When the Voyager 2 spacecraft passed Neptune in 1989, it was 4.5 * 10^9 km from

Question

When the Voyager 2 spacecraft passed Neptune in 1989, it was 4.5 * 10^9 km from the Earth. Its radio transmitter, with which it sent back data and images, broadcast with a mere 21 W of power.

Assuming that the transmitter broadcast equally in all directions,

What are the signal intensity recieved at Earth, (I = ?) and What electric field amplitude was detected (E = ? V/m)? The received signal was somewhat stronger than your result because the spacecraft used a directional antenna, but not by much.

Explanation / Answer

(a) The radio transmitter is radiating energy in all directions at the rate of 21 J per second. The signal intensity received at the earth is given by Equation: I =P/A = P/4*pi*r2 = 21 W/4*pi*(4.5 "10^12)2 = 8.252 * 10^-26 W/m^2 (b) Using Equation, I = c(permittivity of free space)/2 *E0^2= sqrt(2I/c*permittivity of free space) I = sqrt[2 (8.252* 10^-26)/(3*10^8*8.85*10^-12)] => I= 7.89* 10^-12 V/m

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.