# Voyager probes and time dilation

• Andru10
In summary, the Voyager probes are traveling at non-relativistic speeds and at the edge of the solar system, where the gravitational field is about 0, making them a rest frame. Using the formula for time dilation, it can be calculated that 10 years for the probes would mean about 2.8 years for someone on Earth. The gravitational time dilation from the Sun is much higher than that of the Earth, with a factor of 0.999999989933 compared to 0.999999999304, respectively. Taking into account other gravitational fields and the Earth's motion relative to the Sun, the time differential for the Voyager would be around 0.219 seconds after 10
Andru10
The Voyager probes are traveling at non-relativistic speeds, at the edge of the solar system where the gravitational field is about 0 so we could consider them a rest frame. If so, from my calculations using the formula for time dilation:

$$\tau(t) = \frac{c}{g} \operatorname {arsinh} \left( \frac{gt}{c} \right)$$

10 years for the probes would mean about 2.8 years for someone on Earth.

Is this correct? I'm curious if the gravitational acceleration at the Earth's surface is really so significant.

I think I kow what you're doing. You're picking the probes as your inertial frame, and calculate elapsed time on Earth following http://en.wikipedia.org/wiki/Time_dilation#Time_dilation_at_constant_acceleration".

The effect described in the wikipedia article stems from the relative velocity an object would have if - starting from rest - it accelerates for a given time. The acceleration itself contributes exactly nothing to the effect - a circumstance knonw as the "clock hypothesis".

Even if an observer on Earth is accelerating, she's not gaining relative velocity to the probes by this acceleration. So you can't apply this formula. It is not intended for curved spacetime, where acceleration without motion is possible.

What you should do instead is to combine the effects of http://en.wikipedia.org/wiki/Time_dilation#Time_dilation_due_to_relative_velocity".

It's a tiny effect, btw.

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Ich said:
I think I kow what you're doing. You're picking the probes as your inertial frame, and calculate elapsed time on Earth following http://en.wikipedia.org/wiki/Time_dilation#Time_dilation_at_constant_acceleration".

The effect described in the wikipedia article stems from the relative velocity an object would have if - starting from rest - it accelerates for a given time. The acceleration itself contributes exactly nothing to the effect - a circumstance knonw as the "clock hypothesis".

Even if an observer on Earth is accelerating, she's not gaining relative velocity to the probes by this acceleration. So you can't apply this formula. It is not intended for curved spacetime, where acceleration without motion is possible.

What you should do instead is to combine the effects of http://en.wikipedia.org/wiki/Time_dilation#Time_dilation_due_to_relative_velocity".

It's a tiny effect, btw.

Thank you, this cleared things up for me :)

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Indeed the dilation at the surface of the Earth is very small compared to a region without any significant gravitation.

This triggered me to calculate the Pound Rebka experiment using the Schwarzschild solution, and I seem to be stuck.

We have:

Mass of the Earth = 0.004435028 meters
Schwarzschild radius R0 = 0.008870056 meters
R1 = 6375416 meters
R2 = 6375438.6 meters (R1 + 22.6)

Ratio of the two clocks:

$$\sqrt{ {1-R_0/R_1 \over 1-R_0/R_2 } }$$

This gives me: 1.00000000626084025339746259

So now how do I go from here to get 2.5×10−15?

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So now how do I go from here to get 2.5×10−15?
With algebra?

Maybe you made a sign error.

Passionflower said:
Indeed the dilation at the surface of the Earth is very small compared to a region without any significant gravitation.

This triggered me to calculate the Pound Rebka experiment using the Schwarzschild solution, and I seem to be stuck.

We have:

Mass of the Earth = 0.004435028 meters
Schwarzschild radius R0 = 0.008870056 meters
R1 = 6375416 meters
R2 = 6375438.6 meters (R1 + 22.6)

Ratio of the two clocks:

$$\sqrt{ {1-R_0/R_1 \over 1-R_0/R_2 } }$$

This gives me: 1.00000000626084025339746259

So now how do I go from here to get 2.5×10−15?

Independently, I get R0 = 0.008871345 which is close enough to your figure because the figures for the mass and radius of the Earth vary a bit depending on the source and whether you are at the equator or not. Using your figures for R0 and R1 I get:

$$\sqrt{ {1-R_0/R_1 \over 1-R_0/R_2 } } = 0.999999999999998$$

and

$$1-\sqrt{ {1-R_0/R_1 \over 1-R_0/R_2 } } = 2.44E-15$$

Passionflower said:
Indeed the dilation at the surface of the Earth is very small compared to a region without any significant gravitation.

This triggered me to calculate the Pound Rebka experiment using the Schwarzschild solution, and I seem to be stuck.

