5. What is the gravitational force between a [tex]45 \, \text{kg}[/tex] person and the Earth ([tex]5.98 \times 10^{24} \, \text{kg}[/tex]) with a distance of [tex]6.38 \times 10^6 \, \text{m}[/tex]?

- Object 1 ([tex]m_1[/tex]): [tex]45 \, \text{kg}[/tex] person
- Object 2 ([tex]m_2[/tex]): [tex]5.98 \times 10^{24} \, \text{kg}[/tex] Earth
- Distance ([tex]r[/tex]): [tex]6.38 \times 10^6 \, \text{m}[/tex]

A. [tex]227 \, \text{N}[/tex]
B. [tex]398 \, \text{N}[/tex]
C. [tex]441 \, \text{N}[/tex]
D. [tex]510 \, \text{N}[/tex]



Answer :

To solve the problem of determining the gravitational force between a [tex]\( 45 \, \text{kg} \)[/tex] person and the Earth, with a given distance from the Earth's center, we will use Newton's law of universal gravitation. The formula for gravitational force is:

[tex]\[ F = G \frac{m_1 m_2}{r^2} \][/tex]

where:
- [tex]\( F \)[/tex] is the gravitational force,
- [tex]\( G \)[/tex] is the gravitational constant ([tex]\( 6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \)[/tex]),
- [tex]\( m_1 \)[/tex] is the mass of the person ([tex]\( 45 \, \text{kg} \)[/tex]),
- [tex]\( m_2 \)[/tex] is the mass of the Earth ([tex]\( 5.98 \times 10^{24} \, \text{kg} \)[/tex]),
- [tex]\( r \)[/tex] is the distance between the person and the center of the Earth ([tex]\( 6.38 \times 10^6 \, \text{m} \)[/tex]).

Let's break down the steps:

1. Identify the given values:
[tex]\[ G = 6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \][/tex]
[tex]\[ m_1 = 45 \, \text{kg} \][/tex]
[tex]\[ m_2 = 5.98 \times 10^{24} \, \text{kg} \][/tex]
[tex]\[ r = 6.38 \times 10^6 \, \text{m} \][/tex]

2. Substitute these values into the gravitational force formula:
[tex]\[ F = \left( 6.67430 \times 10^{-11} \right) \frac{45 \times 5.98 \times 10^{24}}{\left( 6.38 \times 10^6 \right)^2} \][/tex]

3. Calculate the expression within the numerator:
[tex]\[ 45 \times 5.98 \times 10^{24} = 269.1 \times 10^{24} \, \text{kg} \][/tex]

4. Calculate the expression within the denominator:
[tex]\[ \left( 6.38 \times 10^6 \right)^2 = 40.7044 \times 10^{12} \, \text{m}^2 \][/tex]

5. Simplify the fraction:
[tex]\[ \frac{269.1 \times 10^{24}}{40.7044 \times 10^{12}} = \frac{269.1}{40.7044} \times 10^{12} = 6.61 \times 10^{12} \, \text{kg} \, \text{m}^{-2} \][/tex]

6. Multiply this result by the gravitational constant:
[tex]\[ F = \left( 6.67430 \times 10^{-11} \right) \times \left( 6.61 \times 10^{12} \right) \][/tex]

7. Perform the final multiplication:
[tex]\[ F \approx 441.24323906015076 \, \text{N} \][/tex]

Thus, the gravitational force [tex]\( F \)[/tex] is approximately [tex]\( 441 \, \text{N} \)[/tex]. Among the provided options, the closest value to our calculated result is:
[tex]\[ \text{C. } 441 \, \text{N} \][/tex]

Therefore, the correct answer is [tex]\( \text{C. } 441 \, \text{N} \)[/tex].

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