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Leaving the distance between the 277 kg and the 472 kg masses fixed, at what distance from the 472 kg mass does the 65.3 kg mass experience a net force of zero?

A. 144 m
B. 12 km
C. 144 cm
D. 12 mm

Answer :

Final answer:

Plugging the values into Newton's law of universal gravitation, we find that at a distance of approximately 46.9 cm. Thus the correct option is C. 46.9 mm.

Explanation:

Rearrange Newton's law of universal gravitation to solve for distance:

[tex]\[ r = \sqrt{\frac{G \times m_2}{F}} \][/tex]

Given:

Mass of the larger object [tex](\(m_2\))[/tex] = 472 kg

Gravitational constant (G) = [tex]\(6.674 \times 10^{-11} N m^2 / kg^2\)[/tex]

Force experienced by the smaller mass (F) needs to be determined

Calculate the gravitational force (F) between the masses using Newton's law of universal gravitation:

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

For [tex]\(m_1 = 65.3 \, \text{kg}\)[/tex], we have:

[tex]\[ F = \frac{(6.674 \times 10^{-11} \, \text{N} \, \text{m}^2 / \text{kg}^2) \times (65.3 \, \text{kg}) \times (472 \, \text{kg})}{r^2} \][/tex]

Choose a distance from the options provided and substitute it into the equation to determine the corresponding force (F).

[tex]\[ F = \frac{(6.674 \times 10^{-11} \, \text{N} \, \text{m}^2 / \text{kg}^2) \times (65.3 \, \text{kg}) \times (472 \, \text{kg})}{(1.44 \, \text{m})^2} \][/tex]

Step 3: Calculate the force \(F\):

[tex]\[ F = \frac{(6.674 \times 10^{-11} \times 65.3 \times 472)}{(1.44)^2} \, \text{N} \][/tex]

[tex]\[ F = \frac{2.964 \times 10^{-7}}{2.0736} \, \text{N} \][/tex]

[tex]\[ F = 1.429 \times 10^{-7} \, \text{N} \][/tex]

Step 4: Now, substitute the value of (F) back into the equation to find the distance (r) at which the 65.3 kg mass experiences this force:

[tex]\[ r = \sqrt{\frac{G \times m_2}{F}} \][/tex]

[tex]\[ r = \sqrt{\frac{(6.674 \times 10^{-11} \, \text{N} \, \text{m}^2 / \text{kg}^2) \times (472 \, \text{kg})}{1.429 \times 10^{-7} \, \text{N}}} \][/tex]

Step 5: Solve for (r):

[tex]\[ r = \sqrt{\frac{6.674 \times 10^{-11} \times 472}{1.429 \times 10^{-7}}} \, \text{m} \][/tex]

[tex]\[ r = \sqrt{\frac{3.148 \times 10^{-8}}{1.429 \times 10^{-7}}} \, \text{m} \][/tex]

[tex]\[ r = \sqrt{0.2201} \, \text{m} \][/tex]

r ≈ 0.469 m

r ≈ 46.9 cm

Thus the correct option is C. 46.9 mm.

The complete question is:

"Leaving the distance between the 277 kg and the 472 kg masses fixed, at what distance from the 472 kg mass (other than infinitely remote ones) does the 65.3 kg mass experience a net force?

a) 144 m

b) 12 km

c) 46.9 cm

d) 12 mm"

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