Question

1. If a rod travels with a speed v 0.8 c along its length, how much does it shrink? 2. Assuming that the rest radius of earth is 6,400 km and its orbital speed about the sun is 30 km, how much does earth's diameter appear to be shortened to an observer on the sun, due to earth's orbital motion? 3. A rod of length Lo moves with speed v along the horizontal direction. The rod makes and angle 0o with respect to the x' axis. a. Determine the length of the rod as measured by a stationary observer. b. Determine the angle θ the rod makes with the x axis.

Answer 1

L' = L/ gamma gamma = 1/ squareroot 1 - v^2/c^2 = 1/ squareroot 1 - (0.8)^2 = 1.667 L' = 0.6 times L So, rod by factor of 0.6 \r\n L_x = L_0 cos theta_0 L_y = L_0 sin theta_0 length of rod only in horizontal direction. Lx' = Lx squareroot 1 - v^2/c^2 = L_0 cos theta_0 squareroot 1 - v^2/c^2 Ly' = Ly contracted Length L' = squareroot L x'^2 + L y'^2 L' = squareroot (L_0 cos theta_0)^2 (1 - v^2/c^2) + L^2_0 sin theta^2_0 = L_0 squareroot cos theta^2_0 + sin theta^2_0 - v^2/c^2 cos^2 theta_0 L' = L_0 squareroot 1 - v^2/c^2 cos^2 theta_0 \r\n angle made w. r. t x-axis tan theta = Ly'/Lx' = L_0 sin theta_0/L_0 cos theta_0 squareroot 1 - v^2/c^2 tan theta = tan theta_0/ squareroot 1 - v^2/c^2 r = 6400 km d = 2r = 12800 km = 12.8 times 10^6 m increment L = L - L'