Question

A satellite is launched into an orbit at an altitude 200 km above the surface. Onboard is an exquisitely sensitive atomic clock that is synchronized with an identical clock on Earth. After orbiting for one year, the satellite is captured, returned to Earth, and the clocks compared. What will be the shift in time between the two clocks?

Answer 1

Alright, the given information is,

There are two types of time dilation. One is due to relative velocity, and the other due to proximity to a massive object. First, the time dilation due to velocity is explained in the Special Theory of Relativity, which posits that as you approach near light velocities, the time in the moving frame of reference slows down by a factor , which is the lorentz factor. Using this the time dilation of object moving at a velocity v will experience time dilation as,

t' and t are the changes in time observed in the moving frame and the rest frame of reference respectively.

The minimum velocity that a satellite requires to stay in a Low Earth Orbit at 200 km can be easily determined by comparing forces on the satellite in "circular orbit"

where, F_G and F_C are the gravitational and centripetal forces.

The constant values of the gravitaional constant and the mass of the earth are easily available. R, here is the distance between the centers of the objects. So we have to add 200 km to the radius of the earth. Note, there is no dependence on the mass of the satellite.

Using this velocity in the time dilation equation, where t_0 = 365*24*60*60 seconds = 3.1536 * 10^7 seconds. (seconds in a year)

So the clock on the satellite will be t'-t = 0.0106156 seconds behind when it comes back after a year.

Now, in the General Theory of Relativity, it is postulated that gravity, instead of a force is the bending of 4 dimensional space-time. It is famously said that 'Spacetime tells matter how to move, and matter tells spacetime how to curve'. In that theory, Einstein said that as we get closer to a massive object, there something called gravitational time dilation occurs, which increases the length of a time period compared to a rest frame with some relation to the distance between the clock and the object. Since the satellite is away from the earth, it will move faster compared to the clock on the surface. This means it counteracts the effect of Velocity Time Dilation. The simplified formula can be given as,

Using the same notation we did before and the same values, we get,

Calculating the difference we get, t'-t = 0.0212312 seconds. This is how fast the clock will run at 200 km above the earth were the satellite not moving.

But since it was moving at a high enough velocity, time slows down and still, the gravitational 'quickening' overtakes the velocity time dilation. This leads to the clock on the satellite after a year to have 'ticked' faster and hence be ahead by,