finding gravitational force between objects?

Question

Any help on this problem would be greatly appriciated!

Objects with masses of 290 kg and a 590 kg are separated by 0.420 m.
(a) Find the net gravitational force exerted by these objects on a 36.0 kg object placed midway between them.
Your answer differs from the correct answer by orders of magnitude. (N)

(b) At what position (other than infinitely remote ones) can the 36.0 kg object be placed so as to experience a net force of zero?
(meters from the 590 kg mass)

Answer 1

a) Find the attraction from each of the larger masses on the middle mass. The attractions will point in opposite directions, so subtract one from the other.

b) Set the attractions equal to each other with r as a variable. Solve for r.

Answer 2

hmmm...this a homework question? :-)

(a) the greater mass will produce the greater pull on the middle object..like in a tug-of-war. Since both masses are equidistant from the middle mass, to find the net gravitational pull you just find the "net mass": 590-290 = 300 then determine the Fg between that mass and the 36kg object.

Fg = [Gm1m2] / r

Fg = [6.67x10-11][36][300] / 0.42
Fg = 4.0837 x 10-6
Fg = 4.08 x 10-6 m <-- 3 sig digs

(b) from looking at the formula, we see that mass is directly proportional to r (distance between the centers squared)...therefore, for the 290 and 590 masses to"cancel out" each other we have to look at by what factor the 590 mass is greater than the 290 mass (2.0345) and take the square root of that number (1.426). So...at whatever linear distance the 290 mass is placed on the one side of the central mass (36kg), the 590 mass must be placed at 1.426 times that distance on the *opposite* side. This will cause a net gravitational force of zero on the central mass. (ie: 290kg mass is 1m away...then the 590 mass would have to be 1.426m away...1.43m if you need answer in 3 sig digs)

hope that helped...it would be better to have a white board to answer these questions...

Answer 3

C. will advance if the products are moved closer mutually. The gravitational stress varies by the inverse of the area squared, so it is not constantly the comparable. additionally that's proportional to mass, meaning that it will advance if the two merchandise gains mass, no longer shrink. ultimately, if the products are moved closer mutually, the stress will advance because of the fact the area is interior the denominator of the equation.

Answer 4

F(g) = G(gravitational constant) * m1 * m2 / r^2(distance between them squared).