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 BowenThe 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 Click here to return to the Physics Archives

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