Question

What is the ratio of the sun’s gravitational pull on Mercury to the sun’s gravitational pull on the earth?

Answer 1

Answer:

The answer is $\frac{F_{Sun-Mercury} }{F_{Sun-Earth} } =0,3709$. Let's learn why.

Explanation:

Newton's law of universal gravitation says;

$F_{g} =G.\frac{m_{1}.m_{2}}{r^{2}}$

Here G is a universal gravitational constant and is measured experimentally.

Sun's gravitational pull on mercury is:

$F_{Sun-Mercury} =G.\frac{m_{sun}.3,30.10^{23}}{(5,79.10^{10})^{2} }$

Therefore $F_{Sun-Mercury} = Gm_{sun} 98,4366$

Sun's gravitational pull on Earth is:

$F_{Sun-Earth} =G.\frac{m_{sun} 5,97.10^{24} }{(1,50.10^{11}) ^{2}}$

Therefore $F_{Sun-Earth} =Gm_{sun} 265,33$

As a result;

$\frac{F_{Sun-Mercury}}{F_{Sun-Earth} }=\frac{Gm_{sun}98,4366}{Gm_{sun}265,33 } =0,3709$