Doppler Shift: Communication from a Space Probe

1. The problem statement, all variables and given/known data
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.)

2. Relevant equations
c=3E8
ƒobserved = ƒsourse (1 [itex]\pm[/itex] (rel Velocity/c))

3. 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[itex]\pm[/itex](rel V/c))

ƒo = ƒs(1[itex]\pm[/itex](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[itex]\pm[/itex](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.

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