Answer:
The ratio of the tidal influence of the doctor to the tidal influence of Mercury is 2.0 × 10¹¹.
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).
[tex]T=k.\frac{m}{d^{3} }[/tex]
The tidal influence of the doctor (Td) is:
[tex]Td=k.\frac{m}{d^{3} }= k. \frac{85.00kg}{(1m)^{3} } =k.85kg/m^{3}[/tex]
The tidal influence of mercury (Tm) is:
[tex]Tm=k.\frac{m}{d^{3} }= k. \frac{3.30 \times 10^{23} kg}{(9.2 \times 10^{10})^{3} } =k.4.2 \times 10^{-10}kg/m^{3}[/tex]
The ratio Td/tm is:
[tex]\frac{Td}{Tm} =\frac{k.85kg/m^{3}}{k.4.2 \times 10^{-10}kg/m^{3}} =2.0 \times 10^{11}[/tex]