Merlion Audio

All The Best Sound

Clock Satellite

[6th repost] How much time elapsed according to clock aboard aircraft?

There are two twin old-fashioned pendulum clocks
A and B, which run accurately on equator at rest.

http://en.wikipedia.org/wiki/Pendulum_clock

Clock A is taken aboard supersonic jet, which
cruises along equator around the Earth at low
constant altitude with constant speed v.

The second clock B is kept at rest at observatory,
located at equator.

When the jet roared above observatory the first
time the two clocks were syncronized.

The jet made one full loop and around the Eearth,
and when it passed above the observatory for the
second time, according to clock B the duration of the circumnavigation flight was To.

What was duration of the flight according to clock A?

The speed of low-orbit satellites is c~8km/s.
Ignore roatation of Earth.

The duration of the flight on clock A would appear to be

T_o [1 - (v/_sat)^2]^(1/2) or T_o [1 - (v/c)^2]^(1/2) in your tricky notation.

For a typical, just supersonic speed, this is a reduction in time apparently elapsed by just under 2 minutes.

I notice that no-one seems to have appreciated the relevance of your remark that the speed of a low-orbit satellite is ~ 8 km/s. However, it WAS very NAUGHTY of you to call that ' c ,' even if you thought that might be a subtle hint to look for a " relativistic-looking (v/c)^2" effect involving THAT ' c ' !

I will instead use v_s, for the satellite's speed, subscript ' s ' for satellite. I'll just use v for the plane's speed. I'll also ignore all SR and GR effects --- they are negligible in comparison with the main point at issue, which is the "effective g," g_eff, which is what determines the period of a classical pendulum clock.

[Actually, the "period" of a pendulum clock in such a satellite would NOT be infinite, but rather ~ 84 minutes, if you think carefully about it, though that is NOT for what you'd consider a classical pendulum motion, as it completes a full circuit in that time! Compared to 1/2 - 1 second, the typical period of an Earth-bound pendulum clock, that effectively is "infinite" of course, being > 5000 times longer.]

For anything orbiting the Earth with speed v, the effective gravity is given by

g_eff = g - v^2/r = g [1 - v^2/(rg)]

For an orbiting satellite, of course, (v_s)^2 / r = g, since it is the gravitational acceleration g which produces the centripetal acceleration of the satellite, (v_s)^2 / r.

Since 1/(rg) = 1/(v_s)^2, we can represent g_eff by

g_eff = g [1 - (v/v_s)^2].

Since the period P of a pendulum is given by

P = 2π √(L/g_eff), P is proportional to (g_eff)^(-1/2).

The period will therefore be lengthened by a factor

[1 - (v/v_s)^2]^(-1/2), or [1 - v^2/c^2]^(-1/2), in your tricky notation.

How large will this be? :

Let's take a just supersonic plane with v = 1200km/h. Then with v_sat ~ 8km/s, v/v_sat (your " v/c ") will be ~ 1/24. Then (v/v_s)^2 or (v/c)^2 ~ 0.001736. Since that is fairly small, we could reasonably well approximate the lengthening factor by the simpler expression 1 + 1/2 (v/v_s)^2 or 1 + 1/2 (v/c)^2. I'll compare these two estimates, below.

The accurate expression gives a factor 1.0008692... .

The first term approximation gives 1.0008681... .

Now here's one more slight twist. The elapsed time "shown by a clock" is actually INVERSELY PROPORTIONAL to its period. (It only recognizes that time has advanced by recording how many periods have elapsed.) So the time "showing" on clock A will be SMALLER than that on clock B by the factor given above.

So if the time recorded on the stationary pendulum clock B is T_o, the "time" recorded on the moving clock A on board the plane will be ~ T_o / 1.0008692... .

Since we're just looking at one example to get an order of magnitude, let's also see the order of magnitude by which clock A will apparently be running slow at the end of one plane circuit.

To a fair approximation, Earth's circumference is ~ 40,000km. So the plane would take 40,000/1,200hrs = 100/3 hrs, according to clock B. Approximate this by 33hrs.

Then clock A would show 33hrs / 1.0008692.. ~ 32.97134... hrs or 32h 58m 17s. (I'm keeping more figures because I want the difference.)

Thus clock A would appear to be slow by ~ 0.02866 hrs = 1.720min, or ~1min 43s.

Live long and prosper.

« Previous12345...1415Next »

Galileo - the atomic clock