how does the relative position of mercury and earth affect our calculation for orbital Velocity?

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Expert Solution

Depending on the relative position of the Earth and Mercury, the radar pulse takes between 10 and 30 minutes to travel to Mercury, bounce off, and return.

The frequencies of the returning echoes are different from the frequency of the pulse sent out because they have bounced off the moving surface of Mercury. The part of the signal returned from the approaching edge will come back with an increase in frequency (blueshift) and the part returned from the receding edge will show a decrease in frequency (redshift). Any time a source of radiation is moving radially (towards or away from the observer) there will be a Doppler shift in the received frequency (or wavelength; recall λ = c/f) that is proportional to the velocity along the line of sight. This can be expressed like so:

v_o / c = f / f

where c is the speed of light (3×108 m/s); ∆f is the observed change or shift in frequency (∆f = f – fobs; f is the original frequency that the source would emit were it not moving; and vo is the speed of the source with respect to you along that line of sight. There are two motions of the planet that can produce such a shift: its orbital velocity as a whole around the Sun, and its rotation about its axis. The first echo, from the sub-radar point, is shifted in frequency only by the orbital velocity of the planet as a whole.

In case of any doubt, please do comment sir.

genius_generous answered 1 year ago

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