A spaceship leaves Earth at a steady forward speed of 70% the speed of light. How fast relative to the ship must the spaceship launch a shuttlecraft in the forward direction so that it will be moving
Away from Earth at 98.7% the speed of light?
A) 0.84c
B) 0.97c
C) 0.91c
D) 0.99c
E) 0.93c

Question

Quizplus · Accepted Answer

To find the relative speed of the shuttlecraft with respect to Earth, given the spaceship's speed and the desired final speed of the shuttlecraft, we use the formula for relativistic addition of velocities: $v = \frac{v_1 + v_2}{1 + \frac{v_1 \cdot v_2}{c^2}}$, where $v$ is the resultant velocity, $v_1$ and $v_2$ are the velocities being added, and $c$ is the speed of light. Here, $v_1 = 0.7c$ (spaceship's speed relative to Earth), and we want $v = 0.987c$ (shuttlecraft's speed relative to Earth). Solving for $v_2$ (shuttlecraft's speed relative to the spaceship), we get $v_2 = \frac{v - v_1}{1 - \frac{v \cdot v_1}{c^2}}$. Plugging in the numbers, we find $v_2 \approx 0.93c$, which means the shuttlecraft must be launched at 93% the speed of light relative to the spaceship to achieve a speed of 98.7% the speed of light relative to Earth.

A Spaceship Leaves Earth at a Steady Forward Speed of 70