Is the Pioneer Anomaly Really Solved ?

Bjarne

Right I wrote "measure", but I meant, how is the distance calculated?

mfb

The distance can be evaluated via timing - not very precise, but probably good enough as the probes are far away from any massive object and their position is not so crucial.

Bjarne

It must be important that we know which distance we expect the space probes to be
Also a way of calculating this ?

Drakkith

Try this link: http://www.au.af.mil/au/awc/awcgate/...ic/bsf13-1.htm
And here: http://www2.jpl.nasa.gov/basics/bsf13-1.php

SHISHKABOB

We do that by sending a message out and waiting for a response. The time that it takes the message to get back will tell us how far away the probe is.

Bjarne

The gravitational field of the Earth is causing the signal to blueshift and the speed to the sigal to redshift, so this is how we measure the speed . OK

Let us say that right now we expect the speed of the space probe to be12000m/s

Let us say that time is ticking 1 nanosecond faster at the space probe (due to gravitational time dilation) as 1 second on the Earth. .

Do we expect the space probe to move 12000 m/s (per 1 Earth second) or 12000 m/s (per 1 Space probe second) ?

(To keep it simple ignore the influence of SR, so only GR influence is the question )

mfb

In which reference frame? ;)

Depends on the answer to my question.

The measurement is done on earth, but you can transform it to other systems as well of course.

Bjarne

In the frame of the space probe
Let's say right after launch the speed seen from that reference frame is exactly 12000 m/s
After some decades when leaving the solar system times ticks faster compared to time on Earth, seen from the space probe reference frame.

How fast will the speed of the space probe be after reaching the edge of the solar system, - still seen from the perspective of the space probe ?

I guess still 12000 m/s , also even though that time now is ticking faster as before??

And seen from the perspective of Earth it will look like the probe is acceleration, proportional to the time dilation, - or ?

We will ignore the backwards pulls due to gravity of the solar system

mfb

If we expect the probe to move 1200m/s relative to the sun in the system of the probe, we expect it to move 1200m/s in the system of the probe and with its own clock. Well, it has no precise atomic clock, but we can imagine one. I don't get the point.

Gravity will decelerate the probe (relative to the sun), this effect is always dominant unless you are close to a black hole. You have to take both effects into account if you want to compute the velocity of the probe in the system of the probe.

No.

Bjarne

So what you say is, if we assume it was a atomic clock on bours the speed would be constant, - = 12000 m/s the whole way of of the solar system, - if we ignore the deceleration due the gravity of the solar system ?

Because the clock is ticking slower on Earth compared to on board the space probe, - the speed of the space probe cannot be understood also as 12000 m/s seen from the reference frame (hereafter RF) of the Earth too. (seen from the reference frame of the space probe)

Do you mean the speed also is 12000 m/s seen from the RF of the Earth?, - that wouldn’t be possible.'

Because time is ticking comapable slower on Earth, we on Earth cannot agree that speed is the same as seen from the perspective of the space probe, - (or ?)

What is the solution to that?

mfb

No. In addition, you cannot ignore it. It is like asking "ignoring gravity, how quick does an apple fall?".
If we ignore deceleration, we have to ignore gravitational time dilation as well, as they are both parts of the same effect (gravity). You cannot get one without the other. Without gravity, both earth and the probe will measure 12000m/s all the time.


Right, we and the probe measure different relative velocities probe<->sun.

Bjarne

I am not ignoring it but just trying to keep the question simple.
So what you say is that the speed out of the solar system, seen from the perspective of the space prove is "constant" - but MINUS the deceleration due to gravity ?

mfb

If you calculate the deceleration in the system of the probe.
If you calculate the deceleration in the system of earth, you get a different value due to gravitational time dilation.

You can calculate the system as a whole, but it is quite pointless to split it in arbitrary components to compare them.