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

Step-by-step 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(m_(B) m_(A) )/(r_(AB) ^(2) ) + G(m_(C) m_(A) )/(r_(AC) ^(2) )`

`= 6.67*10^(-11)(517*363)/(0.5^(2) ) +6.67*10^(-11) (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(m_(B) m_(A) )/(r_(AB) ^(2) ) + G(m_(C) m_(B) )/(r_(BC) ^(2) )`

`= -6.67*10^(-11)(517*363)/(0.5^(2) ) +6.67*10^(-11) (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(m_(B) m_(C) )/(r_(CB) ^(2) ) + G(m_(C) m_(A) )/(r_(AC) ^(2) )`

`= -6.67*10^(-11)(517*154)/(0.25^(2) ) +6.67*10^(-11) (154*363)/(0.75^(2) )`

=
`-7.8*10^(-5) N`