Physics Help Please!!! : Two objects attract each other gravitationally with a force...

Two objects attract each other gravitationally with a force of 3.3 10-10 N when they are 0.27 m apart. Their total mass is 4.0 kg. Find their individual masses.
larger mass___ kg
smaller mass__ kg

Thank you!

use F = G m1 m2/r^2 where F is the force, G the newtonian grav cst = 6.67 x 10^-11 Nm^2/kg^2, m1, m2 the masses and r the distance

we know the value of G, F = 3.3 x 10^-10N, r = 0.27m

if the masses sum to 4, we can call the mass of one x and the other has mass 4-x

therefore:

F r^2/G = x(4-x)

3.3 x 10^-10 Nx (0.27m)^2/6.67x10^-11Nm^2/kg^2 = x(4-x)

0.36 = x(4-x)

this yields the qudratic

x^2 - 4x + 0.36 = 0

solve for x: and the masses are 0.091kg and 3.91kg

Fg = Gm1*m2 / r^2
solving for m1*m2 = Fg*r^2/G
G is 6.674 * 10^-11 N*m^2/kg^2; r is .27 m; Fg is 3.3E-10.

m1*m2 = (3.3E-10)(.27^2)/(6.674 * 10^-11)=.360 kg.

Let the larger mass equal x, and the smaller mass equal y.
(x)(y)=.360 kg
x+y =4 kg
y=4-x
(x)(4-x)=.360
You can either solve the quadratic equation or graph the equation on a graphic calculator. As it's been a while since I took an actual math class, I used a graphic calculator.
x = 3.91 kg.
y = 4-3.91 = .09 kg.

F = G(m1*m2)/r and m1+m2 = 4 therefore m1 = 4 -m2

using substitution:

F = G((4-m2)(m2)/r

F*r/G = 4(m2) - (m2)^2

(m2)^2 - 4(m2) + Fr/G = 0

use the quadratic equation to solve for m2.

4 - m2 = m1

gravitational constant G = 6.67300 × 10-11 m3 kg-1 s-2

F=G*m1*m2/d^2
3.3 10-10=6.67300 × 10-11*m1*m2/(0.27^2).............(a)

m1+m2=4..............(b)

Solve the simultaneous equations...