Question

2. A space shuttle and a space station are both in the same circular, counter-clockwise orbit around the Earth. The

orbital radius is 20,000 km.
a. What is the orbital velocity (in m/s) and period (in s) for this orbit?
Suppose the space station is 30 degrees ahead of the space shuttle
as shown in the diagram to the right.
b. How many seconds will it take for the space station to move from
its current position to where the shuttle is currently?
(i.e. the 12 o’clock position.)
The shuttle needs to catch up with the space station in order
to dock, so it fires its retro rockets to slow down when it’s in
the position shown in the diagram above. It moves to a new,
elliptical orbit as shown to the right with a dark line. The period
of this new orbit must equal your answer to “b” so that both
spacecraft will rendezvous at the 12 o’clock position as the shuttle
completes its new orbit.
c. What is the semi-major axis “a” of the new orbit in km?
d. How much speed must the space shuttle lose (Δv) to achieve
the new orbit?
e. What must the space shuttle do when it reaches the 12 o’clock
position after one orbit so that it stays in the same orbit as the space
station for docking?

Answer 1

The orbital velocity is 7,907 m/s and the period is 16,862 s for the circular orbit. It will take 2,827 seconds for the space station to move to the current position of the shuttle. The new elliptical orbit has a semi-major axis of 30,000 km. The space shuttle must lose 2,642 m/s of speed to achieve the new orbit. When the shuttle reaches the 12 o'clock position after one orbit, it must fire its rockets to match the velocity of the space station for docking.

To calculate the orbital velocity and period of a circular orbit, we can use the formulas:

Orbital velocity (v) = sqrt(GM/r)

Period (T) = 2πr/v

Where G is the gravitational constant (6.67 x 10^-11 N(m/kg)^2), M is the mass of Earth (5.97 x 10^24 kg), and r is the orbital radius (20,000 km).

By plugging in the values into the formulas, we get:

Orbital velocity = sqrt((6.67 x 10^-11 N(m/kg)^2)(5.97 x 10^24 kg)/(20,000,000 m)) = 7,907 m/s

Period = 2π(20,000,000 m)/(7,907 m/s) = 16,862 s

b. To calculate the time it takes for the space station to move from its current position to where the shuttle is currently, we need to find the angular distance (θ) between the two positions and use the formula:

Time (t) = θ/ω

Where ω is the angular velocity, which is equal to 2π/T.

The angle between the two positions is 30 degrees, which is equal to π/6 radians. The angular velocity is 2π/16,862 s.

By plugging in the values into the formula, we get:

Time = (π/6 radians)/(2π/16,862 s) = 2,827 s

c. The semi-major axis (a) of the new elliptical orbit can be calculated using the formula:

a = r + (Δr/2)

Where r is the original orbital radius (20,000 km) and Δr is the change in orbital radius, which is equal to the distance the space station traveled (20,000 km) minus the distance it needs to travel to reach the 12 o'clock position (0 km).

By plugging in the values into the formula, we get:

a = 20,000 km + (20,000 km/2) = 30,000 km.

d. The change in velocity (Δv) can be calculated using the equation:

Δv = √(2μ/ra - 2μ/2a)

Where μ is the gravitational parameter, which is equal to GM, G is the gravitational constant, and M is the mass of Earth.

By plugging in the values into the equation, we get:

Δv = √((2(6.67 x 10^-11 N(m/kg)^2)(5.97 x 10^24 kg))/(20,000 km) - (2(6.67 x 10^-11 N(m/kg)^2)(5.97 x 10^24 kg))/(2(30,000 km))) = 2,642 m/s.

e. To stay in the same orbit as the space station for docking, the space shuttle needs to match the velocity of the space station when it reaches the 12 o'clock position. This can be done by firing its rockets to increase its velocity to match that of the space station.