[Bonus] Two spaceships, both travelling at $0.9 \mathrm{c}$ relative to Earth, approach Earth from opposite sides. What speed would an observer on one of the spaceships measure for the other spaceship? $1.8 c$ C less than $\mathrm{c}$

Step 1 ：Two spaceships, both travelling at $0.9 \mathrm{c}$ relative to Earth, approach Earth from opposite sides. We are asked to find the speed that an observer on one of the spaceships would measure for the other spaceship.

Step 2 ：The speed of light is the maximum speed at which all information in the universe can travel. It is the speed of all massless particles and associated fields—including electromagnetic radiation such as light—in vacuum. It is also the speed of gravity (assumed by current theories). For this reason, it is the maximum speed at which information or matter can travel.

Step 3 ：In the context of this problem, even though each spaceship is moving at 0.9c relative to the Earth, the speed of one spaceship as observed from the other cannot exceed the speed of light (c). This is a consequence of the theory of relativity.

Step 4 ：To calculate the relative speed of the two spaceships as observed from one of them, we can use the formula for addition of velocities in special relativity: \(v' = \frac{{v1 + v2}}{{1 + \frac{{v1*v2}}{{c^2}}}}\) where v1 and v2 are the velocities of the two spaceships relative to the Earth, and c is the speed of light.

Step 5 ：Substituting the given values into the formula, we get: \(v1 = 0.9\), \(v2 = 0.9\), and \(c = 1\).

Step 6 ：Calculating the above expression, we get \(v' = 0.994475138121547\).

Step 7 ：The result is approximately 0.9945c, which is less than the speed of light (c). This is consistent with the theory of relativity, which states that the speed of light is the maximum possible speed in the universe.

Step 8 ：Final Answer: The speed of one spaceship as observed from the other is \(\boxed{0.9945c}\), which is less than the speed of light.