Answered

Westonci.ca offers fast, accurate answers to your questions. Join our community and get the insights you need now. Connect with a community of experts ready to provide precise solutions to your questions on our user-friendly Q&A platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.

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 ?

Sagot :

MDKPfromIDN

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

Gravity is a thing has depends on ... https://brainly.com/question/26485200

We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.