Answer :
Answer:
0.00354 (N)
Explanation:
Convert to metric system:
[tex] x_1 = -100 cm = 1 m[/tex]
[tex] x_2 = 420 cm = 4.2 m[/tex]
Formula for gravitational force:
[tex] F_g = G\frac{mM}{s^2}[/tex]
where s is the distance between 2 bodies masses m and M
Substitute the number to the formula above and since the 2 forces are acting in opposite direction, the total net gravitational force on the mass of origin be:
[tex] F_g = F_{g1} - F_{g2}[/tex]
[tex] F_g = G\frac{m_1M}{x_1^2} - G\frac{m_2M}{x_2^2}[/tex]
[tex] F_g = GM(\frac{m_1}{x_1^2} - \frac{m_2}{x_2^2})[/tex]
[tex] F_g = 6.67*10^{-11} * 7500 (\frac{7500}{1^2} - \frac{7500}{4.2^2}) [/tex]
[tex] F_g = 5*10^{-7}(7500 - 425.17) [/tex]
[tex] F_g = 5*10^{-7} * 7074.83 [/tex]
[tex] F_g = 0.00354 (N) [/tex]
The net force on the object at the origin is 0.00396 N.
The given parameters;
- mass of each object, m = 7500 kg
- position of the first object, x₁ = - 100 cm = - 1 m
- position of the second object, x₂ = 420 cm = 4.2 m
- position of the third object, x₃ = 0
The gravitational force on the third object due to the first object;
distance between the two object, r₁ = 0 - (-1) = 1 m
[tex]F_1 = \frac{Gm^2}{r_1^2} \\\\F_1 = \frac{6.67\times 10^{-11} \times (7500)^2}{(1)^1} \\\\F_1 = 0.00375 \ N[/tex]
The gravitational force on the third object due to the second object;
distance between the two object, r₂ = 4.2 - 0 = 4.2 m
[tex]F_2 = \frac{Gm^2}{r_2^2} \\\\F_2 = \frac{(6.67\times 10^{-11}) (7500)^2}{(4.2)^2} \\\\F_2 = 0.000213 \ N[/tex]
The net force on the object at the origin (third object) is calculated as;
[tex]F_{net} = F_1 + F_2\\\\F_{net} = 0.00375 + 0.000213\\\\F_{net} = 0.00396 \ N[/tex]
Thus, the net force on the object at the origin is 0.00396 N.
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