Answer to Question #266249 in Mechanics | Relativity for Dinesh priyananda

Question

The orbital speed of the Earth around the Sun is 30 km/s. In one year, how many seconds do the clocks on the Earth lose with respect to the clocks of an inertial reference frame at rest relative to the Sun

Answer 1

If we are to consider an inertial frame at rest relative to the sun, it must be the one keeping the proper time interval ∆t_p. Over the course of one year, we have

$∆t_p = 1 \; yr ≈ 3.156 \times 10^7 \;s$

The observers on earth are in motion relative to the sun, and therefore they measure a dilated time

interval $∆t^* =γ∆t_p$ . We are asked for the difference between the two clocks after one year as measured in the sun’s inertial frame, or

$difference = ∆t^* − ∆t_p = γ∆t_p − ∆t_p = (γ − 1) ∆t_p$

Since in this case the relative velocity of earth is small with respect to c, v = 30 km/s = 3 × 10⁴ s so v/c= 10⁻⁴ , we can use the second approximation given.

$γ -1 = \frac{1}{\sqrt{1 -\frac{v^2}{c^2}}} -1 ≈ 1 + \frac{1}{2} \frac{v^2}{c^2} -1 = \frac{1}{2} \frac{v^2}{c^2 } \\ difference ≈ \frac{1}{2} \frac{v^2}{c^2} Δt_p \\ = \frac{1}{2} (10^{-4})^2 (3.156 \times 10^7 \;s) \\ = (5 \times 10^{-9}) (3.156 \times 10^7 \; s) \\ = 0.16 \;s$