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Sagot :
So, the force of gravity that the asteroid and the planet have on each other approximately
[tex] \boxed{\sf{5.43 \times 10^{10} \: N}} [/tex]
Introduction
Hi ! Now, I will help to discuss about the gravitational force between two objects. We already know that gravitational force occurs when two or more objects interact with each other at a certain distance and generally orbit each other to their center of mass. 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 :
- F = gravitational force (N)
- G = gravity constant ≈ [tex] \sf{6.67 \times 10^{-11}} [/tex] N.m²/kg²
- [tex] \sf{m_1} [/tex] = mass of the first object (kg)
- [tex] \sf{m_2} [/tex] = mass of the second object (kg)
- r = distance between two objects (m)
Problem Solving
We know that :
- G = gravity constant ≈ [tex] \sf{6.67 \times 10^{-11}} [/tex] N.m²/kg²
- [tex] \sf{m_1} [/tex] = mass of the first object = [tex] \sf{8.4 \times 10^8} [/tex] kg.
- [tex] \sf{m_2} [/tex] = mass of the second object = [tex] \sf{6.2 \times 10^{23}} [/tex] kg.
- r = distance between two objects = [tex] \sf{8 \times 10^5} [/tex]
What was asked :
- F = gravitational force = ... N
Step by step :
[tex] \sf{F = G \times \frac{m_1 \times m_2}{r^2}} [/tex]
[tex] \sf{F = 6.67 \times 10^{-11} \times \frac{8.4 \cdot 10^8 \times 6.2 \cdot 10^{23}}{(8 \times 10^5)^2}} [/tex]
[tex] \sf{F = \frac{347.374 \times 10^{-11 + 8 + 23}}{64 \times 10^10}} [/tex]
[tex] \sf{F \approx 5.43 \times 10^{20 - 10}} [/tex]
[tex] \boxed{\sf{F \approx 5.43 \times 10^{10} \: N}} [/tex]
Conclusion
So, the force of gravity that the asteroid and the planet have on each other approximately
[tex] \boxed{\sf{5.43 \times 10^{10} \: N}} [/tex]
See More
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