Question

An aseroid with a mass of 8.4x10^8 kg and a planet with a mass of 6.2x10^23 kg come close to each other by a distance of 8x10^5 m. what is the force of gravity that the asteroid and the planet have on each other ?

Answer 1

So, the force of gravity that the asteroid and the planet have on each other approximately

$\boxed{\sf{5.43 \times 10^{10} \: N}}$

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 :

$\boxed{\sf{\bold{F = G \times \frac{m_1 \times m_2}{r^2}}}}$

With the following condition :

• F = gravitational force (N)
• G = gravity constant ≈ $\sf{6.67 \times 10^{-11}}$ N.m²/kg²
• $\sf{m_1}$ = mass of the first object (kg)
• $\sf{m_2}$ = mass of the second object (kg)
• r = distance between two objects (m)

Problem Solving

We know that :

• G = gravity constant ≈ $\sf{6.67 \times 10^{-11}}$ N.m²/kg²
• $\sf{m_1}$ = mass of the first object = $\sf{8.4 \times 10^8}$ kg.
• $\sf{m_2}$ = mass of the second object = $\sf{6.2 \times 10^{23}}$ kg.
• r = distance between two objects = $\sf{8 \times 10^5}$

What was asked :

• F = gravitational force = ... N

Step by step :

$\sf{F = G \times \frac{m_1 \times m_2}{r^2}}$

$\sf{F = 6.67 \times 10^{-11} \times \frac{8.4 \cdot 10^8 \times 6.2 \cdot 10^{23}}{(8 \times 10^5)^2}}$

$\sf{F = \frac{347.374 \times 10^{-11 + 8 + 23}}{64 \times 10^10}}$

$\sf{F \approx 5.43 \times 10^{20 - 10}}$

$\boxed{\sf{F \approx 5.43 \times 10^{10} \: N}}$

Conclusion

So, the force of gravity that the asteroid and the planet have on each other approximately

$\boxed{\sf{5.43 \times 10^{10} \: N}}$

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