find the time dilation ts/to for Earth's orbit about the sun if vEarth=10889 km/s. Derive the equation for ts.

Question

ts: time seen by the stationary observer on Earth
to: time seen by the observer on spacecraft

Answer 1

The special relativity effects dominate over any general relativity (gravitational) effects. Regardless of the velocity being small, it is still finite. One problem with your question is the the Earth's mean orbital velocity around the sun is 29.8 km/s not 10889km/s. I will answer both ways, just in case this is a fictional earth with an absurd orbital velocity.

The gamma correction to the passage of time is 1/SQRT(1-v^2/c^2). Expressing the Earth's velocity as a fraction of the speed of light,
its actual orbital velocity of 29.8 km/s is .0001c . The earth observer will experience time passing 1.00000005 times faster than the observer on Earth (ts/to) or about 1.5 extra seconds/year.

In the fictional example of 10889 km/s, v = 0.036297C and the gamma factor is 1.00066. Over a one year time period, the earthobserver will end up about 5 hours and 40 minutes ahead of the earth bound observer.

It is probably more reasonable to have the stationary observer in space and the Earth moving. Since this is a special relativity problem as worded, it does not matter much which observer you choose to make stationary, just remember that whoever isn't moving experiences time faster than the one who is.

Answer 2

O.K., for starters-if the Earth stopped rotating on its axis, there could be little motion within the surroundings as good because the oceans. This could affect each our climate and the temperature of the surroundings-plus its capacity to take away impurities (smog, volcanic ash, and many others.). If the Earth have been to give up in its orbit across the Sun, it could no longer have sufficient ahead pace=centrifugal drive, to maintain an outstanding direction. Eventually gravity (both from the Sun or outer area), could take over, at which factor it is a one-method price tag to both of the ones. So a long way, I didn't see any point out of the Moon in those solutions. Since it's held in anxiety guess. the Earth and the Sun, it could surely hit Earth in each instances. That's a nasty hair day. Afterthought-it could even have an impact on plate motion underneath the Earth's crust, too. Not that it could topic-due to the fact that every body could be long gone. This (plate tectonic component), is a further query...

Answer 3

The Earth moves in an elliptical orbit at an average frequency of 3.168895541 x10^-8 Hertz.As observed from a fixed star.Its Gravitational acceleration varies inversely to the radius vector square. The Earth is moving very slow compared to the speed of light.therefore rlativistic time dilation does not apply to the Earth. If we apply gravitational time dilation =the difference between the minimum time and the max dilated time we have a gravitational time dilation of 17.52 percent of the minimum gravitational time as observed from a fixed star.The Equation for Gravitational time is T^2 = R/a^2 where R is the radius vector and a is the value of the gravitational field at that point.(Gravitational Time ^2 is inversely proportional to the Gravitional field.)
This is a much greater time dilation than the average relativitic time dilation which is a neglible .000005 percent.

Therfore, Relativistic Time dilation does not measure the same time as Gravitational Time dilation.
Note from Earth the Star is observed as stationary . From a Star the Earth is observed as moving.