So, the force of gravity that the asteroid and the planet have on each other approximately [tex] \boxed{\sf{2.9 \times 10^{17} \: N}} [/tex]
Hi ! Now, I will help to discuss about the gravitational force between two objects. The force of gravity is not affected by the radius of an object, but radius between two object. Moreover, if the object is a planet, the radius of the planet is only to calculate the "gravitational acceleration" on the planet itself,does not determine the gravitational force between the two planets. For the gravitational force between two objects, it can be calculated using the following formula :
[tex] \boxed{\sf{\bold{F = G \times \frac{m_1 \times m_2}{r^2}}}} [/tex]
With the following condition :
We know that :
What was asked :
Step by step :
[tex] \sf{F = G \times \frac{m_X \times m_Y}{r^2}} [/tex]
[tex] \sf{F = 6.67 \cdot 10^{-11} \times \frac{1.55 \cdot 10^{22} \cdot 3.95 \times 10^{28}}{(3.75 \times 10^{11})^2}} [/tex]
[tex] \sf{F \approx \frac{40.84 \times 10^{-11 + 22 + 28}}{14.0625 \times 10^{22}}} [/tex]
[tex] \sf{F \approx 2.9 \times 10^{39 - 22}} [/tex]
[tex] \sf{F \approx 2.9 \times 10^{17} \: N} [/tex]
So, the force of gravity that the asteroid and the planet have on each other approximately
[tex] \boxed{\sf{2.9 \times 10^{17} \: N}} [/tex]
