Two small spaceships, each with mass m = 2000 kg, are in the circular Earth orbit of the figure, at an altitude h of 450 km. Igor, the commander of one of the ships, arrives at any fixed point in the orbit 90.0 s ahead of Picard, the commander of the other ship. At point P in the figure, Picard fires an instantaneous burst in the forward direction, reducing his ship's speed by 1.10%. After this burst, he follows the elliptical orbit shown dashed in the figure. How much earlier than Igor will Picard return to P?
To solve this problem, we need to consider the conservation of angular momentum and energy in circular orbits, as well as the effect of the burst on Picard's speed.
1. Find the radius of the circular orbit:
The altitude of the orbit, h, is given as 450 km. The radius of the circular orbit can be calculated by adding the radius of the Earth, RE, to the altitude, h.
RE = 6371 km (average value for the Earth's radius)
Radius of circular orbit = RE + h = 6371 km + 450 km = 6821 km
2. Calculate the initial speed of the spaceships in the circular orbit:
The total energy of an object in a circular orbit is given by the sum of its kinetic energy and gravitational potential energy. The kinetic energy is given by (1/2)mv^2, where m is the mass of the spaceship and v is its speed. The gravitational potential energy is given by (-GmM)/(2r), where G is the gravitational constant, M is the mass of the Earth, and r is the radius of the orbit.
Setting the total energy equal to zero, we can solve for v:
(1/2)mv^2 - (GmM)/(2r) = 0
v^2 = (GM)/r
v = sqrt[(GM)/r], where sqrt denotes square root
Plugging in the values, where G = 6.67430 × 10^-11 m^3 kg^−1 s^−2 and M = 5.972 × 10^24 kg:
v = sqrt[(6.67430 × 10^-11 m^3 kg^−1 s^−2 * 5.972 × 10^24 kg) / (6821 km * 1000 m/km)]
v ≈ 7643.85 m/s
3. Calculate the time period for the circular orbit:
The time period, T, for a circular orbit can be calculated using the formula:
T = (2πr) / v
Plugging in the values:
T = (2π * 6821 km * 1000 m/km) / 7643.85 m/s
T ≈ 5322.14 s
4. Calculate the time difference between Igor's and Picard's arrival at point P:
Given that Igor arrives 90.0 s ahead of Picard, the time difference between their arrivals is 90.0 s.
5. Calculate the change in Picard's speed due to the burst:
Picard's speed is reduced by 1.10% due to the burst. Let's denote his initial speed as v0. After the burst, his speed becomes v' = v0 - (1.10% * v0). Substituting the value for v, we have v' = v - (1.10% * v).
6. Calculate the time period for Picard's elliptical orbit after the burst:
Since the speed is reduced, the orbital radius will also change. Let's denote the new radius as r'. We can calculate the new time period, T', using the formula:
T' = (2πr') / v'
7. Calculate the new radius of the elliptical orbit:
The new radius is given by the formula:
r' = (h + RE) / (1 - Δr / r), where Δr is the change in radius due to the burst, and r is the initial radius of the circular orbit.
Let's calculate the values:
Δr = (1.10% * r) = (1.10% * 6821 km * 1000 m/km) = 75050 m
r' = (450 km + 6371 km * 1000 m/km) / (1 - (75050 m / 6821000 m))
r' ≈ 7162459.9 m
8. Calculate the new time period, T':
T' = (2π * 7162459.9 m) / (7643.85 m/s - (1.10% * 7643.85 m/s))
T' ≈ 5408.02 s
9. Calculate the time difference for Picard to return to point P:
Picard's return time can be calculated by subtracting his new time period, T', from Igor's arrival time difference at point P, which is 90.0 s:
Return time difference = 90.0 s - 5408.02 s
Return time difference ≈ -5318.02 s
Since the time difference is negative, it means that Picard returns to point P approximately 5318.02 seconds before Igor.