Question

Objects with masses of 145 kg and a 445 kg are separated by 0.450 m.

(a) Find the net gravitational force exerted by these objects on a 39.0-kg object placed midway between them.

magnitude _________N direction

(b) At what position (other than infinitely remote ones) can the 39.0-kg object be placed so as to experience a net force of zero?

_____________m from the 445-kg mass

Answer 1

a) at the middle

distance between them , d = 0.450/2 = 0.225 m

net gravitational force = 6.673 *10^-11 * 39 * (445 - 145)/(0.225^2)

net gravitational force = 1.54 *10^-5 N

the net gravitational force is 1.54 *10^-5 N direction is towards 445 Kg

b)

let the distance from 445 kg is x

for the net force to be zero

G * 39 * 145/(0.45 - x)^2 = G * 39 * 445/x^2

145/(0.45 - x)^2 = 445/x^2

solving for x

x = 0.286 m

the position of third mass must be 0.286 m from 445 Kg mass