What is the net gravitational force and at what position does it have net force of zero?

Question

Objects with masses of 150 kg and a 450 kg are separated by 0.380 m.
(a) Find the net gravitational force exerted by these objects on a 59.0 kg object placed midway between them.
N
(b) At what position (other than infinitely remote ones) can the 59.0 kg object be placed so as to experience a net force of zero?
m from the 450 kg mass

Answer 1

a) The formula to be used is F = G*m1*m2/d^2. G is the Universal Gravitational Constant, 6.67*10^-11 N*m^2/kg^2. Figure with just 2 particles at a time, the 59 kg object with first one, then the other of the 2 outside masses. The first time you use the formula, you'll get a force on the 59 kg object towards one, next time, a force towards the other. Subtract them.

b) In a), the 59.0 kg object was midway between the other 2. For this problem you'll need to set up an equation to find the d (it'll still be between the 2 but close to the 150 kg mass) such that the force towards each of the outside masses is equal and opposite. You'll need a bit of algebra.

Answer 2

we are assuming the spheres have negligible length. Gravitational rigidity is (G*m1*m2)/(d^2), the position: G is the gravitational consistent of the universe m1 is the mass of one merchandise m2 is the mass of the different merchandise d is the gap between them assume the "small mass" is x kg, and this is y cm from the 21kg sphere and z cm from the 12kg sphere. The rigidity of gravity between the small mass and 21 kg sphere is: (G * 21 kg * x kg) / (y^2) The rigidity of gravity between the small mass and 12 kg sphere is: (G * 12 kg * x kg) / (z^2) To make the internet rigidity 0, those 2 above might want to be equivalent (to cancel one yet another out), so: (G * 21 kg * x kg) / (y^2) = (G * 12 kg * x kg) / (z^2) 21 / (y^2) = 12 / (z^2) 21z^2 = 12y^2 also, all of us comprehend that y + z = the gap between the spheres = 18cm consequently, z = 18 - y 21(18 - y)^2 = 12y^2 21(324 - 36y + y^2) = 12y^2 6804 - 756y + 21y^2 = 12y^2 6804 - 756y + 21y^2 - 12y^2 = 0 9y^2 - 756y + 6804 = 0 y^2 - 84y + 756 = 0 y = (-(-80 4) +/- sqrt((-80 4)^2 - 4(a million)(756))) / 2(a million) y = (80 4 +/- sqrt(7056 - 3024)) / 2 y = (80 4 +/- sqrt(4032)) / 2 y = 40 2 +/- sqrt(1008) y =~ seventy 3.seventy 5 or 10.25 If y =~ seventy 3.seventy 5, then z =~ -fifty 5.seventy 5, yet when z is adverse, then the small mass isn't between both spheres, so both gravitational forces can't cancel one yet another out, so as it is ineffective. consequently, y =~ 10.25, so z=~ 7.seventy 5 So the mass might want to be about 10.25 cm from the 21 kg sphere, and about 7.seventy 5 cm from the 12 kg sphere.