We have:

Mass of the Earth = 0.004435028 meters
Schwarzschild radius R0 = 0.008870056 meters
R1 = 6375416 meters
R2 = 6375438.6 meters (R1 + 22.6)

Ratio of the two clocks:

$$\sqrt{ {1-R_0/R_1 \over 1-R_0/R_2 } }$$

This gives me: 1.00000000626084025339746259

So now how do I go from here to get 2.5×10−15?

Edited: Actually I have to divide by c^2 so I get:

$$\sqrt{ {1-{R_0 \over R_1 c^2} \over 1-{R_0 \over R_2 c^2}}}$$

1.000000000000000000000000069661241837728484515687

Not really any closer :(

Your figure for R0 has already been divided by c^2, since R0 = 2GM/c^2 = 0.008870056 meters

yuiop said:
Independently, I get R0 = 0.008871345 which is close enough to your figure because the figures for the mass and radius of the Earth vary a bit depending on the source and whether you are at the equator or not. Using your figures for R0 and R1 I get:

$$\sqrt{ {1-R_0/R_1 \over 1-R_0/R_2 } } = 0.999999999999998$$

and

$$1-\sqrt{ {1-R_0/R_1 \over 1-R_0/R_2 } } = 2.44E-15$$
Got it! I missed one single digit somewhere, it's too late here (4:14 AM).

2.46596 E-15

2.46632 E-15

yuiop said:
Your figure for R0 has already been divided by c^2, since R0 = 2GM/c^2 = 0.008870056 meters
Yes that's right, I took it already out.

If we do not take into account other gravitational fields or the velocity of the Voyager which is obviously totally unrealistic the time differential with respect to the Voyager far away from the Earth should be somewhere like:

0.99999999930435

So for 10 years that accumulates to about 0.219 seconds.

Do we know the current velocity of the Voyager?
In we have it we can include that in the calculation.

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Do we know the current velocity of the Voyager?
http://voyager.jpl.nasa.gov/mission/weekly-reports/" But Earth is moving, too.

To make a reasonably accurate guess, you have to factor in the sun's gravitational potential, as well as Earth's motion relative to the sun.

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Ich said:
To make a reasonably accurate guess, you have to factor in the sun's gravitational potential, as well as Earth's motion relative to the sun.
I totally agree.

In fact the gravitational time dilation from the Sun is much higher than the time dilation of the Earth even for a clock on the Earth's surface!

While gravitational time delay factor due to the Earth is: 0.999999999304 the Sun's gravitational time delay factor, roughly 143 million km removed is: 0.999999989933 two orders of magnitude larger!

At least according to my calculations.

Yes - MTW has a table of "metric correction factors" - which is just Gm/rc^2 - for various objects. (pg 459)

Basically larger-scale structures contribute more to time dilation.

a statue (Venus de Milo) (m=2e5 g, r = 30cm) -> 5e-25
at the surface of the Earth (m=6e27g, r= 6.4e8cm) -> 6e-10
at Earth's distance from sun (m=we33g, r=1.5e13mc) ->1e-8
at Sun's distance from center of galaxy (m=1e44g, r=2.5e23cm) -> 6e-7
at distance of galaxy from Virgo cluster (m=6e47g, r= 3e25cm) -> 1e-6

## 1. What are the Voyager probes and why are they important?

The Voyager probes are two robotic spacecrafts, Voyager 1 and Voyager 2, launched by NASA in 1977 to explore the outer planets of our solar system. They have since provided valuable data and images of these planets, their moons, and other celestial bodies. The Voyager probes are important because they have greatly advanced our understanding of our solar system and have continued to send data even as they travel beyond our solar system.

## 2. How fast are the Voyager probes traveling?

The Voyager probes are currently traveling at a speed of about 38,000 miles per hour. However, this speed is constantly decreasing as they move further away from the sun and its gravitational pull.

## 3. What is time dilation and how does it affect the Voyager probes?

Time dilation is a phenomenon in which time appears to pass at a slower rate for an object moving at a high speed. This effect is a result of Einstein's theory of relativity. As the Voyager probes travel at high speeds, time dilation affects the accuracy of their clocks and the timing of their transmissions. This has to be taken into consideration when interpreting data from the probes.

## 4. How far have the Voyager probes traveled and how long will they continue to send data?

As of 2020, Voyager 1 has traveled over 14.2 billion miles from Earth, while Voyager 2 has traveled over 11.7 billion miles. They are both expected to continue sending data until at least 2025, when their power sources are estimated to run out. However, it is possible that they may continue to send data for several more years after that.

## 5. Have the Voyager probes reached any other star systems?

No, the Voyager probes have not yet reached any other star systems. They are currently traveling through interstellar space, but it will take tens of thousands of years for them to reach the nearest star system, Proxima Centauri.

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