A navigational beacon in deep space broadcasts at a radio frequency of 50 MHz. A spaceship approaches the beacon with a relative velocity of 0.40c. What is the frequency of the beacon radio signal that is detected on the ship? Answer in MHz with no decimal places. A) 55 MHz
B) 60 MHz
C) 66 MHz
D) 71 MHz
E) 76 MHz

To first order  f' = f (1 - u/c) where observer and source are separating

f' = (1 + .4) = 70 MHz   so (d) would be correct

The next term would be 1/2 (u/c)^2 = 1/2 * ,4^2 = .08 and the correction would be .08 * 50 = 4 Mhz

Most approximations would use (d).

potential energy is directly proportional to height.

more height = more potential energy.

greatest increase in height = greatest increase in potential energy

where are the points with the greatest increase in height ?

has to be c-e or d-e .