Answer:
The ratio of the tidal influence of the doctor to the tidal influence of Mercury is 2.0 × 10¹¹.
Step-by-step explanation:
Tidal influence (T) is proportional to the mass of a disturbing body (m) and is inversely proportional to the cube of its distance (d).
![T=k.(m)/(d^(3) )](https://img.qammunity.org/2020/formulas/chemistry/college/ff6jythczrtdhuagb5nia1nrqxdh41tabs.png)
The tidal influence of the doctor (Td) is:
![Td=k.(m)/(d^(3) )= k. (85.00kg)/((1m)^(3) ) =k.85kg/m^(3)](https://img.qammunity.org/2020/formulas/chemistry/college/wtzy4qbdytc3ppomlxap4xg6x1qx26d74f.png)
The tidal influence of mercury (Tm) is:
![Tm=k.(m)/(d^(3) )= k. (3.30 * 10^(23) kg)/((9.2 * 10^(10))^(3) ) =k.4.2 * 10^(-10)kg/m^(3)](https://img.qammunity.org/2020/formulas/chemistry/college/w16ojjpxqkx914595xjylm5rrmyvpo5pij.png)
The ratio Td/tm is:
![(Td)/(Tm) =(k.85kg/m^(3))/(k.4.2 * 10^(-10)kg/m^(3)) =2.0 * 10^(11)](https://img.qammunity.org/2020/formulas/chemistry/college/qn461c62jh1qlu2otq14h8zyh725nlzmvg.png)