Velocity Time dilation on International Space Station

VELOCITY TIME DILATION ON THE INTERNATIONAL SPACE STATION

INTRODUCTION

International Space Station [ISS] is a satellite orbiting the Earth at a height of 250 miles from the surface. It completes one revolution around the Earth in approximately 90 Earth minutes. Consider the ISS which revolves at a certain velocity, an observer [observer 1] on Earth’s surface who also performs circular motion since Earth is rotating and another observer [observer 2] at the center of Earth who will not perform circular motion. According to the Special Theory of Relativity, a clock on ISS would run slower relative to the stationary observer at the center of Earth. We’ll find the time gained by the ISS and observer 1 with respect to observer 2.

ASSUMPTIONS

1. The revolution path is exactly circular.
2. Earth is a perfect homogenous sphere.
3. The effect of Gravitational time dilation on ISS is negligible.

CALCULATION

The Angular velocity of Earth

ω_e = 2π/T_e [rad/s]

T_e – Rotational time period of Earth [T ≈ 24 hours = 86400s]

ω_e = 2*3.14/86400

ω_e = 7.2722*10^-5 rad/s

The angular velocity is same at all points on earth since it doesn’t exhibit differential rotation. But the tangential velocity on surface varies with the distance from the center. The greater the distance, higher is the velocity. It is important to note that since ground level observer is also performing a circular motion, he will also experience time dilation relative to observer at the center of Earth.

The Angular velocity of ISS

ω_i = 2π/T_i [rad/s]

T_i – Revolution time period [T ≈ 90 minutes = 5400s]

ω_i = (2*3.14)/5400

ω_i = 1.1629*10^-3 rad/s

The tangential velocity of observer at center of Earth is given by,

V₀ = [R]*ω_e [m/s]

R – Average radius of earth [R = 0 m]

V₀ = 0 m/s

The tangential velocity of ground observer is given by,

V_e = [R]*ω_e [m/s]

R – Average radius of earth [R = 6371000m]

V_e = 6371000*7.2722*10^-5

V_e = 463.3118 m/s

The tangential velocity of ISS is given by,

V_i = [R + h]*ω_i [m/s]

R – Average radius of earth [R = 6371000m]

h – Height of ISS from ground [h = 400 km]

V_i = 6771*1000*1.1629*10^-3

V_i = 7873.995 m/s

According to the Special Theory of Relativity, the time dilation equation is,

t’ = t/γ [s]

t’– ISS time. [s]

t₁ – Ground level observer’s time. [s]

t₂ - Proper time or center of earth observer’s time. [s]

γ – Relativistic gamma factor, γ = 1/√ [1-(v/c) ²]

c - Velocity of light [c = 3*10⁸ m/s]

t’ = t*√ [1-(v/c) ²]

Center of Earth observer time

t₂ = t since v = 0

Ground level observer time

t₁ = t*√ [1-2.3850*10^-12]

t₁ = t*√ [0.999999999997615]

t₁ = t* 0.999999999998808

ISS time

t’ = t*√ [1-6.4448*10^-10]

t’ = t*√ [0.99999999935552]

t’ = t* 0.99999999967776

CONCLUSION

We can observe that proper and actual time isn’t the same which proves that time dilates on revolving ISS relative to the stationary one. We’ll consider 5 different t’ values and calculate t value. The larger the t₂ the more is the difference between t₂ and t’. Thus ISS will gain time over both observer1 and observer2. In fact in one year it gains 0.01 second with respect to observer2.

Time	t2 [Center of Earth observer] (s)	t1 [Ground level observer] (s)	t’ [Revolving ISS] (s)
1 minute	60	59.9999999999285	59.9999999806656
1 hour	3600	3599.99999999571	3599.99999883994
1 day	86400	86399.999999897	86399.9999721585
1 month	2592000	2591999.99999691	2591999.99916475
1 year	31536000	31535999.9999624	31535999.9898378