02= 1.0
𝑟2=√(0.8 − 1 + 3 × 10−6)2+ 0.62+ 02= 0.28284
Substituting into the equation for 𝑈:
𝑈 = − 0.999997
1.0 −3 × 10−6
0.28284 −1
2(0.82+ 0.62)= −1.4999
