Answer :
To determine the magnitude of the force of gravity acting on the spaceship, we can use Newton's law of universal gravitation. The formula is given by:
[tex]\[ F = G \frac{m_1 m_2}{r^2} \][/tex]
where:
- [tex]\( F \)[/tex] is the force of gravity,
- [tex]\( G \)[/tex] is the gravitational constant [tex]\((6.673 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2)\)[/tex],
- [tex]\( m_1 \)[/tex] is the mass of the first object (Earth, which has a mass of [tex]\(5.98 \times 10^{24} \, \text{kg}\)[/tex]),
- [tex]\( m_2 \)[/tex] is the mass of the second object (the spaceship, which has a mass of [tex]\(3000 \, \text{kg}\)[/tex]),
- [tex]\( r \)[/tex] is the distance between the centers of the two masses (in this case, [tex]\(50,000 \, \text{km}\)[/tex]).
First, convert the distance from kilometers to meters since the standard unit for distance in these calculations is meters:
[tex]\[ 50,000 \, \text{km} = 50,000 \times 1000 \, \text{m} = 50,000,000 \, \text{m} \][/tex]
Next, substitute the values into the formula:
[tex]\[ F = 6.673 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2 \times \frac{(5.98 \times 10^{24} \, \text{kg}) \times (3000 \, \text{kg})}{(50,000,000 \, \text{m})^2} \][/tex]
Calculate the denominator:
[tex]\[ (50,000,000 \, \text{m})^2 = 2.5 \times 10^{15} \, \text{m}^2 \][/tex]
Now, calculate the numerator:
[tex]\[ (5.98 \times 10^{24} \, \text{kg}) \times (3000 \, \text{kg}) = 1.794 \times 10^{28} \, \text{kg}^2 \][/tex]
Combine these into the formula:
[tex]\[ F = 6.673 \times 10^{-11} \times \frac{1.794 \times 10^{28}}{2.5 \times 10^{15}} \][/tex]
Divide the numbers within the fraction:
[tex]\[ \frac{1.794 \times 10^{28}}{2.5 \times 10^{15}} = 7.176 \times 10^{12} \][/tex]
Finally, multiply by the gravitational constant:
[tex]\[ F = 6.673 \times 10^{-11} \times 7.176 \times 10^{12} \, \text{N} \][/tex]
Now calculate:
[tex]\[ F \approx 478.85448 \, \text{N} \][/tex]
Therefore, the magnitude of the force of gravity acting on the spaceship is approximately 478 newtons, which corresponds to option B.
[tex]\[ F = G \frac{m_1 m_2}{r^2} \][/tex]
where:
- [tex]\( F \)[/tex] is the force of gravity,
- [tex]\( G \)[/tex] is the gravitational constant [tex]\((6.673 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2)\)[/tex],
- [tex]\( m_1 \)[/tex] is the mass of the first object (Earth, which has a mass of [tex]\(5.98 \times 10^{24} \, \text{kg}\)[/tex]),
- [tex]\( m_2 \)[/tex] is the mass of the second object (the spaceship, which has a mass of [tex]\(3000 \, \text{kg}\)[/tex]),
- [tex]\( r \)[/tex] is the distance between the centers of the two masses (in this case, [tex]\(50,000 \, \text{km}\)[/tex]).
First, convert the distance from kilometers to meters since the standard unit for distance in these calculations is meters:
[tex]\[ 50,000 \, \text{km} = 50,000 \times 1000 \, \text{m} = 50,000,000 \, \text{m} \][/tex]
Next, substitute the values into the formula:
[tex]\[ F = 6.673 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2 \times \frac{(5.98 \times 10^{24} \, \text{kg}) \times (3000 \, \text{kg})}{(50,000,000 \, \text{m})^2} \][/tex]
Calculate the denominator:
[tex]\[ (50,000,000 \, \text{m})^2 = 2.5 \times 10^{15} \, \text{m}^2 \][/tex]
Now, calculate the numerator:
[tex]\[ (5.98 \times 10^{24} \, \text{kg}) \times (3000 \, \text{kg}) = 1.794 \times 10^{28} \, \text{kg}^2 \][/tex]
Combine these into the formula:
[tex]\[ F = 6.673 \times 10^{-11} \times \frac{1.794 \times 10^{28}}{2.5 \times 10^{15}} \][/tex]
Divide the numbers within the fraction:
[tex]\[ \frac{1.794 \times 10^{28}}{2.5 \times 10^{15}} = 7.176 \times 10^{12} \][/tex]
Finally, multiply by the gravitational constant:
[tex]\[ F = 6.673 \times 10^{-11} \times 7.176 \times 10^{12} \, \text{N} \][/tex]
Now calculate:
[tex]\[ F \approx 478.85448 \, \text{N} \][/tex]
Therefore, the magnitude of the force of gravity acting on the spaceship is approximately 478 newtons, which corresponds to option B.