Answer:
The answer is [tex]\frac{F_{Sun-Mercury} }{F_{Sun-Earth} } =0,3709[/tex]. Let's learn why.
Explanation:
Newton's law of universal gravitation says;
[tex]F_{g} =G.\frac{m_{1}.m_{2}}{r^{2}}[/tex]
Here G is a universal gravitational constant and is measured experimentally.
Sun's gravitational pull on mercury is:
[tex]F_{Sun-Mercury} =G.\frac{m_{sun}.3,30.10^{23}}{(5,79.10^{10})^{2} }[/tex]
Therefore [tex]F_{Sun-Mercury} = Gm_{sun} 98,4366[/tex]
Sun's gravitational pull on Earth is:
[tex]F_{Sun-Earth} =G.\frac{m_{sun} 5,97.10^{24} }{(1,50.10^{11}) ^{2}}[/tex]
Therefore [tex]F_{Sun-Earth} =Gm_{sun} 265,33[/tex]
As a result;
[tex]\frac{F_{Sun-Mercury}}{F_{Sun-Earth} }=\frac{Gm_{sun}98,4366}{Gm_{sun}265,33 } =0,3709[/tex]
