Question

The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles are mA = 363 kg, mB = 517 kg, and mC = 154 kg. Find the magnitude and direction of the net gravitational force acting on (a) particle A, (b) particle B, and (c) particle C.

Answer 1

Explanation:

For Particle A:

The net gravitational force is the sum of the two forces due to particle B and due to particle C. Both of these particles have positive direction.

$F_{A} = F_{BA} + F_{CA} = G\frac{m_{B} m_{A} }{r_{AB} ^{2} } + G\frac{m_{C} m_{A} }{r_{AC} ^{2} }$

    $= 6.67*10^{-11}\frac{517*363}{0.5^{2} } +6.67*10^{-11} \frac{154*363}{0.75^{2} }$

    = $5.7*10^{-5}$ N

For Particle B:

The net gravitational force is the sum of the two forces: due to particle A  and due to particle C. The force due to particle A is negative as its direction is to left. The force due to particle C is positive.

$F_{B} =- F_{BA} + F_{CB} = G\frac{m_{B} m_{A} }{r_{AB} ^{2} } + G\frac{m_{C} m_{B} }{r_{BC} ^{2} }$

$= -6.67*10^{-11}\frac{517*363}{0.5^{2} } +6.67*10^{-11} \frac{154*517}{0.25^{2} }$

= $3.48*10^{-5}$N

For particle C:

The net gravitational force is the sum of two forces: due to particle A and due to particle B. Both forces are negative then,

$F_{C} =- F_{BC} + F_{CA} = -G\frac{m_{B} m_{C} }{r_{CB} ^{2} } + G\frac{m_{C} m_{A} }{r_{AC} ^{2} }$

     $= -6.67*10^{-11}\frac{517*154}{0.25^{2} } +6.67*10^{-11} \frac{154*363}{0.75^{2} }$

     = $-7.8*10^{-5} N$