How long would it take radio signals to travel from Earth to Mercury
and back if Mercury was at its farthest point from Earth?

Hi novellahub

It would take electromagnetic waves (radio, visible light, x-ray etc)
23 minutes and 12 seconds to travel from Earth to the planet Mercury
and back again. This is how to perform the calculation.

Mercury is at its furthest point from earth when its orbit takes it
directly opposite the sun from our planet. Mercury orbits at a mean
radius of 57,910,000 km from the sun, our planet has a mean orbit of
149,600,000 km. The sun itself has a diameter of 1,390,000 km, so when
they are furthest apart the two planets have a separation of (57.9 x
10^6 km + 149.6 x 10^6 km + 1.390 x 10^6 km) -- that's the size of
earth's orbit, plus the size of mercury's orbit, plus the diameter of
the sun, and gives a grand total of 208.9 x 10^6 km.
http://seds.lpl.arizona.edu/nineplanets/nineplanets/mercury.html
http://seds.lpl.arizona.edu/nineplanets/nineplanets/earth.html
http://seds.lpl.arizona.edu/nineplanets/nineplanets/sun.html

This figure can be represented in metres as 208.9 x 10^9 (208 billion
metres), and the speed of light in a vacuum, in metres per second, is
300 x 10^6 m/s.
http://www.what-is-the-speed-of-light.com/

So now it's time to do some simple arithmetic. (208.9 x 10^9) / (300 x
10^6) = 696.3 seconds. However we want a round-trip time, so
multiplying by 2 we get 1392.6. As we know that there are 60 seconds
in a minute this tells us that our radio waves would take 23 minutes
and 12 seconds to complete their journey from Earth to Mercury and
back again.


I hope this answers the question to your satisfaction; if you'd like
anything explained in greater detail then please ask for
clarification.


Regards
iaint-ga


Google searches performed:
orbit planet mercury
"speed of light"


For the pedants, I have to point out that of course when mercury and
earth are at their greatest separation, then the sun would lie between
them and so radio waves wouldn't make it from one planet to the other.
The calculations above have completely ignored this problem -- if it
troubles you, imagine the calculation is actually for a neutrino or
some other particle which travels at near enough light speed and has
very weak interaction with matter!

I thought the orbit radius was calculated around the Sun's center, not
the surface. Not that is matters too much in the overall time... about
9s ( 1.39 x 10^6 Km x 2 / (300 x 10^3 km/s). Any ideas ?

Curepi

There are many aspects of the posted answer which are misguided, or
simply wrong.  This is an unfortunate departure from the usual high
quality of answers provided by this service.

First, of course, curepi is correct: the diameter of the sun has no
place in this calculation, because orbits are measured by the size of
the ellipse they trace out, not by distance from the surface of the
sun.  Mercury and the Earth are at their absolute greatest distance
when each is at aphelion, and they are on opposite sides of the sun. 
Thus using mean distance from sun introduces errors of order 10
million km, so doing calculations with four significant figures is
meaningless.  Further, if one did inexplicably insist on retaining the
extra digits in such an approximate calculation as this, the same
number of digits should be kept in all quantities.  Thus, for example,
the speed of light is 299.8 million m/s, not 300.  The note at the end
is also nonsense, because due to the slight obliquity of the orbits of
the planets, it is rather rare for Mercury to pass across the disk of
the sun (either between the earth and sun, or on the far side).

The true maximum distance, given, for example, at
http://www.tcaep.co.uk/astro/planets/mercury/ 

is 221.9 million km.  That's 24 min 40 sec there-and-back travel time.

another comment, is that your radio signal may not even make it, if
the direction is too close to the sun, as there would likely be
intereference caused by the sun's activity and mass.

What an interesting example of how a few minor details can increase
the answer by about 6%. Perhaps, I can justify rounding the answer to
a full 25 minutes. Everyone is assuming this is a 2 dimentional
problem, but the plain of Mercury's orbit is tilted about 9 degrees
with respest to the eccliptic = the plane of the Earth's orbit. To a
minor extent both elipses have some minor 3 dimenional distortions due
mostly to the position of Jupiter and Venus, so still longer paths (by
a few parts per million) occur several times per billion years. The
aphelion of Earth and especially Mercury rotate with respect to each
other signifcantly in a few hundred years. Since the sun's gravity can
not increase the speed of light, it is reasonable to assume it
decreases it to perhaps an average of 299.7, and the ions streaming
out from the sun decrease the speed of light to 299.6 KMPS. Since the
gravity of the sun bends the radio signal (or light) the path length
likely increases by several parts per million. Likely a knife edge
effect would allow comunications to continue even when Mercury was
centered exactly behind the sun, but the effective radiated power
likely needs to increase a few trillion watts to compensate for
increased path noise. Sorry I don't know how to calculate any of these
effects. It seems to me Mercury at aphelion and Earth a perrihillion
would produce the longest path, but that is likely how racecar
calculated the 23 minutes and 40 seconds.  Neil

I appologize: the longest path will be when Mercury and Earth are
farthest from the sun, simultaneously, but this only occurs (to 3
decmil places) about twice per decade. I also apologize for typing 23
minutes and 40 seconds for racecar's answer. 24 minutes and 40 seconds
was his answer. If Mercury occassionly is 40 million miles from the
sun and Earth is occassionly 93 million miles from the sun: The path
is occassionly 133 million miles, add one milion in longest 3
dimentional solutions = 134 million miles at 186,000 miles per second
= 720 seconds one way = 1440 seconds round trip; divide by 60 = 24
minutes which is in line with the other answers. which is ok as none
of my numbers are likely accurate to 4 decimil places (other than 60
seconds in a minute)  Neil

I must be delerious from my cold. I commited one of my pet peves. I
should have said + or minus one part per thousand instead of decimil
places which I likely spelled wrong besides. I agree, astronomical
distances are almost always measured from the mass center, rather than
the surface, even when the wording infers surface. I wish answers
google allowed us to edit our comments and provided a spell check. 
Mercury is wrongly thought to be a poor place for a human habitat.
Temperature measurements are typically -133 degrees in the shade, even
when 600 degrees warmer in the sun is typical. Sorry, I don't know if
that is degrees F or c, but you get the idea. Also Mercury has almost
no tilt on it's axis, so the bottom of shallow craters, near the north
and south pole, of Mercury, never get hot until we put a colony there
which will warm the crater with the waste heat. The bombardment of
ions is rather sever, especially if the magnetic poles are close to
the geographic poles, but thin shilding would protect the colonists
from radiation almost as well as the same thickness on the Moon of
Earth. Micrometerites may be no grater hazzard than on Earth's Moon
since Mercury is farther from the asteroid belt. Likely it is hot more
than a few meter below the crater surface, but that may be useful as a
geothermal energy source. Clearly it is hot just outside the crater
about half of the time. It would be practical to explore breifly
outside the crater at several month intervals. Fast ions would be at
least a minor radiation hazard. The extreme temperature differences
are because Mercury only has about one millionth as much atmosphere as
Earth.  The bulk of Mercury would protect the colonist at either the
north or the south pole of Mercury from a super nova that was close
enough to kill all other humans in this solar system. The 9 degrees
difference between the plane of the orbit of Mercury and the plane of
the orbit of Earth allows telescopes to see most of the surface of
Mercury over a few months period.  Neil