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How did Voyager 2 communicate to earth?
Name: Brian Lintz
Status: N/a
Age: N/A
Location: N/A
Country: N/A
Date: Around 1993
Question:
The message sending device that Voyager 2 used to send pictures of
Uranus was using 2 watts of power. Why was the power delivered to the Earth
less than a trillionth of this? Was not the transmitter directed toward the
Earth? Does not the signal dissipation follow the inverse square law? Why do
so many science articles use watts as units of energy instead of power? Why
do not they use joules? For example, Uranus receives 1/400th the solar light
and heat that the Earth gets. Uranus is 20 A.U. from the sun while Earth is
1 A.U. 1/400 = 1/20(squared): So the above makes sense according to the
inverse square law. Then why does not the same reasoning apply to energy
coming from Uranus to Earth? Why is not the radiation decreased by
(1/19)squared? Why is it really decreased by a trillionth instead?
Replies:
Your understanding of the physics is excellent. Most writers do
not understand the difference between energy and power. Much of what is
written is thus not accurate. The inverse square law is accurate and
correctly applied here. Your arguments are OK for comparing the amount of
energy from the sun as a relative ratio. Here we want to know the power we
would see over a certain area at a distance of R from the source. So our
formula would be that the intensity over one square meter would be 4 * pi * (2
watts)/(R*R) where R is the distance in meters. The distance by my rough
estimate should be 2.8 x 10^12. If the antenna has been isotropic the power
should have been down by a factor of 10 to the negative 24 , 10 ^(-24) instead
of only a trillion. If they are accurate in the report, it means that the
directionality of the antenna was pretty great. Your understanding of the
principles is great. I am now surprised that the signal is so large when it
gets back.
Sam Bowen
The first thing to realize about physics is that it does not just
consist of mathematical formulas, but every formula has (or should have) an
understandable meaning or implication. You were quite right to think of the
inverse square law here, but because you apparently did not really understand
what that law means, you applied it incorrectly. I hope the following
discussion will make things clearer! First off, what could an inverse square
law mean? Let us try and think of something that grows as the second power of
distance. The first thing that should come to mind is area, since an area is
always given by a product of two distances. So, does the inverse square law
have anything to do with inverse areas? Imagine a point source of light.
After the light is turned on, it spreads out in all directions at the speed of
light and we can imagine traveling with the light as it goes on its way. At
greater and greater distances from the light source, the same amount of light
is spread out over a bigger and bigger area. Ah-Hah! So, how does that area
change with distance? As the square of the distance of course! (The area of
a sphere of radius r is 4 pi r^2.) So, if we detect light with our eyes, or
with our fixed antenna, or whatever, the area of detection is the same for
every distance from the source, and therefore we see a smaller and smaller
portion of the light as we go further and further away. What is that small
fraction? It is the ratio of the area of our detector (some fixed number) to
the total area that the light is spread over, i.e. the fraction is some
constant divided by the area of the big sphere the light is now spread over,
so that it decreases as 1/r^2 as the distance r gets bigger and bigger. That
the meaning of the inverse square law. Now actually, it does not matter that
the source is radiating equally in all directions, because from far enough
away, light from any kind of source (even a laser) has to spread apart over
some possibly tiny portion of the sphere of radius r. A laser will have light
spread over a very tiny portion of the sphere, but nevertheless, it will (from
far enough away) also obey this inverse square law. So, to go back to your
specific question - with the light from the sun, the earth is at 1/20 the
distance of Uranus from the source (the sun), and therefore the area of the
sphere the light from the sun is spread over is 400 times as big when you are
out as far as Uranus, and therefore the light intensity is only 1/400 as
great. Fine. Now let us turn to the Voyager space craft, sending 2 Watts of
radio waves to the earth. You asked, why does not the earth receive 1/400 of
2 Watts? Well, where would this 1/400 come from? Remember, the 1/400 for the
light from the sun came from the ratio of two areas - the sphere at the
earth's distance from the sun, and the sphere at Uranus' distance from the
sun. What are the two areas we are taking a ratio of for Voyager? Well, we a
not interested in the light intensity (power per unit area) this time, but in
the total received power. So, let us say we have some specific radio detector
in mind on the earth - say the Arecibo one, which is about 1 mile across. The
area of our detector is thus about 1 square mile. What is the other area we
need to divide into this 1 square mile? It is the total area that Voyager's 2
Watts must be spread over by the time it reaches the earth. The distance to
Voyager is about 20 Astronomical units or about 2 Billion miles. (1 AU is 93
million miles, if I recall correctly.) So the area of a sphere at that
distance is somewhere around 10 Billion square miles. Now, Voyager's antenna
does have some directionality, so the radio waves spread out over only a small
fraction of that total sphere. Even if that fraction was only 1 millionth,
however, the area we are talking about is still 10 thousand billion, or ten
trillion square miles, so the power being received by our mile-wide receiver
is going to be perhaps 2 tenths of a trillionth of a Watt.
Arthur Smith
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Update: June 2012
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