Math Problem Statement

What is the magnitude of the attractive force between object A with a mass of 4,750 kg and object B with a mass of 8,568 kg if the distance between the two bodies is 50 m? The Universal Constant of Gravitation (G) is 66.73 * 10^-12 m^3/(kg*s^2). Give your answer in micronewtons (µN) using three significant digits.

Solution

The problem in the image asks for the gravitational force between two objects. We can calculate this using Newton's Law of Universal Gravitation:

F=Gm1m2r2F = G \frac{m_1 m_2}{r^2}

Where:

  • FF is the gravitational force.
  • GG is the gravitational constant, given as 66.73×1012m3/kgs266.73 \times 10^{-12} \, \text{m}^3/\text{kg}\cdot\text{s}^2.
  • m1=4750kgm_1 = 4750 \, \text{kg} (mass of object A).
  • m2=8568kgm_2 = 8568 \, \text{kg} (mass of object B).
  • r=50mr = 50 \, \text{m} (distance between the two bodies).

Let's calculate the force:

F=(66.73×1012)×(4750×8568)502F = \left( 66.73 \times 10^{-12} \right) \times \frac{(4750 \times 8568)}{50^2}

  1. Calculate the masses' product:

4750×8568=40653000kg24750 \times 8568 = 40653000 \, \text{kg}^2

  1. Calculate the square of the distance:

502=2500m250^2 = 2500 \, \text{m}^2

  1. Plug values into the formula:

F=(66.73×1012)×406530002500F = \left( 66.73 \times 10^{-12} \right) \times \frac{40653000}{2500}

  1. Simplify:

406530002500=16261.2\frac{40653000}{2500} = 16261.2

  1. Now multiply by the gravitational constant:

F=66.73×1012×16261.2=1.085×107NF = 66.73 \times 10^{-12} \times 16261.2 = 1.085 \times 10^{-7} \, \text{N}

  1. Convert to micronewtons (μN\mu N):

1.085×107N=108.5μN1.085 \times 10^{-7} \, \text{N} = 108.5 \, \mu N

Thus, the gravitational force between the two objects is approximately 108.5 µN.

Let me know if you'd like more details on any of these steps!

Here are some related questions to explore:

  1. How does the force change if the distance between the objects is doubled?
  2. What would happen to the force if the mass of object A is halved?
  3. How does the gravitational force compare to other fundamental forces?
  4. Why is the gravitational constant so small?
  5. What are the real-world applications of calculating gravitational forces?

Tip: The force of gravity decreases rapidly with distance, as it is inversely proportional to the square of the distance between the two masses.

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Math Problem Analysis

Mathematical Concepts

Gravitational Force
Newton's Law of Universal Gravitation
Unit Conversion

Formulas

F = G * (m1 * m2) / r^2

Theorems

Newton's Law of Universal Gravitation

Suitable Grade Level

High School, College