# Clocks On The Space Shuttle

1. Feb 1, 2004

### sam1967

I just read that the clocks on board the space shuttle go more slowly than the clocks on earth.
does that mean time has speeded up for those on the shutte or slowed down ?
i reckon it means time has slowed down. if they stayed up for a year by their clocks then by our clocks more than a year will have passed.

2. Feb 1, 2004

### Mr. Robin Parsons

Is this a question, or statment?

3. Feb 1, 2004

### sam1967

it was a question as I am really not sure.

4. Feb 1, 2004

### Mr. Robin Parsons

So maybe you would want to type in the words "time dilation" into the search feature, (in these forums) look in titles, or threads, and you will find some 'stuff' to help you start building knowledge towards understanding the how and why of that...stuff.... O.K.?

5. Feb 1, 2004

### Staff: Mentor

The speed of the space shuttle causes the clocks to slow down relative to a clock on earh. The altitude (reduced gravitational force) causes the clocks to speed up. The net effec is that the clocks (time) go faster in space than on earth.

6. Feb 1, 2004

### Mr. Robin Parsons

And hence the relativisation of light's speed(?)

7. Feb 2, 2004

### Phobos

Staff Emeritus
Mr. Robin Parsons - Your second post was uncalled for. There was nothing wrong with the question. Yes, sam1967 could have used the search feature to find some similar topic...but give a new member a break. I'd rather see repeated discussions than rude responses.

8. Feb 2, 2004

### franznietzsche

no, the relativisation of time

9. Feb 2, 2004

### Mr. Robin Parsons

O.K. but I was trying something trying to see if I could help, the rude part, is, perhaps, not as much there, as you might think...But I would apologise to sam1967 if they were, what? "off-ended" by my, well, thought....Sorry sam1967, didn't intend, or mean to be 'rude' just wondered if newbies even looked in the search features, they are rich... My Opinion...

10. Feb 3, 2004

### Phobos

Staff Emeritus
thanks

11. Feb 4, 2004

### Mr. Robin Parsons

No Problemo

12. Aug 3, 2011

### just board

You might be incorrect, the clocks on the shuttle are not your wind up clocks next to your bed.
They are digital and atomically. gravity does not effect the mechanics of the clock because there are no gears, weights,springs in it.

13. Aug 3, 2011

### HallsofIvy

You noticed, didn't you, that this thread was 7 years old?

14. Jun 27, 2012

### thinkoutside

Lol. Space time even exists in forums.

15. Jun 27, 2012

### Staff: Mentor

To observers on the shuttle, the clocks are running at their usual rate, irrespective of whether they are mechanical or digital. However, according to russ waters, to observers on the earth's surface, the clocks on the shuttle are running faster than their own, and it is earth surface clocks that are running at the usual rate.

16. Jun 27, 2012

### Staff: Mentor

I don't think russ's statement was correct for the Space Shuttle. It is correct for satellites in high orbits, such as the GPS satellites, where the effect of altitude (speeding up clocks relative to those on the Earth's surface) outweighs the effect of orbital velocity (slowing down clocks relative to those on the Earth's surface). For satellites in low Earth orbit, such as the Shuttle, the velocity effect outweighs the altitude effect.

An approximate equation for the rate of time flow for an object in Earth's gravity field is

$$\frac{d\tau}{dt} = \sqrt{1 - \frac{2 G M}{c^{2} r} - \frac{v^{2}}{c^{2}}}$$

For an object on Earth's surface, we can eliminate the v^2 term by using the poles as a reference; since the Earth's surface is (approximately) an equipotential surface, clocks on the surface all go at the same rate, and the rate at the poles is the easiest to calculate. The polar radius of the Earth is 6.36 x 10^6 meters according to Wikipedia. G is 6.67 x 10^-11, and M for the Earth is 5.97 x 10^24 kg. This gives a result of $d\tau / dt = 1 - 6.97 * 10^{-10}$.

For objects in orbit, since v^2 = GM/r for an object in a free-fall circular orbit (which we'll assume is a good enough approximation for this problem), we can rewrite the above more simply as

$$\frac{d\tau}{dt} = \sqrt{1 - \frac{3G M}{c^{2} r}}$$

For the Space Shuttle, r is about 200 miles, or 3.2 x 10^5 m, above the Earth's surface. For the GPS satellites, r is about 4.2 Earth radii. The corresponding numbers are:

Shuttle: $d\tau / dt = 1 - 9.95* 10^{-10}$, which is slower than ground clocks by about 3 parts in 10^-10.

GPS satellite: $d\tau / dt = 1 - 2.49 * 10^{-10}$, which is faster than ground clocks by about 4.5 parts in 10^-10.

17. Jun 27, 2012

### D H

Staff Emeritus
And now it is over 8 years old.