Calculating Time Delay for Radio Telescope Pointing

[toothpaste666]

Homework Statement

A radio telescope, whose two antennas are separated by 55 m, is designed to receive 3.0-MHz radio waves produced by astronomical objects. The received radio waves create 3.0-MHz electronic signals in the telescope's left and right antennas. These signals then travel by equal-length cables to a centrally located amplifier, where they are added together. The telescope can be "pointed" to a certain region of the sky by adding the instantaneous signal from the right antenna to a "time-delayed" signal received by the left antenna a time Δt ago. (This time delay of the left signal can be easily accomplished with the proper electronic circuit.)
If a radio astronomer wishes to "view" radio signals arriving from an object oriented at a 12 ∘ angle to the vertical as in the figure (Figure 1) , what time delay Δt is necessary?

Homework Equations

dx/l = m(lambda)

The Attempt at a Solution

i use dx/l = dsin(theta)
x = lsin(theta)
they give me l = 55 m theta = 12 degrees
x = 55sin(12) = 11.4 m
which is the distance ahead that one sound wave is of the other
sound travels 343 m/s so it travels that distance in
1 s/ 343m (11.4 m) = .033 seconds
so that is how much they need to delay the signal.

is this valid?

 

[berkeman]

Homework Statement

A radio telescope, whose two antennas are separated by 55 m, is designed to receive 3.0-MHz radio waves produced by astronomical objects. The received radio waves create 3.0-MHz electronic signals in the telescope's left and right antennas. These signals then travel by equal-length cables to a centrally located amplifier, where they are added together. The telescope can be "pointed" to a certain region of the sky by adding the instantaneous signal from the right antenna to a "time-delayed" signal received by the left antenna a time Δt ago. (This time delay of the left signal can be easily accomplished with the proper electronic circuit.)
If a radio astronomer wishes to "view" radio signals arriving from an object oriented at a 12 ∘ angle to the vertical as in the figure (Figure 1) , what time delay Δt is necessary?

Homework Equations

dx/l = m(lambda)

The Attempt at a Solution

i use dx/l = dsin(theta)
x = lsin(theta)
they give me l = 55 m theta = 12 degrees
x = 55sin(12) = 11.4 m
which is the distance ahead that one sound wave is of the other
sound travels 343 m/s so it travels that distance in
1 s/ 343m (11.4 m) = .033 seconds
so that is how much they need to delay the signal.

is this valid?

Hint -- (don't use the speed of sound...) :-)

 

[toothpaste666]

ok so in that case i would use the given frequency and use dx/l = dsin(theta) = m(lambda) to find the wave length and then use that and the given frequency to find the speed?
is the rest of my method correct? in other words when i find that speed can i use that and the distance x = 11.4 to find the time?
I am having a hard time picturing these optics problems. I am not a very geometric thinker.

 

[berkeman]

ok so in that case i would use the given frequency and use dx/l = dsin(theta) = m(lambda) to find the wave length and then use that and the given frequency to find the speed?
is the rest of my method correct? in other words when i find that speed can i use that and the distance x = 11.4 to find the time?
I am having a hard time picturing these optics problems. I am not a very geometric thinker.

You don't need to find the speed. The incoming radio waves travel not at the speed of sound, but at the speed of ________.

Draw a line from the left telescope to a point on the incoming radio wave line to the right telescope. Make that line perpendicular to both incoming radio wave lines. See how the intersection on the right incoming radio wave line is *above* the telescope by some distance? That's the extra distance that the radio waves have to travel to get to the right antenna, as compared to the left antenna. Traveling that extra distance takes a bit of time, which is found by knowing the distance, and the speed of travel... :-)

 

[toothpaste666]

oh yeah radio waves are electromagnetic waves so they travel at c (the speed of light?
so using
1s/3E8 m (11.4 m) = 3.8E-8 s

 

[toothpaste666]

ah yes i get it now. thank you!

 

[berkeman]

:-)