# NASA Solar Probe Plus to hit 0.067% of c

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1. May 1, 2013

### zincshow

In 2018, NASA will launch "Solar Probe Plus" in an orbit that will swing by close to the sun. It is expected to reach speeds as high as 450,000 mph (0.067 percent the speed of light). At what speed would you (if you were inside the spacecraft) start to see relativistic effects inside the space craft and what would they be? - ie. do transistors and computers continue to work no matter what the speed?

2. May 1, 2013

### phinds

The person, and all the stuff, inside a spaceship is at rest regardless of the speed of the ship. Google "frame of reference".

3. May 1, 2013

### lpetrich

For that spacecraft, the size of relativistic effects that we would observe is about (1/2)*(v/c)2, or about 2*10-7 in this case. That implies a time dilation of about 7 seconds per (Earth) year. So we won't be able to see its time dilation unless it has a very good onboard clock.

But one can use the spacecraft as a transponder for precision timing, and one may be able to see not only the signals' time delay near the Sun, but also the spacecraft's relativistic orbit precession.

Solar Probe Plus: A NASA Mission to Touch the Sun
Missions - Solar Probe Plus - NASA Science

It will get to about 9.5 solar radii from the Sun, and it will be traveling at about 200 km/s when there.

From http://solarprobe.jhuapl.edu/spacecraft/mission-design.php [Broken], its closest approach will be about 0.04 AU with an aphelion of 0.73 AU and a period of about 88 days. Its GR precession is about 200 seconds of arc per century, or 0.5 seconds per orbit. That's about 300 km at aphelion or 900 microseconds of light travel time. That's more than the relativistic time delay for a radio signal passing near the Sun, so it should be detectable.

Last edited by a moderator: May 6, 2017
4. May 1, 2013

### zincshow

Hey thanks, lets say it did have a good clock, which of the statements would be true about what observers here on earth will see:

When going towards the sun
1. the radio signal and visual image will be blue shifted
2. we assume the clock on the spacecraft will be going slower than our clocks

When coming back towards earth
1. the radio signal and visual image will be red shifted
2. we assume the clock on the spacecraft will be going faster than our clocks
3. when (and if of course) the spacecraft gets back close to earth the spacecraft clock will show exactly the same time have as ours, having slowed down on the way there and sped up coming back.

5. May 2, 2013

### lpetrich

Seems like a case of the Twin Paradox. In general, if one goes from space-time point A to space-time point B, then the faster and more back-and-forth one's path is, the less proper time one will see between A and B.

One's time will be slower than coordinate time at infinity by

dt(proper)/dt(coordinate) = 1 - (1/2)*v2 + V

where V is the gravitational potential = - GM/r

6. May 2, 2013

### lpetrich

Oops, should have divided by c2:

dt(proper)/dt(coordinate) = 1 - (1/2)*(v/c)2 + V/c2

For a circular orbit, we get

dt(proper)/dt(coordinate) = 1 - (3/2)*(GM/rc2)