Ciruclar motion and gravitation question

[chazgurl4life]

Q:Two objects attract each other gravitationally with a force of 3.3 10-10 N when they are 0.25 m apart. Their total mass is 4.0 kg. Find their individual masses.

Now my professor only told us to use F= Gm1m2/r2
Force in this case would be the 3.3e-10 Newtons, the radius squared = 0.0625 m and G is a the gravitational constant 6.67e-11 N so when i solve for m1 and m2 the larger mass come out to 4.2 kg and that's impossible

 

[arildno]

Hence, you haven't done your arithmetic right.

 

[Hootenanny]

How did you solve for m1 and m2?

 

[arildno]

You have:
$$m_{1}m_{2}=\frac{r^{2}F}{G}$$
withe the condition $$m_{1}=4-m_{2}$$
you get a quadratic equation to solve for $m_{2}$

 

[Hootenanny]

That's how I would do it.

 

[chazgurl4life]

ok so what i come up with is :

4m-m2^2=.25^2(3.3e-10)
---------------
6.67e-11

so when i solve i get :(Mass 2)^2-4(mass2)-.309 m=0
but when i solve for mass to i get 4.02 kg.. what am i doing wrong?

 

[Hootenanny]

You should get:
$$4m_2 - m_{2}^{2} = \frac{r^2 \cdot F}{G}$$

 

[arildno]

First of all:
DO NOT ENTER NUMBERS BEFORE THE VERY LAST MOMENT!

You have:
$$m_{2}(4-m_{2})=\frac{r^{2}F}{G}$$
which you rewrite as:
$$m_{2}^{2}-4m_{2}+\frac{r^{2}F}{G}=0$$

This has the solutions:
$$m_{2}=\frac{4\pm\sqrt{16-\frac{4r^{2}F}{G}}}{2}$$
This yields ALWAYS positive answer(s), in so far as the solution exists.

Note that the sum of your two solutions equals 4. What does that mean?

 

[chazgurl4life]

thank you very much.:-)..i realize where i was going wrong.

So after my calculations mass 1 =.078 kg and mass 2= 3.92 kg

 

[arildno]

I haven't the slightest idea what plug&chug would yield. Your numbers sort of add up to 4, though, which isn't bad at all..

 

[Hootenanny]

plug&chug?

 

[arildno]

Plug into the formula, chug out the answer..

 

[Hootenanny]

ahhhh... It's all starting to make sense...