To determine the velocity required for a clock to appear slower by a specific amount of time, we can use the concept of time dilation in special relativity. The time dilation equation for relative velocity is given by:
Δt' = Δt / √(1 - (v^2/c^2))
Where: Δt' is the measured time interval for the moving clock, Δt is the proper time interval (time measured by a stationary clock), v is the relative velocity between the two frames of reference, c is the speed of light.
In this case, we want the moving clock to appear slower by 30 seconds compared to the stationary clock. Therefore, we have Δt' = Δt + 30 seconds.
We can rearrange the time dilation equation to solve for v:
v = c * √(1 - (Δt/Δt')^2)
Let's assume that Δt is 24 hours (which is equivalent to 86,400 seconds) for simplicity. Plugging these values into the equation, we get:
v = c * √(1 - (86,400 / (86,400 + 30))^2)
v ≈ c * √(1 - (86,400 / 86,430)^2)
v ≈ c * √(1 - 0.999653^2)
v ≈ c * √(1 - 0.999306)
v ≈ c * √(0.000694)
v ≈ c * 0.02634
v ≈ 7,902,000 meters per second
Therefore, in order for a clock traveling at a velocity of approximately 7,902,000 meters per second (or about 7,902 kilometers per second) to appear slower by 30 seconds in one day compared to an observer, it would need to travel at that velocity.