Answer :
To determine the force of gravity acting on a 1-kilogram box located [tex]\(1.3 \times 10^7\)[/tex] meters from the center of the Earth, we will use the formula for gravitational force:
[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]\(6.673 \times 10^{-11}\)[/tex] N·m[tex]\(^2\)[/tex]/kg[tex]\(^2\)[/tex]),
- [tex]\( m_1 \)[/tex] is the mass of the Earth ([tex]\(5.98 \times 10^{24}\)[/tex] kg),
- [tex]\( m_2 \)[/tex] is the mass of the box (1 kg),
- [tex]\( r \)[/tex] is the distance from the center of the Earth ([tex]\(1.3 \times 10^7\)[/tex] m).
Let's break down the calculation:
1. Calculate the numerator [tex]\( G \cdot m_1 \cdot m_2 \)[/tex]:
[tex]\[ 6.673 \times 10^{-11} \, \text{N·m}^2/\text{kg}^2 \times 5.98 \times 10^{24} \, \text{kg} \times 1 \, \text{kg} \][/tex]
2. Calculate the denominator [tex]\( r^2 \)[/tex]:
[tex]\[ (1.3 \times 10^7 \, \text{m})^2 \][/tex]
3. Divide the numerator by the denominator to find [tex]\( F \)[/tex]:
[tex]\[ F = \frac{6.673 \times 10^{-11} \, \text{N·m}^2/\text{kg}^2 \times 5.98 \times 10^{24} \, \text{kg} \times 1 \, \text{kg}}{(1.3 \times 10^7 \, \text{m})^2} \][/tex]
Upon performing the calculations, we find that:
[tex]\[ F \approx 2.3612153846153845 \, \text{newtons} \][/tex]
Thus, the correct answer is approximately 2.36 newtons.
So, the correct answer is:
A. 2.36 newtons
[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]\(6.673 \times 10^{-11}\)[/tex] N·m[tex]\(^2\)[/tex]/kg[tex]\(^2\)[/tex]),
- [tex]\( m_1 \)[/tex] is the mass of the Earth ([tex]\(5.98 \times 10^{24}\)[/tex] kg),
- [tex]\( m_2 \)[/tex] is the mass of the box (1 kg),
- [tex]\( r \)[/tex] is the distance from the center of the Earth ([tex]\(1.3 \times 10^7\)[/tex] m).
Let's break down the calculation:
1. Calculate the numerator [tex]\( G \cdot m_1 \cdot m_2 \)[/tex]:
[tex]\[ 6.673 \times 10^{-11} \, \text{N·m}^2/\text{kg}^2 \times 5.98 \times 10^{24} \, \text{kg} \times 1 \, \text{kg} \][/tex]
2. Calculate the denominator [tex]\( r^2 \)[/tex]:
[tex]\[ (1.3 \times 10^7 \, \text{m})^2 \][/tex]
3. Divide the numerator by the denominator to find [tex]\( F \)[/tex]:
[tex]\[ F = \frac{6.673 \times 10^{-11} \, \text{N·m}^2/\text{kg}^2 \times 5.98 \times 10^{24} \, \text{kg} \times 1 \, \text{kg}}{(1.3 \times 10^7 \, \text{m})^2} \][/tex]
Upon performing the calculations, we find that:
[tex]\[ F \approx 2.3612153846153845 \, \text{newtons} \][/tex]
Thus, the correct answer is approximately 2.36 newtons.
So, the correct answer is:
A. 2.36 newtons
