Doppler Shift: Communication from a Space Probe

[blue_lilly]

Homework Statement

Space probes communicate with controllers on Earth using radio waves. Suppose a probe has a radio-transmitter that operates at a frequency of 516.0 MHz when at rest relative to the observer. This probe is now wandering the solar system. The signal you pick up from the probe is shifted 5.955E+4 Hz higher in frequency than the rest frequency given above. How fast is the probe traveling and is it traveling towards or away from Earth? (assume it is either traveling directly towards or directly away from Earth. You must get both answers correct at the same time.)

Homework Equations

c=3E8
ƒobserved = ƒsourse (1 \pm (rel Velocity/c))

The Attempt at a Solution

I know that this is a one way Doppler Shift, I am the observer and the source is the probe. Because it a Doppler Shift I am going to use this formula, ƒobserved=ƒsourse(1\pm(rel V/c))

ƒo = ƒs(1\pm(rel V/c))
ƒs: I am given the ƒs in MHz so I need to convert to Hz.

516MHz = 516000000Hz = 5.16E8Hz​

ƒo: In the problem, it says "The signal you pick up from the probe is shifted 5.955E+4Hz (59550Hz) higher in frequency". This means that the ƒrequency that is obsorbed is the ƒrequency of the source plus the amount it is shifted.

ƒo = ƒs + shift
ƒo = 516000000Hz + 59550Hz = 516059550Hz = 5.1605955E8Hz​

c: The speed of light is 3E8.
relV: We are solving for this.

ƒo = ƒs(1\pm(relV /c))
516059550Hz = 516000000Hz (1-( relV / 3E8 ))
59550Hz = 1-( relV / 3E8 )
59549Hz = relV / 3E8
(59549Hz)(3E8) = relV
relV = 1.78647E13 m/s

So the relV is 1.78647E13 m/s and the direction would be towards Earth because the ƒo is louder then the ƒs.

But the answer is incorrect and I am not sure what I did wrong.
Any help would be GREATLY appreciated.

 

[paisiello2]

How can anything travel faster than light? 1.78647E13 m/s is a huge number.

I think you simply did an arithmetic mistake in step 3 of solving for relV. You would have caught this had you put in all the units.

Lastly, a frequency isn't "louder" than another frequency. It is simply higher or lower.

 

[berkeman]

Check the algebra in this step:

516059550Hz = 516000000Hz (1-( relV / 3E8 ))
59550Hz = 1-( relV / 3E8 )

As paisiello2 says, you should carry units along in your calculations, to help you catch mistakes like this...

 

[blue_lilly]

Check the algebra in this step:

As paisiello2 says, you should carry units along in your calculations, to help you catch mistakes like this...

I keep trying to fix the carry over numbers but its not working very well, so sorry if it looks really confusing.

ƒo = ƒs(1±(relV /c))
516059550Hz = 516000000Hz (1-( relV / 3E8 ))
-516000000Hz -516000000Hz
59550Hz = 1-( relV / 3E8 )
-1 -1 [subtract 1 from both sides]
59549Hz = - (relV / 3E8)
x(3E8) x(3E8) [multiply both sides by 3E8]
(59549Hz)(3E8) = -relV
-relV = (59549Hz)(3E8)
relV = -1.78647E13 m/s [reverse neg sign]

I rewrote the problem in my note book and the answer is still wrong. Am I not seeing the algebra mistake?

 

[berkeman]

The algebra mistake is still here:

516059550Hz = 516000000Hz (1-( relV / 3E8 ))
59550Hz = 1-( relV / 3E8 )

The units error is that you have Hz on the LHS of the equation, and the RHS is unitless.

That's because you subtracted 516000000Hz from both sides, instead of...

 

[blue_lilly]

The algebra mistake is still here:



The units error is that you have Hz on the LHS of the equation, and the RHS is unitless.

That's because you subtracted 516000000Hz from both sides, instead of...

Does this fix the unit error? 516059550Hz = 516000000Hz (1-( relV / 3E8m/s^2 ))

Instead of subtracting 516000000Hz, should I have added 516000000Hz to both sides?

 

[berkeman]

Does this fix the unit error? 516059550Hz = 516000000Hz (1-( relV / 3E8m/s^2 ))

Instead of subtracting 516000000Hz, should I have added 516000000Hz to both sides?

Nope.

If you want to solve this equation: 15x = 30

What do you do to isolate x? You certainly don't subtract 15 from both sides...

 

[blue_lilly]

Nope.

If you want to solve this equation: 15x = 30

What do you do to isolate x? You certainly don't subtract 15 from both sides...

ohhh, i feel like an idiot

ƒo = ƒs(1±(relV /c))
516059550Hz = 516000000Hz (1-( relV / 3E8 ))
/516000000Hz /516000000Hz
1.000115407Hz = 1-( relV / 3E8m/s^2 )
1.154067E-4 = - relV / 3E8m/s^2
x(3E8m/s^2) x(3E8m/s^2)
(1.154067E-4)(3E8m/s^2)
34622.09 = - relV
- relV = 34622.09
relV = - 34622.09 m/s^2

SO does this look right then? Also would the neg sign mean that it is traveling away from you?

 

[berkeman]

ohhh, i feel like an idiot

ƒo = ƒs(1±(relV /c))
516059550Hz = 516000000Hz (1-( relV / 3E8 ))
/516000000Hz /516000000Hz

1.000115407[STRIKE]Hz[/STRIKE] = 1-( relV / 3E8m/s[STRIKE]^2[/STRIKE] )

1.154067E-4 = - relV / 3E8m/s[STRIKE]^2[/STRIKE]
x(3E8m/s[STRIKE]^2[/STRIKE]) x(3E8m/s[STRIKE]^2[/STRIKE])

(1.154067E-4)(3E8m/s[STRIKE]^2[/STRIKE])=34622.09 = - relV

- relV = 34622.09

relV = - 34622.09 m/s[STRIKE]^2[/STRIKE]

SO does this look right then? Also would the neg sign mean that it is traveling away from you?

Closer. The units are not correct, though. See my edits above...

The units of velocity are m/s, not m/s^2. And when you divided Hz/Hz in the first step, that makes the LHS dimensionless, just like the RHS. The units of the LHS and RHS of any equation have to match.