Question

Three identical very dense masses of 7500 kg each are placed on the x axis. One mass is at x1 = -100 cm , one is at the origin, and one is at x2 = 420 cm . Part A What is the magnitude of the net gravitational force Fgrav on the mass at the origin due to the other two masses? Take the gravitational constant to be G = 6.67×10−11 N⋅m2/kg2 .

Answer 1

Answer:

0.00354 (N)

Explanation:

Convert to metric system:

$x_1 = -100 cm = 1 m$

$x_2 = 420 cm = 4.2 m$

Formula for gravitational force:

$F_g = G\frac{mM}{s^2}$

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:

$F_g = F_{g1} - F_{g2}$

$F_g = G\frac{m_1M}{x_1^2} - G\frac{m_2M}{x_2^2}$

$F_g = GM(\frac{m_1}{x_1^2} - \frac{m_2}{x_2^2})$

$F_g = 6.67*10^{-11} * 7500 (\frac{7500}{1^2} - \frac{7500}{4.2^2})$

$F_g = 5*10^{-7}(7500 - 425.17)$

$F_g = 5*10^{-7} * 7074.83$

$F_g = 0.00354 (N)$