Question 26.2: Speedy Plunge GOAL Apply the concept of length contraction t...
Speedy Plunge
GOAL Apply the concept of length contraction to a distance.
PROBLEM (a) An observer on Earth sees a spaceship at an altitude of 4 350 km moving downward toward Earth with a speed of 0.970c. What is the distance from the spaceship to Earth as measured by the spaceship’s captain? (b) After firing his engines, the captain measures her ship’s altitude as 267 km, whereas the observer on Earth measures it to be 625 km. What is the speed of the spaceship at this instant?
STRATEGY To the captain, Earth is rushing toward her ship at 0.970c; hence, the distance between her ship and Earth is contracted. Substitution into Equation 26.9
K E=\gamma m c^{2}-m c^{2} [26.9]
yields the answer. In part (b) use the same equation, substituting the distances and solving for the speed.
(a) Find the distance from the ship to Earth as measured by the captain.
Substitute into Equation 26.4, getting the altitude as measured by the captain in the ship:
L=L_{p}{\sqrt{1-v^{2}/c^{2}}}=(4~350\,{\mathrm{km}}){\sqrt{1-(0.970c)^{2}/c^{2}}}
=~1.06\times10^{3}{\mathrm{~km}}
L=\frac{L_p}{γ}=L_{p}{\sqrt{1-v^{2}/c^{2}}} [26.4]
(b) What is the subsequent speed of the spaceship if the Earth observer measures the distance from the ship to Earth as 625 km and the captain measures it as 267 km?
Apply the length-contraction equation:
L=L_{p}{\sqrt{1-v^{2}/c^{2}}}
Square both sides of this equation and solve for υ:
L^{2}=L_{p}^{\;2}(1-v^{2}/c^{2})\;\;\;\rightarrow\;\;\;1-v^{2}/c^{2}=\left({\frac{L}{L_{p}}}\right)^{2}
v=\,c\sqrt{1-(L/L_{p})^{2}}=\,c\sqrt{1-(267\mathrm{~km/625~km})^{2}}
υ = 0.904c
REMARKS The proper length is always the length measured by an observer at rest with respect to that length