Question

Two atomic clocks are synchronized while on Earth before one is launched (within a GPS satellite) into a medium-altitude Earth orbit. The satellite orbits the Earth at an altitude of 22,000 km, with an orbital velocity of 15,000 km/hr. After 1000 orbits, estimate the time difference between the two atomic clocks? (Assume that only special relativity effects are present.)

Answer 1

Answer on Question #43043, Physics, Relativity

Two atomic clocks are synchronized while on Earth before one is launched (within a GPS satellite) into a medium-altitude Earth orbit. The satellite orbits the Earth at an altitude of 22,000 km, with an orbital velocity of 15,000 km/hr. After 1000 orbits, estimate the time difference between the two atomic clocks? (Assume that only special relativity effects are present.)

Solution:

Given:

$$\begin{array}{l} h = 22000 \text{ km} = 22 \cdot 10^{6} \text{ m}, \\ v = 15000 \text{ km/hr} = 4166.67 \text{ m/s}, \\ N = 1000 \text{ orbits}, \\ \delta t = ? \end{array}$$

According to special relativity, the rate of a clock is greatest according to an observer who is at rest with respect to the clock. In a frame of reference in which the clock is not at rest, the clock runs more slowly, as expressed by the Lorentz factor.

The Lorentz factor is defined as:

$$\gamma = \frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}} = \frac{1}{\sqrt{1 - \beta^{2}}}$$

where:

- $v$ is the relative velocity between inertial reference frames,

- $\beta$ is the ratio of $v$ to the speed of light $c$.

- $c = 3 \cdot 10^{8} \text{ m/s}$ is the speed of light in a vacuum.

Thus,

$$\gamma = \frac{1}{\sqrt{1 - \frac{4166.67^{2}}{9 \cdot 10^{16}}}} = 1.000000000096451$$

Time dilation

The time $(\Delta t')$ between two ticks as measured in the frame in which the clock is moving, is longer than the time $(\Delta t)$ between these ticks as measured in the rest frame of the clock:

$$\Delta t' = \gamma \Delta t$$

For time $\Delta t$ we have:

$$\Delta t = \frac{\text{Distance}}{\text{velocity}} = \frac{2\pi(R + h)N}{v}$$

where $R = 6.371 \cdot 10^{6} \text{ m}$ is the Earth's mean radius.

Thus,

$$\Delta t = \frac{2\pi(6.371 \cdot 10^{6} + 22 \cdot 10^{6}) \cdot 1000}{4166.67} = 4.27824 \cdot 10^{7} \text{ s}$$

The time difference between the two atomic clocks is

$$\delta t = \Delta t' - \Delta t = \gamma \Delta t - \Delta t = \Delta t (\gamma - 1)$$

Thus,

$$\delta t = 4.27824 \cdot 10^{7}(1.000000000096451 - 1.0) = 0.00413 \text{ s} = 4.13 \cdot 10^{-3} \text{ s}$$

Answer. $\delta t = 4.13 \cdot 10^{-3} \text{ s}$.

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